Additional Questions - Chapter 8 - Ionic Equilibrium - 12th Chemistry Guide Samacheer Kalvi Solutions

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Multiple Additional Questions
1 Marks Questions and Answers
I. Choose the best answer.
Question 1.
Which one of the following buffer is present in blood?
(a) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$
(b) $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$
(c) $\mathrm{H}_2 \mathrm{CO}_3+\mathrm{NaHCO}_3$
(d) $\mathrm{HCI}+\mathrm{NaCl}$
Answer:
(c) $\mathrm{H}_2 \mathrm{CO}_3+\mathrm{NaHCO}_3$
Question 2.
Which of the following is mostly used in fertilizer industry?
(a) Lactic acid
(b) Sulphuric acid
(c) Tannic acid
(d) Carbonic acid
Answer:
(b) Sulphuric acid
Question 3.
Which of the following is present in an antacid tablet?
(a) $\mathrm{NaOH}$
(b) $\mathrm{Mg}(\mathrm{OH})_2$
(c) $\mathrm{Al}(\mathrm{OH})_3$
(d) either (b) or (c)
Answer:
(d) either (b) or (c)
Question 4.
The acid present in milk is
(a) Lactic acid

(b) Tannic acid
(c) Tartaric acid
(a) Acetic acid
Answer:
(a) Lactic acid
Question 5.
Consider the following statements.
(i) Acid tastes sour
(ii) Acid turns red litmus to blue
(iii) Acid reacts with metals and liberates hydrogen gas
Which of the above statement is I are correct?
(a) (i) only
(b) (i) \& (ii)
(c) (i) \& (iii)
(d) (ii) only
Answer:
(c) (i) \& (iii)
Question 6.
Consider the following statements.
(i) Acid tastes sour.
(ii) Acid turns blue litmus to red
(iii) Acid has a tendency to accept a proton from other substances.
Which of the above statement is I are not correct?
(a) (i) \& (ii)
(b) (ii) \& (iii)
(c) (iii) only
(a) (ii) only
Answer:
(c) (iii) only
Question 7.
Which of the following can act as an acid as well as base by Lowry - Bronsted theory?
(a) $\mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{NH}_3$
(c) $\mathrm{NH}_4 \mathrm{OH}$
(d) $\mathrm{Ca}(\mathrm{OH})_2$
Answer:
(a) $\mathrm{H}_2 \mathrm{O}$

Question 8.
In the reaction $\mathrm{HCI}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}+\mathrm{Cl}^{-}$which one of the acid-base pair?
(a) $\mathrm{HCl}+\mathrm{H}_3 \mathrm{O}^{+}$
(b) $\mathrm{HCI}+\mathrm{Cl}^{-}$
(c) $\mathrm{H}_3 \mathrm{O}+\mathrm{Cl}$
(d) $\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}^{-}$
Answer:
(b) $\mathrm{HCI}+\mathrm{CI}^{-}$
Question 9.
Which of the following is considered as Lewis acid?
(a) $\mathrm{NH}_3$
(b) $\mathrm{BF}_3$
(c) $\mathrm{HF}$
(d) $\mathrm{HCl}$
Answer:
(b) $\mathrm{BF}_3$
Question 10 .
Which of the following is considered as Lewis base?
(a) $\mathrm{BF}_3$
(b) $\mathrm{AICI}_3$
(c) $\mathrm{HCI}$
(d) $\mathrm{NH}_3$
Answer:
(d) $\mathrm{NH}_3$

Question 11.
Consider the following statements.
(i) A Lewis acid is a species that accepts an electron pair.
(ii) A Lewis acid is a species that donates an electron pair.
(iii) The ligand act as Lewis base.
Which of the above statement is / are not correct?
(a) (i) only
(b) (ii) only
(c) (I) \& (ii)
(d) (ii) \& (iii)
Answer:
(b) (ii) only
Question 12 .
In $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ which one of the following acts as Lewis acid?
(a) $\mathrm{Cr}$
(b) $\mathrm{Cr}^{3+}$
(c) $(\mathrm{HO})_6$
(d) $\mathrm{Cr}^{3-}$
Answer:
(b) $\mathrm{Cr}^{3+}$
Question 13.
In $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ which one of the following acts as Lewis base?
(a) $\mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{H}_3 \mathrm{O}^{+}$
(c) $\mathrm{Cr}^{3+}$
(d) $\mathrm{Cr}$
Answer:
(a) $\mathrm{H}_2 \mathrm{O}$

Question 14.
Among the following which is the strongest acid?
(a) Formic acid
(b) Acetic acid
(c) Hydrochloric acid
(d) Lactic acid
Answer:
(c) Hydrochloric acid
Question 15 .
Which of the following is the weak acid?
(a) $\mathrm{HCl}$
(b) $\mathrm{H}_2 \mathrm{SO}_4$
(c) $\mathrm{HNO}_3$
(d) $\mathrm{CH}_3 \mathrm{COOH}$
Answer:
(d) $\mathrm{CH}_3 \mathrm{COOH}$
Question 16.
Identify the weakest acid?
(a) $\mathrm{H}_3 \mathrm{O}^{+}$
(b) $\mathrm{H}_2 \mathrm{SO}_4$
(c) $\mathrm{OH}^{-}$
(d) $\mathrm{CH}_3 \mathrm{COOH}$
Answer:
(c) $\mathrm{OH}^{-}$
Question 17.
Which one of the following is the very weak base?
(a) $\mathrm{NO}_2^{-}$
(b) $\mathrm{NO}_3^{-}$
(c) $\mathrm{NH}_2^{-}$
(d) $\mathrm{O}^{2-}$
Answer:
(b) $\mathrm{NO}_3^{-}$

Question 18.
Which one of the following is the strong base?
(a) $\mathrm{ClO}_4^{-}$
(b) $\mathrm{HSO}_4$
(c) $\mathrm{O}^{2-}$
(d) $\mathrm{F}^{-}$
Answer:
(c) $\mathrm{O}^{2-}$
Question 19.
Which of the following is the weak base?
(a) $\mathrm{H}^{-}$
(b) $\mathrm{OH}^{-}$
(c) $\mathrm{HSO}_4$
(d) $\mathrm{F}^{-}$
Answer:
(d) $\mathrm{F}^{-}$
Question 20.
The value of ionic product of water at $25^{\circ} \mathrm{C}$ is
(a) $1 \times 10^{-7}$
(b) $1 \times 10^7$
(c) $1 \times 10^{-14}$
(d) $1 \times 10^{14}$
Answer:
(c) $1 \times 10^{-14}$
Question 21.
Consider the following statements.
(i) The dissociation of water is an exothermic reaction.
(ii) With the increase in temperature, the ionic product of water value decreases.
(iii) With the increase in temperature, the ionic product of water value increases. Which of the above statement is / are correct?
(a) (i) and (ii)
(b) (ii) only
(c) (iii) only
(d) (ii) \& (iii)

Question 22.
Which of the following is a neutral solution?
(a) Aqueous $\mathrm{NaCl}$ solution
(b) Aqueous $\mathrm{NaOH}$ solution
(c) Aqueous $\mathrm{HCl}$ solution
(d) Aqueous $\mathrm{NH}_3$
Answer:
(a) Aqueous $\mathrm{NaCl}$ solution
Question 23.
The $\mathrm{pH}$ of a neutral solution is
(a) less than 7
(b) more than 7
(c) equal to 7
(d) 14
Answer:
(a) less than 7
Question 24.
If the $\mathrm{pH}$ of a solution is less than 7, it is called solution.
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric
Answer:
(b) Acidic
Question 25.
If the $\mathrm{pH}$ of a solution is more than 7 , it is called solution.
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric
Answer:
(a) Basic

Question 26.
The $\mathrm{pH}$ value of water is
(a) 14
(b) 7
(c) 3
(d) 1
Answer:
(b) 7
Question 27.
The $\mathrm{pH}$ of drain cleaner is
(a) 7
(b) 1
(c) 14
(d) 0
Answer:
(c) 14
Question 28.
The $\mathrm{pH}$ of the battery acid is
(a) 7
(b) 1
(c) 14
(d) 0
Answer:
(d) 0
Question 29.
The $\mathrm{pH}$ of $0.001 \mathrm{MHCI}$ solution is
(a) 3
(b) 2
(c) 1
(d) 11
Answer:

(a) 3
Solution:
$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=0.001 \mathrm{~mol} \mathrm{dm}^{-3}} \\
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10}[0.001] \\
& =-1 \log _{10}\left[10^{-3}\right]=3 \\
& \mathrm{pH}=3
\end{aligned}
$
Question 30.
The $\mathrm{pH}$ of $0.01 \mathrm{M} \mathrm{HCl}$ solution is
(a) 3
(b) 2
(c) 1
(d) 10
Answer:
(b) 2
Solution:
$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=0.01 \mathrm{M}} \\
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10}[0.01] \\
& =-\log _{10}\left[10^{-3}\right]=3 \\
& \mathrm{pH}=3
\end{aligned}
$
Question 31.
What is the $\mathrm{pH}$ of $0.1 \mathrm{M} \mathrm{HCI}$ solution?
(a) 1
(b) 2
(c) 13
(d) 3
Answer:
(a) 1
Solution:
$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=0.1 \mathrm{M}} \\
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10}[0.1] \\
& =-\log _{10}\left[10^{-1}\right] \\
& \mathrm{pH}=1
\end{aligned}
$

Question 32.
Consider the following statements.
(i) Degree of dissociation (a) is the fraction of the total number of moles of a substance that dissociates at equilibrium.
(ii) When the dilution increases by 100 times, the dissociation increases by 100 times.
(iii) When the dilution increases by 100 times, the dissociation increases by 10 times.
Which of the above statement is I are correct?
(a) (ii) only
(b) (i) \& (iii)
(c) (iii) only
(d) (I) only
Answer:
(a) (ii) only
Question 33.
Which of the following is not a buffer solution?
(a) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$
(b) $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$
(c) $\mathrm{H}_2 \mathrm{CO}_3+\mathrm{NaHCO}_3$
(d) $\mathrm{NaOH}+\mathrm{NaCI}$
Answer:
(d) $\mathrm{NaOH}+\mathrm{NaCI}$
Question 34.
The mathematical expression of buffer capacity is
(a) $\boldsymbol{\beta}=\frac{\mathrm{dB}}{\mathrm{d}(\mathbf{p H})}$
(b) $\beta=\frac{d(\mathrm{pH})}{\mathrm{dB}}$
(c) $\beta=-\frac{\mathrm{dB}}{\mathrm{d}(\mathrm{pH})}$
(d) $\beta=-\frac{\mathrm{d}(\mathrm{pH})}{\mathrm{dB}}$
Answer:
(a) $\boldsymbol{\beta}=\frac{\mathrm{dB}}{\mathrm{d}(\mathbf{p H})}$

Question 35.
Which one of the following is not correct?
(a) $\mathrm{pH}+\mathrm{pOH}=14$
(b) $\mathrm{pH}=7+\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}$
(c) $\mathrm{pH} \times \mathrm{pOH}=1 \times 10^{14}$
(d) $\mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$
Answer:
(c) $\mathrm{pH} \times \mathrm{pOH}=1 \times 10^{14}$
Question 36.
The chemical present in kidney as stones is
(a) $\mathrm{CaCl}_2$
(b) $\mathrm{Ca}\left(\mathrm{CO}_3\right)_2$
(c) Calcium nitrate
(d) Calcium oxalate
Answer:
(d) Calcium oxalate
Question 37.
If the $\mathrm{pH}$ of an aqueous solution is 7 , the solution is
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(c) neutral
Question 38.
$\mathrm{Cl}^{-}$is the conjugate base of
(a) $\mathrm{HClO}_4$
(b) $\mathrm{HCI}$
(c) $\mathrm{ClO}_4^{-}$
(d) $\mathrm{HClO}_3$
Answer:
(b) $\mathrm{HCI}$

Question 39.
The conjugate base of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{SO}_4$ are
(a) $\mathrm{OH}^{-}$and $\mathrm{HSO}_4$
(b) $\mathrm{H}_4 \mathrm{O}$ and $\mathrm{SO}_4^{2-}$
(c) $\mathrm{OH}$ and $\mathrm{SO}_4^{2-}$
(d) $\mathrm{H}_3 \mathrm{O}$ and $\mathrm{HSO}_4$
Answer:
(a) $\mathrm{OH}^{-}$and $\mathrm{HSO}_4$
Question 40.
The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{AgI}$ is $1.5 \times 10^{-16}$. On mixing equal volume of the following solutions, precipitation will occur only with
(a) $10^{-7} \mathrm{MAg}^{+}$and $10^{-19} \mathrm{M} \mathrm{I}^{-}$
(b) $10^{-8} \mathrm{MAg}^{+}$and $10^{-8} \mathrm{M} \mathrm{I}^{-}$
(c) $10^{-16} \mathrm{M} \mathrm{Ag}^{+}$and $10^{-16} \mathrm{M} \mathrm{I}^{-}$
(d) $10^{-9} \mathrm{M} \mathrm{Ag}^{+}$and $10^{-9} \mathrm{Mr}$
Answer:
(b) $10^{-8} \mathrm{MAg}^{+}$and $10^{-8} \mathrm{M} \mathrm{I}^{-}$
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{AgI}=1.5 \times 10^{-16}$
$10^{-8} \mathrm{M} \mathrm{Ag}^{+}$and $10^{-8} \mathrm{M} \mathrm{I}$
Ionic product $=10^{-16}=\mathrm{K}_{\mathrm{sp}}$
Question 41.
The strongest Bronsted base in the following anion is
(a) $\mathrm{ClO}^{-}$
(b) $\mathrm{ClO}^{2-}$
(c) $\mathrm{ClO}^{3-}$
(d) $\mathrm{ClO}^{4-}$
Answer:
(a) $\mathrm{ClO}^{-}$
Solution:
$\mathrm{HClO}$ is the weakest acid and its conjugate base $\mathrm{ClO}^{-}$is the strongest base.

Question 42.
Calculate the hydrolysis constant of the salt containing $\mathrm{NO}_2$.
Given the $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{HNO}_2=4.5 \times 10^{-10}$
(a) $2.22 \times 10^{-5}$
(b) $2.02 \times 10^5$
(c) $4.33 \times 10^4$
(d) $3.03 \times 10^{-5}$
Answer:
(a) $2.22 \times 10^{-5}$
Solution:
$
\mathrm{h}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\frac{1 \times 10^{-14}}{4.5 \times 10^{-10}}=2.22 \times 10^{-5}
$
Question 43.
Electrophiles are usually
(a) Lewis acid
(b) Lewis base
(c) Bronsted acid
(d) Bronted base
Answer:
(a) Lewis acid
Solution:
Lewis acid are electrophile because they accept electron pair.
Question 44.
Which one is a Lewis acid?
(a) $\mathrm{CIF}_3$
(b) $\mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{NH}_3$
(d) $\mathrm{OH}$
Answer:
(a) $\mathrm{CIF}_3$
Solution:
$\mathrm{CIF}_3$ have vacant d-orbital in central atom.

Question 45.
An aqueous solution of ammonium acetate is
(a) faintly acidic
(b) faintly basic
(c) fairly acidic
(d) Almost neutral
Answer:
(d) Almost neutral
Solution:
It is a salt of weak acid and weak base.
Question 46.
The dissociation constant of a weak acid is $1.0 \times 10^{-10}$. The equilibrium constant for the reaction with strong
base is
(a) $1.0 \times 10^{-5}$
(b) $1.0 \times 10^{-9}$
(c) $1.0 \times 10^9$
(d) $1.0 \times 10^{14}$
Answer:
(c) $1.0 \times 10^9$
Solution:
$
\begin{aligned}
\mathrm{HA} & \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-} \\
\mathrm{K}_{\mathrm{a}} & =\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \\
\mathrm{HA}+\mathrm{OH}^{-} & \rightleftharpoons \mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O} \\
\mathrm{K} & =\frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]\left[\mathrm{OH}^{-}\right]}
\end{aligned}
$
Dividing (1) by (2)
$
\begin{aligned}
\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}} & =\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}=10^{-14} \\
\mathrm{~K} & =\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{w}}}=\frac{10^{-5}}{10^{-14}}=10^9
\end{aligned}
$

Question 47.
Arrange the acids
(i) $\mathrm{H}_2 \mathrm{SO}_3$
(ii) $\mathrm{H}_3 \mathrm{PO}_3$ and
(iii) $\mathrm{HClO}_3$ in the decreasing order of acidity.
(a) (i) $>$ (iii) $>$ (ii)
(b) (i) $>$ (ii) $>$ (iii)
(c) (ii) $>$ (iii) $>$ (I)
(d) (iii) $>$ (i) $>$ (ii)
Answer:
(d) (iii) $>$ (i) $>$ (ii)
Solution:
Acidity is directly proportional to oxidation number. As the oxidation number of $\mathrm{S}, \mathrm{P}$ and $\mathrm{Cl}_{\text {in }} \mathrm{H}_2 \mathrm{SO}_3$, $\mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{HCIO}_3$ is $+4,+3,+5$ respectively. So decreasing order of acidity will be (iii) $>$ (I) $>$ (ii)
Question 48.
The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of a weak monoprotic acid $1 \%$ ionised is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3
Solution:
Conc $=0.1 \mathrm{M}$
$
\begin{aligned}
& \alpha=1 \\
& =0.1 \times \frac{1}{100}=10^{-3} \\
& {\left[\mathrm{H}^{+}\right]=10^{-3}} \\
& \therefore \mathrm{pH}=3
\end{aligned}
$

Question 49.
$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{Cr}(\mathrm{OH})_3$ is $2.7 \times 10^{-3}$. What is the solubility in moles / litre?
(a) $1 \times 10^{-8}$
(b) $8 \times 10^{-8}$
(c) $1.1 \times 10^{-8}$
(d) $0.18 \times 10^{-8}$
Answer:
(b) $8 \times 10^{-8}$
Solution:
$
\begin{aligned}
\mathrm{Cr}(\mathrm{OH})_3 \rightarrow & \mathrm{Cr}^{3+}+3 \mathrm{OH}^{-} \\
\mathrm{K}_{\mathrm{sp}} & =x \cdot(3 x)^3=27 x^4 \\
\therefore x & =4 \sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{27}}=4 \sqrt{\frac{2.7 \times 10^{-31}}{27}} \\
\therefore x & =1 \times 10^{-8}
\end{aligned}
$

Question 50.
$\mathrm{pK}_{\mathrm{a}}$ of acetic acid is 4.74 . The concenttation of $\mathrm{CH}_3 \mathrm{COONa}$ is $0.01 \mathrm{M}$. The $\mathrm{pH}$ of $\mathrm{CH}_3 \mathrm{COONa}$ is
(a) 3.37
(b) 4.37
(c) 4.74
(d) 0.474
Answer:
(a) 3.37
Solultion:
$
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\mathrm{c} . \alpha=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}} \\
& \mathrm{pH}=-\log \left(\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}\right)^{1 / 2} \\
& =\frac{1}{2}\left[-\log \mathrm{K}_{\mathrm{a}}-\log \mathrm{c}\right] \\
& =\frac{1}{2}\left[4.74-\log 10^{-2}\right] \\
& =\frac{1}{2}[4.74+2]=3.37 \\
& \mathrm{pH}=3.37
\end{aligned}
$
Question 51.
One litre of water contains $10 \mathrm{~mol}$ hydrogen ions. The degree of ionisation in water will be
(a) $8 \times 10^{-7}$
(b) $0.8 \times 10^{-9}$
(c) $3.6 \times 10^{-7}$
(d) $3.6 \times 10^{-9}$
Answer:
(a) $8 \times 10^{-7}$
Solution:
1 litre of water contains $1000 / 18$ mole.
So, degree of ionisation $=\frac{10^{-7} \times 18}{1000}=1.8 \times 10^{-7}$

Question 52.
If the solubility product of lead iodide $\left(\mathrm{PbI}_2\right)$ is $3.2 \times 10^{-8}$. Then its solubility in moles / litre will be
(a) $2 \times 10^{-3}$
(b) $4 \times 10^{-4}$
(c) $1.6 \times 10^{-5}$
(d) $1.8 \times 10^{-5}$

Answer:

(a) $2 \times 10^{-3}$
Solution:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3 \\
& 4 \mathrm{~s}^3=3.2 \times 10^{-8} \\
& \mathrm{~s}=2 \times 10^{-3} \mathrm{M}
\end{aligned}
$
Question 53.
The $\mathrm{pH}$ of a soft drink is 3.82 . Its hydrogen ion concentration will be
(a) $1.96 \times 10^{-2} \mathrm{~mol} / \mathrm{L}$
(b) $1.96 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$
(c) $1.5 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$
(d) $1.96 \times 10^{-1} \mathrm{~mol} / \mathrm{L}$
Answer:
(c) $1.5 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$
Solution:
$
\begin{aligned}
& \mathrm{pH}=3.82=-\log _{10}\left[\mathrm{H}^{+}\right] \\
& \therefore\left[\mathrm{H}^{+}\right]=1.5 \times 10^{-4} \mathrm{~mol} / \text { litre }
\end{aligned}
$
Question 54.
The $\mathrm{pH}$ of a solution at $25^{\circ} \mathrm{C}$ containing $0.10 \mathrm{M}$ sodium acetate and $0.03 \mathrm{M}$ acetic acid is $\left(\mathrm{pK}_{\mathrm{a}}\right.$ for $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right)$
(a) 4.09
(b) 5.09
(c) 6.10
(d) 7.09
Answer:
(b) 5.09
Solution:
$
\begin{aligned}
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& =4.57+\log \frac{0.10}{0.03}=5.09
\end{aligned}
$
Question 55
A weak acid is $0.1 \%$ ionised in $0.1 \mathrm{M}$ solution. Its $\mathrm{pH}$ is
(a) 2

Answer:
(d) $\mathrm{H}_2 \mathrm{SO}_4<\mathrm{H}_2 \mathrm{CO}_3<\mathrm{CH}_3 \mathrm{COOH}$
Question 59.
What is the $\mathrm{pH}$ of $1 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}$ solution?. Ka of acetic acid is $1.8 \times 10^{-5} \cdot \mathrm{K}=10^{-14} \mathrm{~mol}^2$ litre 2 .
(a) 9.4
(b) 4.8
(c) 3.6
(d) 2.4
Answer:
(a) 9.4
Solution:
$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COO}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}+\mathrm{OH}^{-} \\
& {\left[\mathrm{OH}^{-}\right]=\mathrm{c} \times \mathrm{h}} \\
& \mathbf{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}} \times c}=\sqrt{\frac{10^{-14}}{1.8 \times 10^{-5}} \times 1} \\
& =2.35 \times 10^{-5} \\
& \mathrm{pOH}=4.62 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{pH}=14-4.62=9.38
\end{aligned}
$

Question 60 .
$
\begin{aligned}
& 4 \mathrm{Na}+\mathrm{O}_2 \rightarrow 2 \mathrm{Na}_2 \mathrm{O} \\
& \mathrm{Na}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}
\end{aligned}
$
In the given reaction, the oxide of sodium is
(a) Acidic
(b) Basic
(c) Amphoteric
(d) Neutral
Answer:
(b) Basic
Solution.
$\mathrm{Na}_2 \mathrm{O}$ form $\mathrm{NaOH}$ so that it is basic oxide.
Question 61.
The $\mathrm{pH}$ of $0.001 \mathrm{M} \mathrm{NaOH}$ will be
(a) 3
(b) 2
(c) 11
(d) 12
Answer:
(c) 11
Solution:
$0.001 \mathrm{M} \mathrm{NaOH}$ means $\left[\mathrm{OH}^{-}\right] 0.001$.
$
\begin{aligned}
& 10^{-3} \mathrm{pOH}=3 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{pH}=14-3=11
\end{aligned}
$
Question 62.
The addition of pure solid sodium carbonate to pure water causes
(a) an increase in hydronium ion concentration
(b) an increase in alkalinity
(c) No change in acidity
(d) A decrease in hydroxide ion
Answer:
(b) an increase in alkalinity
Hint.
Adding $\mathrm{Na}_2 \mathrm{CO}_3$ to water makes the solution basic and hence $\mathrm{pH}$ increases from 7 .
Question 63.
When solid potassium cyanide is added in water then
(a) $\mathrm{pH}$ will increase

(b) $\mathrm{pH}$ will decrease
(c) $\mathrm{pH}$ will remain the same
(d) electrical conductivity will not change
Answer:
(a) $\mathrm{pH}$ will increase
Hint:
$\mathrm{KCN}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{KOH}+\mathrm{HCN}$.
$\mathrm{KOH}$ is a strong base and $\mathrm{HCN}$ is a weak acid. So $\mathrm{pH}$ will increase.
Question 64.
$\mathrm{pH}$ of a solution is 5 . Its hydroxyl ion concentration is
(a) 5
(b) 10
(c) $10^{-5}$
(d) $10^{-9}$
Answer:
(d) $10^{-9}$
Solution:
$
\begin{aligned}
& \mathrm{pH}=5 \text { means }\left[\mathrm{H}^{+}\right]=10^{-5} \\
& \mathrm{pOH}=14-\mathrm{pH}=14-5=9 \\
& {\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{pOH}}=10^{-9}}
\end{aligned}
$
Question 65.
Which one of the following is a buffer?
(a) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$
(b) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONH}_4$
(c) $\mathrm{NaOH}+\mathrm{NaCI}$
(d) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NH}_4 \mathrm{CI}$
Answer:
(a) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$
Question 66.
Which will have maximum $\mathrm{pH}$ ?
(a) Distilled water
(b) $1 \mathrm{M} \mathrm{NH}_3$
(c) $1 \mathrm{M} \mathrm{NaOH}$
(d) Water saturated by chlorine
Answer:
(c) $1 \mathrm{M} \mathrm{NaOH}$
Hint:
$\mathrm{NaOH}$ has maximum $\left[\mathrm{OH}^{-}\right]$and minimum of $\left[\mathrm{H}^{+}\right]$and so maximum $\mathrm{pH}$ value.

Question 67.
$\mathrm{pH}$ of a solution is 9.5 . The solution is
(a) Neutral
(b) Acidic
(c) Basic
(d) Amphoteric
Answer:
(c) Basic
Solution:
If $\mathrm{pH}=7$ solution is neutral
$\mathrm{pH}<7$ solution is acidic
$\mathrm{pH}>7$ solution is basic
Question 68 .
A solution has $\mathrm{pH}=5$, it is diluted 100 times, then it will become
(a) Neutral
(b) Basic
(c) unaffected
(d) more acidic
Answer:
(a) Neutral
Solution:
$\mathrm{pH}=5$ means $\left[\mathrm{H}^{+}\right]=10^{-5}$
After dilution $\left[\mathrm{H}^{+}\right]=10^{-5} / 100=10^{-7} \mathrm{M}$
$\left[\mathrm{H}^{+}\right]$from $\mathrm{H}_2 \mathrm{O}$ cannot be neglected.
Total $\left[\mathrm{H}^{+}\right]=10^{-7}+10^{-7}=2 \times 10^{-7}$
$\mathrm{pH}=7-0.3010=6.6990=7$
$\mathrm{pH}=7$ (Neutral)
Question 69.
By adding a strong acid to the buffer solution, the $\mathrm{pH}$ of the buffer solution
(a) remains constant
(b) increases
(c) decreases
(d) becomes zero
Answer:
(a) remains constant

Question 70.
$\mathrm{pH}$ of human blood is 7.4. Then $\mathrm{H}^{+}$concentration will be
(a) $4 \times 10^{-8}$
(b) $2 \times 10^{-8}$
(c) $4 \times 10^{-4}$
(d) $2 \times 10^{-4}$
Answer:
(a) $4 \times 10^{-8}$
$
\begin{aligned}
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& 7.4=-\log \left[\mathrm{H}^{+}\right] \\
& 7.4=\log 1-\log \left[\mathrm{H}^{+}\right]
\end{aligned}
$
$
\begin{aligned}
& \log \left[\mathrm{H}^{+}\right]=\log 1-7.4 \\
& \log \left[\mathrm{H}^{+}\right]=8.6 \\
& {\left[\mathrm{H}^{+}\right] \text {Antilog of } 8.6} \\
& =4 \times 10^{-8}
\end{aligned}
$
Question 71.
The highest $\mathrm{pH} 14$ is given by
(a) $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
(b) $0.1 \mathrm{M} \mathrm{NaOH}$
(c) $1 \mathrm{~N} \mathrm{NaOH}$
(d) $1 \mathrm{NHCl}$
Answer:
(a) $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
Solution:
$
\begin{aligned}
& {\left[\mathrm{OH}^{-}\right]=1} \\
& \mathrm{pOH}=0 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{pH}=14-0=14
\end{aligned}
$
Question 72 .
Which of the following is not a Bronsted acid?
(a) $\mathrm{CH}_3 \mathrm{NH}_4$
(b) $\mathrm{CH}_3 \mathrm{COO}^{-}$
(c) $\mathrm{H}_2 \mathrm{O}$
(d) $\mathrm{HSO}_4$

Answer:
(b) $\mathrm{CH}_3 \mathrm{COO}^{-}$
Hint:
Those substances which give a proton is called Bronsted acid, while $\mathrm{CH}_3 \mathrm{COO}^{-}$doesn't have a proton. So it is not a Bronsted acid.
Question 73.
Pure water is kept in a vessel and it remains exposed to atmospheric $\mathrm{CO}_2$ which is absorbed, then its $\mathrm{pH}$ will be
(a) greater than 7
(b) less than 7
(c) equal to 7
(d) depends on ionic production of water

Answer:

(b) less than 7
Hint:
$\mathrm{CO}_2$ is acidic oxide which on dissolution in water develops acidic nature.
Question 74.
As the temperature increases, the $\mathrm{pH}$ of $\mathrm{KOH}$ solution
(a) will decrease
(b) will increase
(c) remains constant
(d) depends upon concentration of $\mathrm{KOH}$ solution
Answer:
(a) will decrease

Question 75.
The $\mathrm{pH}$ of millimolar $\mathrm{HCl}$ is
(a) 1
(b) 3
(c) 2
(d) 4
Answer:
(b) 3
$
\begin{aligned}
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& {\left[\mathrm{H}^{+}\right]=10^{-3}} \\
& \mathrm{pH}=\log 1-\log \left[\mathrm{H}^{+}\right] \\
& =\log 1-\log 10^{-3}=3
\end{aligned}
$
Question 76.
The unit of ionic product of water $\mathrm{K}$ is
(a) $\mathrm{mol}^{-1} \mathrm{~L}^{-1}$
(b) $\mathrm{mol}^{-2} \mathrm{~L}^{-2}$
(c) $\mathrm{mol}^{-2} \mathrm{~L}^{-1}$
(d) $\mathrm{mol}^2 \mathrm{~L}^{-2}$
Answer:
(d) $\mathrm{mol}^2 \mathrm{~L}^{-2}$
Question 77.
Review the equilibrium and choose the correct statement.
$
\mathrm{HClO}_4+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{ClO}_4^{+}
$
(a) $\mathrm{HClO}_4$ is the conjugate acid of $\mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{H}_3 \mathrm{O}$ is the conjugate base of $\mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{H}_2 \mathrm{O}$ is the conjugate acid of $\mathrm{H}_3 \mathrm{O}$
(d) $\mathrm{ClO}_4$ is the conjugate base of $\mathrm{HCIO}_4$
Answer:
(d) $\mathrm{ClO}_4$ is the conjugate base of $\mathrm{HCIO}_4$
Question 78 .
Which of the following is the strongest conjugate base?
(a) $\mathrm{CI}^{-}$
(b) $\mathrm{CH}_3 \mathrm{COO}$
(c) $\mathrm{SO}_4^{2-}$
(d) $\mathrm{NO}_2^{-}$
Hint:
$\mathrm{CH}_3 \mathrm{COO}$ is a conjugate base of a weak acid.

$
\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}+\mathrm{H}^{+}
$
Question 79.
Which one of the following substance has the highest proton affinity?
(a) $\mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{H}_2 \mathrm{~S}$
(c) $\mathrm{NH}_3$
(d) $\mathrm{PH}_3$
Answer:
(c) $\mathrm{NH}_3$
Question 80
Which of the following is the strongest Lewis acid?
(a) $\mathrm{BI}_3$
(b) $\mathrm{BBr}_3$
(c) $\mathrm{BCI}_3$
(d) $\mathrm{BF}_3$

Answer:

(a) $\mathrm{BI}_3$
Hint:
Larger the size of the halogen atom less is the back donation of electrons into empty $2 \mathrm{p}$ orbital of $\mathrm{B}$.
Question 81.
Which of the following is the weakest acid?
(a) $\mathrm{HF}$
(b) $\mathrm{HCI}$
(c) $\mathrm{HBr}$
(d) $\mathrm{HI}$
Answer:
(a) $\mathrm{HF}$
Hint:
HF does not give proton easily.

Question 82.
Among the following, the weakest Lewis base is
(a) $\mathrm{H}^{-}$
(b) $\mathrm{OH}^{-}$
(c) $\mathrm{CI}^{-}$
(d) $\mathrm{HClO}_3^{-}$

Answer:

(c) $\mathrm{CI}^{-}$
Hint:
$\mathrm{CI}^{-}$is a conjugate base of strong acid $\mathrm{HCI}$.
Question 83 .
Which of the following is not a Lewis acid?
(a) $\mathrm{BF}_3$
(b) $\mathrm{AlCI}_3$
(c) $\mathrm{HCl}$
(d) $\mathrm{LiAIH}_4$
Hint:
It is a nucleophile and capable of donating electron pair and so it can act as Lewis base.
Question 84
Which one of the following is called amphoteric solvent?
(a) Ammonium hydroxide
(b) Chloroform
(c) Benzene
(d) Water
Answer:
(d) Water
Hint:
$\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}$. Acid due to donation of proton.
Acid
$\mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}$. Base due to accepting a proton. So water is an amphoteric solvent.
Base

Question 85 .
Which of the following is non-electrolyte?
(a) $\mathrm{NaCl}$
(b) $\mathrm{CaCl}_2$
(c) $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$
(d) $\mathrm{CH}_3 \mathrm{COOH}$

Answer:

(c) $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$
Hint:
$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ is a sugar and non-electrolyte.
Question 86.
At infinite dilution, the percentage ionisation for both strong and weak electrolyte is
(a) $1 \%$
(b) $20 \%$
(c) $50 \%$
(d) $100 \%$
Answer:
(d) $100 \%$
Hint:
According to Ostwald's dilution law, degree of ionisation is directly proportional to the dilution.
Question 87.
An acid $\mathrm{HA}$ ionises as $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$The $\mathrm{pH}$ of $1.0 \mathrm{M}$ solution is 5 . its dissociation constant would be
(a) $1 \times 10^{-5}$
(b) $1 \times 10^{-10}$
(c) 5
(d) $5 \times 10^8$
Answer:
(b) $1 \times 10^{-10}$
Question 88.
Three reactions involving $\mathrm{H}_2 \mathrm{PO}_4$ are given below.
(i) $\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-}$
(ii) $\mathrm{H}_2 \mathrm{PO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HPO}_4^2+\mathrm{H}_3 \mathrm{O}^{+}$
(iii) $\mathrm{H}_2 \mathrm{PO}_4+\mathrm{OH}^{-} \rightarrow \mathrm{H}_3 \mathrm{PO}_4+\mathrm{O}_2^{-}$

In which of the above does $\mathrm{H}_2 \mathrm{PO}_4{ }^{-}$act as an acid.
(a) (i) only
(b) (ii) only
(c) (i) \& (iii)
(d) (iii) only
Answer:
(b) (ii) only
Question 89.
Which of the following is not a Lewis acid?
(a) $\mathrm{CO}$
(b) $\mathrm{SiCl}_4$
(c) $\mathrm{SO}_3$
(d) $\mathrm{Zn}^{2+}$
Answer:
(c) $\mathrm{SO}_3$
$\mathrm{CO}$ does not contain vacant d-orbital.
Question 90.
A chemist dissolves an excess of $\mathrm{BaSO}_4$ in pure water at $25^{\circ} \mathrm{C}$ if its $\mathrm{K}_{\mathrm{sp}}=1 \times 10^{-10}$ What is the concentration of Barium in the water?
(a) $10^{-14} \mathrm{M}$
(b) $10^{-5} \mathrm{M}$
(c) $10^{-15} \mathrm{M}$
(d) $10^{-6} \mathrm{M}$
Answer:
(d) $10^{-6} \mathrm{M}$
Question 91.
On addition of ammonium chloride to a solution of ammonium hydroxide
(a) dissociation of $\mathrm{NH}_4 \mathrm{OH}$ increases
(b) concentration of $\mathrm{OH}^{-}$increases
(c) concentration of $\mathrm{OH}^{-}$decreases
(d) concentration of $\mathrm{NH}_4$ and $\mathrm{OH}^{-}$increases

Answer:

(a) dissociation of $\mathrm{NH}_4 \mathrm{OH}$ increases
Hint:
Due to common ion effect.

Question 92.
The solubility product of a salt having a general formula $\mathrm{MX}_2$ in water is $4 \times 10^{-2}$. The concentration of $\mathrm{M}^{2+}$ ions in the aqueous solution of the salt is
(a) $2.0 \times 10^{-6} \mathrm{M}$
(b) $1.0 \times 10^{-4} \mathrm{M}$
(c) $1.6 \times 10^{-4} \mathrm{M}$
(d) $4.0 \times 10^{-2} \mathrm{M}$
Answer:
(b) $1.0 \times 10^{-4} \mathrm{M}$
Solution:
$
\begin{aligned}
& \mathrm{MX}_2 \rightleftharpoons \mathrm{M}^{2+}+2 \mathrm{X} \\
& \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^2(\mathrm{~s})=4 \mathrm{~s}^3 \\
& \mathrm{~s}=2 \sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=\sqrt[3]{\frac{4 \times 10^{-12}}{4}} \\
& =1.0 \times 10^{-4} \mathrm{M}
\end{aligned}
$
Question 93.
The solubility of an aqueous solution of $\mathrm{Mg}(\mathrm{OH})_2$ be $\mathrm{x}$ then its $\mathrm{K}_{\mathrm{sp}}$ is
(a) $4 \mathrm{x}^3$
(b) $108 x^5$
(c) $27 x^4$
(d) $9 \mathrm{x}$
Answer:
(a) $4 \mathrm{x}^3$
$
\begin{aligned}
\mathrm{Mg}(\mathrm{OH})_2 & \rightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{OH}^{-} \\
\mathrm{K}_{\mathrm{sp}} & =4 x^3 \quad(2 x)^2
\end{aligned}
$

Question 94.
What is the correct representation of the solubility product constant of $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
(a) $\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{-2}\right]$
(b) $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CrO}_4^{-2}\right]$
(c) $\left[2 \mathrm{Ag}^{+}\right]\left[\mathrm{CrO}_4^{-2}\right]$
(d) $\left[2 \mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{-2}\right]$
Answer:
(a) $\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{-2}\right]$ $\mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons\left[2 \mathrm{Ag}^{+}\right]\left[\mathrm{CrO}_4^{-2}\right]$ Hence, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2+\left[\mathrm{CrO}_4^{-2}\right]$
Question 95.
What is the $\mathrm{pH}$ value of $\frac{N}{100} \mathrm{KOH}$ solution?
(a) 10
(b) 3
(c) 2
(d) 11
Answer:
(d) 11
Solution.
$10^{-3} \mathrm{~N} \mathrm{KOH}$ will give $\left[\mathrm{OH}^{-}\right]=10^{-3} \mathrm{M}$
$
\begin{aligned}
& \mathrm{pOH}=3 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{pH}=14-3=11
\end{aligned}
$
Question 96.
Which pair will show common ion effect?
(a) $\mathrm{BaCI}_2+\mathrm{Ba}\left(\mathrm{NO}_3\right)_2$
(b) $\mathrm{NaCI}+\mathrm{HCI}$
(c) $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{CI}$
(d) $\mathrm{AgCN}+\mathrm{KCN}$
Answer:
(c) $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{CI}$

Question 97.
The sotubility of $\mathrm{AgCI}$ will be minimum in
(a) $0.001 \mathrm{M} \mathrm{AgNO}_3$
(b) pure water
(c) $0.01 \mathrm{M} \mathrm{CaCI}_2$
(d) $0.01 \mathrm{M} \mathrm{NaCl}$
Answer:
(c) $0.01 \mathrm{M} \mathrm{CaCI}_2$
Solution.
$0.01 \mathrm{M} \mathrm{CaCI}_2$ gives maximum $\mathrm{CI}^{-}$ions to keep $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{AgCl}$ constant, decrease in $\left[\mathrm{Ag}^{+}\right]$will be maximum.
Question 98.
Ionic product of water increases if
(a) pressure is reduced
(b) $\mathrm{H}^{+}$is added
(c) $\mathrm{OH}^{-}$is added
(d) temperature increases
Answer:
(d) temperature increases
Solution:
$\mathrm{K}_{\mathrm{W}}$ increases with increase $\mathrm{in}$ temperature.
Question 99.
$\mathrm{pH}$ of water is 7 . When a substance $\mathrm{Y}$ is added in water, the $\mathrm{pH}$ becomes 13 . The substance $\mathrm{Y}$ is a salt of
(a) strong acid and strong base
(b) weak acid and weak base
(c) strong acid and weak base
(d) weak acid and strong base
Answer:
(d) weak acid and strong base
Question 100.
Sodium chloride is purified by passing $\mathrm{HCl}$ gas in a impure solution of sodium chloride. It is based on
(a) Buffer action
(b) Common ion effect
(c) Association of salt
(d) Hydrolysis of salt

Answer:
(b) Common ion effect
II. Fill in the blanks.
1. ....................... theory does not explain the behaviour of acids and base in non aqueous solvents.
2. According to Lowry Bronsted theory, an acid is defined as a substance that has a tendency to .................. a proton and base is a substance that has a tendency to ................... a proton.
3. $\mathrm{HCI}$ and $\mathrm{Cl}$ are called - ......................... pairs.
4. A ....................... acid is a positive ion or an electron deficient molecule.
5. ........ is an anion or neutral molecule that donates one lone pair of electrons.
6. The ligands act as ....................... and the central metal atoms that accepts a pair of electrons behave as a ........................
7. Carbonium ion act as and carbanion act as .........................
8. Acids with ............................. greater than ten are considered as strong acids and less than one are called weak acids.
9. $\mathrm{OH}^{-}$and $\mathrm{H}_2$ are considered as ............................
10. $\mathrm{ClO}_4^{-}, \mathrm{Cl}^{-}, \mathrm{HSO}_4, \mathrm{NO}_3^{-}$are considered as
11. .......... can act as an acid as well as base.
12. At $25^{\circ} \mathrm{C}$, the value of $\mathrm{K}_{\mathrm{W}}$ is equal to ...........................
13. With the increase in temperature, $\mathrm{K}_{\mathrm{W}}$ value is .................
14. The dissociation of water is an ............................. reaction.
15. Aqueous solution of $\mathrm{HCl}$ is whereas aqueous solution of $\mathrm{NH}_3$ is ......................
16. For neutral solutions, the concentration of $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$as well as $\left[\mathrm{OH}^{-}\right]$is equal to at $25^{\circ} \mathrm{C}$.
17. The $\mathrm{pH}$ of battery acid is equal to ..................
18. The $\mathrm{pH}$ of drain cleaner is equal to ,,,,,,,,,,,,,,,,,,,,
19. ......... is the fraction of the total number of moles of a substance that dissociates at equilibrium.
20. When the dilution increases by 100 times, the dissociation increases by .......................
21. When dilution. .......................the degree of dissociation of weak electrolyte also increases.
22. The buffer present in the blood is...........................
23. ......... introduced a quantity called buffer index ? as a quantitative measure of the
24. When an acid reacts with a base, a salt and water are formed and the reaction is called .........................
25 . ........ is the conjugate base of the weak acid $\mathrm{CH}_3 \mathrm{COOH}$.
26. Kidney stones are developed over a period of time due to the precipitation of ..................
27. The $\mathrm{pH}$ of sea water is .............. than 7.
28. $\mathrm{O}^{2-}$ and $\mathrm{H}^{-}$are ...............
29. All metal ions (or) atoms are ...............
30. All anions are ....................

Answers:
1. Arrhenius
2. donate, accept
3. Conjugate acid-base
4. Lewis
5. Lewis base
6. Lewis base, Lewis acid
7. Lewis acid, Lewis base
8. Ka value
9. very weak acids
10. very weak base
11. Water
12. $1 \times 10^{-14}$
13. increases
14. endothermic
15. acidic, basic
16. $1 \times 10^{-17}$
17. zero
18. 14
19. degree of dissociation?
20. 10 times
21. increases
22. $\mathrm{H}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$
23. Vanslyke, buffer capacity
24. Neutralization
25. $\mathrm{CH}_3 \mathrm{COO}^{-}$
26. Calcium oxalate
27. greater
28. strong base
29. Lewis acids
30. Lewis bases
III. Match the following column - I \& II using the correct code given below that.

Answer:
(a) 4312
Question 2.

Answer:
(b) 3142
Question 3.

Answer:
(c) 2341
Question 4.

Answer:
(d) 4321
Question 5.

Answer:
(a) 3142
Question 6.

Answer:
(a) 2341
Question 7.

Answer:
(b) 2413
Question 8.

Answer:
(c) 3142
Question 9.

Answer:
(d) 3412
Question 10.

Answer:
(a) 2143
Question 11.

Answer:
(a) 3421
IV. Assertion and reasons.
Question 1.
Assertion(A): In the process of dissolution of $\mathrm{HCl}$ in water, $\mathrm{HCl}$ act as acid and $\mathrm{H}_2 \mathrm{O}$ act as base.
Reason ( $\mathrm{R}$ : When $\mathrm{HCl}$ is dissolved in water, it donates a proton to water.
(a) Both $A$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $A$
(b) Both $A$ and $R$ are wrong
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$.
(d) A is wrong but $\mathrm{R}$ is correct.
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ explains $A$
Question 2.
Assertion(A): When ammonia dissolved in water, $\mathrm{H} 20$ acts as an acid.
Reason (R): When ammonia is dissolved in water, it accepts a proton from water. According to Lowry Bronsted theory, proton donor is acid and so water act as an acid.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) Both $A$ and $R$ are wrong
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$.
(d) $A$ is wrong but $\mathrm{R}$ is correct. .
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 3.
Assertion (A): In the reaction $\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}, \mathrm{HCl}$ and $\mathrm{Cl}^{-}$are conjugate acid - base pair.
Reason (R): By Lowry - Bronsted theory, chemical species that differ only by a proton are called conjugate acid - base pair.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) Both $A$ and $R$ are wrong
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$
(d) A is wrong but $\mathrm{R}$ is correct.
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$

Question 4.
Assertion(A): $\mathrm{BF}_3$ is a Lewis acid.
Reason (R): Boron has a vacant $2 \mathrm{p}$ orbital to accept the lone pair of electrons donated by any substance to form a new coordinate covalent bond.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) Both $\mathrm{A}$ and $\mathrm{R}$ are wrong .
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$.
(d) A is wrong but $\mathrm{R}$ is correct.
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 5.
Assertion(A): in coordination compounds, the ligands acts as Lewis acid aid the central metal atom or ion act as Lewis base.
Reason (R): Ligands are capable of accepting of a pair of electrons donated by the central metal atom or ion.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) Both $\mathrm{A}$ and $\mathrm{R}$ are wrong
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$.
(d) $A$ is wrong but $\mathrm{R}$ is correct.
Answer:
(b) Both $A$ and $\mathrm{R}$ are wrong
Question 6.
Assertion(A): $\mathrm{SiF}_4$ can act as Lewis acid.
Reason (R): In $\mathrm{SiF}_4$, the central atom can expand its octet due to the availability of empty d-orbitais and can accept a pair of electrons.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) Both $A$ and $R$ are wrong
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 7.
Assertion(A): $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{ROH}$ all are examples of Lewis bases.
Reason (R): Molecules with one or more lone pairs of electrons act as Lewis bases.
(a) Both $A$ and $R$ are wrong
(b) $\mathrm{A}$ is correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.
(c) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct.
(d) $A$ is correct but $\mathrm{R}$ is wrong
Answer:
(b) $\mathrm{A}$ is correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.

Question 8.
Assertlon(A): $\mathrm{HCl}$ is an strong acid while $\mathrm{HCOOH}$ is a weak acid. Reason (R): $\mathrm{HCI}$ is completely ionised in water whereas $\mathrm{HCOOH}$ is paritally ionised in water.
(a) Both $A$ and $\mathrm{R}$ are wrong
(b) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Answer:
(d) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 9.
Assertion(A): With the increase in temperature, the ionic product of water also increases.
Reason (R): The dissociation of water is an endothermic reaction.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(c) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
(d) Both $\mathrm{A}$ and $\mathrm{R}$ are wrong
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 10.
Assertion(A): When dilution increases, the degree of dissociation of weak electrolyte also increases.
Reason (R): The degree of dissociation a is inversely proportional to concentration c. When the dilution increases by 100 times, the dissociation increases by 10 times.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) Both $A$ and $R$ are wrong
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 11.
Assertion(A): The addition of sodium acetate to acetic acid solution leads to the suppression in the dissociation of acetic acid.
Reason (R): This is due to common ion effect. i.e., $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COONa}$ both contains $\mathrm{CH}_3 \mathrm{COO}^{-}$ ion as common.
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) BothA and $\mathrm{R}$ are wrong
(c) A is correct but $\mathrm{R}$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct

Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 12.
Assertion(A): The solution of $\mathrm{NH}_4 \mathrm{CI}$ has $\mathrm{pH}$ value less than 7 .
Reason (R): The salt of weak base $\left(\mathrm{NH}_4 \mathrm{OH}\right)$ and strong acid (HCl) is acidic in nature, when dissolved in water. So $\mathrm{pH}$ value is less than 7 .
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) Both $A$ and $\mathrm{R}$ are wrong
(c) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(d) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
Question 13.
Assertion(A): $\mathrm{pH}=7$ signifies pure water.
Reason (R): $\mathrm{pH}=7$ means it is a neutral solution where $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]$
(a) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong
(b) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
(c) Both $\mathrm{A}$ and $\mathrm{R}$ are wrong
(d) $\mathrm{A}$ is correct and $\mathrm{R}$ does not explain $\mathrm{A}$
Answer:
(b) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct
Question 14.
Assertion(A): A mixture of $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COONH}_4$ is an acidic buffer.
Reason (R): An acidic buffer contains a weak acid and the salt of weak acid with strong base.
(a) $\mathrm{A}$ is correct but $\mathrm{R}$ is wrong.
(b) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct.
(c) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(d) Both $\mathrm{A}$ and $\mathrm{R}$ are wrong
Answer:
(b) $\mathrm{A}$ is wrong but $\mathrm{R}$ is correct.

Question 15.
Assertion(A): Buffer mixture is the one whose $\mathrm{pH}$ remains constant even by addition of strong acid or strong base.
Reason (R): To resist changes in its $\mathrm{pH}$ on the addition of an acid or base, the buffer solution should contain both acidic as well as basic components so as to neutralise the effect of added acid or base.
(a) Both $A$ and $R$ are correct and $R$ is the correct explanation of $\mathrm{A}$
(b) Both $A$ and $R$ are wrong
(c) $A$ is correct but $R$ is wrong
(d) $A$ is wrong but $R$ is correct
Answer:
(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
V. Find the odd one out and give the reasons.
Question 1.
(a) $\mathrm{HNO}_3$
(b) $\mathrm{Ba}(\mathrm{OH})_2$
(c) $\mathrm{H}_3 \mathrm{PO}_4$
(d) $\mathrm{CH}_3 \mathrm{COOH}$
Answer:
(b) $\mathrm{Ba}(\mathrm{OH})_2$
Reason: $\mathrm{Ba}(\mathrm{OH})_2$ is the base whereas the others are acids.
Question 2.
(a) $\mathrm{NH}_3$
(b) $\mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{RNH}_2$
(d) $\mathrm{BF}_3$
Answer:
(d) $\mathrm{BF}_3$
Reason: $\mathrm{BF}_3$ is a Lewis acid whereas others are Lewis base.

Question 3.
(a) $\mathrm{SiF}_4$
(b) $\mathrm{SF}_4$
(c) $\mathrm{FeCl}_3$
(d) $\mathrm{NH}_3$
Answer:
(d) $\mathrm{NH}_3$
Reason: $\mathrm{NH}_3$ is a Lewis base whereas others are Lewis acid.
Question 4.
(a) $\mathrm{HCl}$
(b) $\mathrm{H}_2 \mathrm{SO}_4$
(c) $\mathrm{CH}_3 \mathrm{COOH}$
(d) $\mathrm{HNO}_3$
Answer:
(c) $\mathrm{CH}_3 \mathrm{COOH}$
Reason: $\mathrm{CH}_3 \mathrm{COOH}$ is a weak acid whereas others are strong acids.
Question 5.
(a) $\mathrm{HCOOH}$
(b) $\mathrm{CH}_3 \mathrm{COOH}$
(c) Lactic acid
(d) $\mathrm{HCI}$
Answer:
(d) $\mathrm{HCI}$
Reason: $\mathrm{HCl}$ is a strong acid whereas others are weak acids.
Question 6.
(a) $\mathrm{HClO}_4$
(b) $\mathrm{HCI}$
(c) $\mathrm{HSO}_4$
(d) $\mathrm{H}_2 \mathrm{SO}_4$
Answer:
(c) $\mathrm{HSO}_4$
Reason: $\mathrm{HSO}_4$ is a very weak base whereas others are strong acid.

Question 7.
(a) $\mathrm{NH}_2^{-}$
(b) $\mathrm{O}_2^{-}$
(c) $\mathrm{H}^{-}$
(d) $\mathrm{OH}^{-}$
Answer:
(d) $\mathrm{OH}^{-}$
Reason: $\mathrm{OH}^{-}$is a very weak acid whereas others are strong bases.
Question 8.
(a) $\mathrm{HNO}_2$
(b) $\mathrm{HF}$
(c) $\mathrm{H}_2 \mathrm{SO}_4$
(d) $\mathrm{CH}_3 \mathrm{COOH}$
Answer:
(c) $\mathrm{H}_2 \mathrm{SO}_4$
Reason: $\mathrm{H}_2 \mathrm{SO}_4$ is a strong acid whereas others are weak acids.
Question 9.
(a) $\mathrm{F}^{-}$
(b) $\mathrm{CH}_3 \mathrm{COO}$
(c) $\mathrm{O}_2$
(d) $\mathrm{NO}_2^{-}$
Answer:
(c) $\mathrm{O}^{2-}$
Reason: $\mathrm{O}^{2-}$ is a strong base whereas others are weak bases.
Question 10.
(a) Vinegar
(b) Black coffee
(c) Sea water
(d) Orange juice
Answer:
(c) Sea water
Reason: Sea water is basic and has $\mathrm{pH}>7$ whereas others are acidic and have $\mathrm{pH}<7$.

Question 11.
(a) Baking soda
(b) Tomato
(c) Soapy water
(d) Drain cleaner
Answer:
(b) Tomato
Reason: Tomato has $\mathrm{pH}$ less than 7 and it is acidic whereas others have $\mathrm{pH}$ greater than 7 and they are basic.
Question 12
(a) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$
(b) $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{CI}$
(c) $\mathrm{H}_2 \mathrm{CO}_3+\mathrm{NaHCO}_3$
(d) $\mathrm{NaOH}+\mathrm{NaCl}$
Answer:
(d) $\mathrm{NaOH}+\mathrm{NaCl}$
Reason: $\mathrm{NaOH}+\mathrm{NaCl}$ is not a buffer mixture whereas others are buffer mixtures.
VI. Find out the incorrect pair.
Question 1.
(a) $\mathrm{HNO}_3, \mathrm{H}_2 \mathrm{SO}_4$
(b) $\mathrm{Al}(\mathrm{OH})_3, \mathrm{Mg}(\mathrm{OH})_2$
(c) $\mathrm{CH}_3 \mathrm{COOH}, \mathrm{HCOOH}$
(d) $\mathrm{H}_2 \mathrm{O}, \mathrm{OH}^{-}$
Answer:
(d) $\mathrm{H}_2 \mathrm{O}, \mathrm{OH}$
Question 2.
(a) $\mathrm{HCl}, \mathrm{Cl}^{-}$
(b) $\mathrm{H}_2 \mathrm{O}, \mathrm{H}_3 \mathrm{O}^{+}$
(c) $\mathrm{HNO}_3, \mathrm{HNO}_2$
(d) $\mathrm{H}_2 \mathrm{SO}_4, \mathrm{HSO}_4^{-}$
Answer:
(c) $\mathrm{HNO}_3, \mathrm{HNO}_2$

Question 3.
(a) $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{ROH}, \mathrm{ROR}$
(c) $\mathrm{CN}^{-}, \mathrm{SCN}^{-}$
(d) $\mathrm{BF}_3, \mathrm{H}_2 \mathrm{O}$
Answer:
(d) $\mathrm{BF}_3, \mathrm{H}_2 \mathrm{O}$
Question 4 .
(a) $\mathrm{BF}_3, \mathrm{BF}_2$
(b) $\mathrm{Fe}_2, \mathrm{Fe}_3$
(c) $\mathrm{CaO}, \mathrm{Mg}(\mathrm{OH})_2$
(d) $\mathrm{SiF}_4, \mathrm{SF}_4$
Answer:
(c) $\mathrm{CaO}, \mathrm{Mg}(\mathrm{OH})_2$
Question 5 .
(a) Orange juice, Tomato juice
(b) Soapy water, Sea water
(c) Water, $\mathrm{H}_3 \mathrm{O}$
(d) Bleach, Ammonia solution
Answer:
(c) Water, $\mathrm{H}_3 \mathrm{O}$
VII. Find out the correct pair.
Question 1.
(a) $\mathrm{HNO}_3, \mathrm{Ba}(\mathrm{OH})_2$
(b) $\mathrm{CH}_3 \mathrm{COOH}, \mathrm{HCI}$
(c) $\mathrm{H}_3 \mathrm{O}^{+}, \mathrm{Cl}^{-}$
(d) $\mathrm{HCl}+\mathrm{H}_2 \mathrm{SO}_4$
Answer:
(d) $\mathrm{HCl}+\mathrm{H}_2 \mathrm{SO}_4$.

Question 2.
(a) $\mathrm{BF}_3, \mathrm{NH}_4^{+}$
(b) $\mathrm{CH}_2^{-}, \mathrm{CH}_3^{+}$
(c) $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}$
(d) $\left(\mathrm{CH}_3\right) 3 \mathrm{C}^{+}, \mathrm{CH}_2=\mathrm{CH}_2$
Answer:
(c) $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}$
Question 3.
(a) $\mathrm{H}_3 \mathrm{O}_{+}, \mathrm{HCI}$
(b) $\mathrm{HSO}_4, \mathrm{NO}_2$
(c) $\mathrm{HNO}_2, \mathrm{H}_2$
(d) $\mathrm{HCl}, \mathrm{Cl}^{-}$
Answer:
(d) $\mathrm{HCI}, \mathrm{CI}$
Question 4.
(a) Orange, Black coffee
(b) Baking soda, Water
(c) Ammonia, Stomach acid
(d) Bleach, Tomato
Answer:
(a) Orange, Black coffee
Question 5.
(a) $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NaOH}$
(b) $\mathrm{NaOH}+\mathrm{NaCl}$
(c) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$
(d) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONH}_4$
Answer:
(c) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$

VIII. Answer the following.
Question 1.
What are the general characteristics of acid and base?
Answer:
1. Acid tastes sour, turns the blue litmus to red and reacts with metals such as zinc and produces hydrogen gas.
2. Base tastes bitter, turns the red litmus to blue and soapy to touch.
Question 2.
Explain the Arrhenius concept of acid and base with example.
Answer:
1. According to Arrhenius concept, an acid is a substance that dissociates to give hydrogen ions in water. For example,
$
\mathrm{HCl}_{(\mathrm{g})} \stackrel{\mathrm{H}_2 \mathrm{O}}{\rightleftharpoons} \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{-(\mathrm{aq})}^{-}
$
2. Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example,
$
\mathrm{Ca}(\mathrm{OH})_2 \stackrel{\mathrm{H}_2 \mathrm{O}}{\rightleftharpoons} \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-} \text {(aq) }
$
Base
Question 3.
What are the limitations of Arrhenius concept?
Answer:
1. Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan.
2. This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.
Question 4.
What is meant by strong acid and weak acid? Explain with example.
Answer:
1. A strong acid is the one that is almost completely dissociated in water.
$
\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}
$

2. A weak acid is the one that is partially dissociated in water.
$
\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}+\mathrm{CH}_3 \mathrm{COO}^{-}
$
Question 5.
Give two examples for
1. Strong acid
2. Strong base
Answer:
1. Strong acid: $\mathrm{HCIO}_4, \mathrm{H}_2 \mathrm{SO}_4$
2. Strong base: $\mathrm{NH}_2{ }^{-}, \mathrm{O}^{2-}$
Question 6 .
Give two examples for
1. Very weak acid
2. Very weak base
Answer:
1. Very weak acid: $\mathrm{OH}, \mathrm{H}_2$
2. Very weak base: $\mathrm{Cl}^{-}, \mathrm{ClO}_4^{-}$
Question 7.
Give two examples for
1. Weak acid
2. Weak base
Answer:
1. Weak acid: $\mathrm{HF}, \mathrm{CH}_3 \mathrm{COOH}$
2. Weak base: $\mathrm{F}, \mathrm{CH}_3 \mathrm{COO}$

Question 8.
What is meant by auto ionisation of water?
Answer:
Pure water has a little tendency to dissociate. i:e., one water molecule donates a proton to another water molecule. This is known as auto ionisation of water.

Question 9.
Define - ionic product of water.
Answer:
$
\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}+\mathrm{OH}
$
The dissociation constant for the above ionisation is given as,
$
\mathrm{K}_{\mathrm{w}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}\right]^2}
$
The concentration of pure water is one.
i.e., $\left[\mathrm{H}_2 \mathrm{O}\right]^2=1$
$\mathrm{K}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]$
$\mathrm{K}_{\mathrm{W}}=$ ionic product of water.
$\mathrm{K}_{\mathrm{W}}=1 \times 10^{-14}$ at $25^{\circ} \mathrm{C}$
Question 10.
$\mathrm{K}_{\mathrm{W}}=1 \times 10^{-14}$ at $25^{\circ} \mathrm{C}$. Justify this statement.
Answer:
1. Experimentally found that the concentration of $\mathrm{H}_3 \mathrm{O}$ in pure water is $1 \times 10^{-7}$ at $25^{\circ} \mathrm{C}$.
2. Since the dissociation of water produces equal number of $\mathrm{H}_3 \mathrm{O}_4$ and $\mathrm{OH}^{-}$, the concentration of $\mathrm{OH}^{-}$is also equal to $1 \times 10^{-}$at $25^{\circ} \mathrm{C}$. The ionic product of water at $25^{\circ} \mathrm{C}$ is
$
\begin{aligned}
& \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \\
& =\left[1 \times 10^{-7}\right]\left[1 \times 10^{-7}\right] \\
& \mathrm{K}_{\mathrm{W}}=\left[1 \times 10^{-14}\right]
\end{aligned}
$

Question 11.
With increase in temperature, $\mathrm{K}_{\mathrm{W}}$ also increases. Why?
Answer:
1. All equilibrium constant $\mathrm{K}_{\mathrm{W}}$. is also a constant at a particular temperature.
2. The dissociation of water is an endothermic reaction.
3. With the increase in temperature, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$and $\mathrm{OH}^{-}$also increases and hence the ionic product also increases.
Question 12.
Aqueous $\mathrm{HCl}$ is an acidic solution whereas aqueous $\mathrm{NH}_3$ is a basic solution. Justify this statement.
Answer:
$\mathrm{HCI}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}_{+}+\mathrm{Cl}^{-}$in this case, in addition to auto ionisation of water, $\mathrm{HCI}$ molecule also produces $\mathrm{H}_3 \mathrm{O}$ ion by donating a proton to water and hence $[\mathrm{H} 3 \mathrm{OJ}>[\mathrm{OH}]$. It means that the aqueous $\mathrm{HCI}$ solution is acidic. Similarly in basic solution such as aqueous $\mathrm{NH}_3,\left[\mathrm{OH}^{-}\right]>\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$and it is basic.
Question 13.
What is the statement of Ostwald's dilution law.
Answer:
When dilution increases, the degree of dissociation of weak electrolyte also increases.
$
\mathrm{K}_{\mathrm{a}}=\frac{\alpha^2 \mathrm{c}}{1-\alpha}
$
This statement is known as Ostwald's dilution law.
Question 14.
Define - Salt hydrolysis.
Answer:
Salts completely dissociate in aqueous solution to give their constituent ions. The ions so produced are hydrated in water. in certain cases, the cation, anion ?r both react with water and the reaction is called salt hydrolysis. e.g.,
$
\mathrm{NaOH}_{(\mathrm{aq})}+\mathrm{HNO}_{3(\mathrm{aq})} \rightarrow \mathrm{NaNO}_{3(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}_{(1)}
$
Question 15.
What is meant by conjugate acid-base pair? Find the conjugate acid / base for the following species

Answer:
$
\mathrm{HNO}_2, \mathrm{CH}^{-}, \mathrm{HCIO}_4, \mathrm{OH}^{-}, \mathrm{CO}_3^{2-}, \mathrm{S}^{2-}
$
An acid-base pair which differs by a proton only $\left(\mathrm{HA} \rightleftharpoons \mathrm{A}^{-}+\mathrm{H}^{+}\right)$is known as conjugate acid-base pair. Conjugate acid: $\mathrm{HCN}, \mathrm{H}_2 \mathrm{O}, \mathrm{HCO}_3^{-}, \mathrm{HS}^{-}$.
Conjugate base: $\mathrm{NO}_2^{-}, \mathrm{ClO}_4^{-}, \mathrm{O}_2^{-}$.

Question 16.
Which of the following are Lewis Acids?
$\mathrm{H}_2 \mathrm{O}, \mathrm{BF}_3, \mathrm{H}^{+}$and $\mathrm{NH}_4^{+}$
Answer:
$\mathrm{BF}_3, \mathrm{H}^{+}$ions are Lewis acids.
Question 17.
What will be the conjugate bases for the Bronsted acids? $\mathrm{HF}, \mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{H}_2 \mathrm{CO}_3$ ?
Answer:
Conjugate bases: $\mathrm{F}^{-}, \mathrm{HSO}_4^{-}, \mathrm{HCO}_3^{-}$.
Question 18 .
Write the conjugate acids for the following Bronsted bases:
$\mathrm{NH}_2^{-}, \mathrm{NH}_3$ and $\mathrm{HCOO}^{-}$
Answer:
$\mathrm{NH}_3, \mathrm{NH}_4$ and $\mathrm{HCOOH}$
Question 19.
The species $\mathrm{H}_2 \mathrm{O}, \mathrm{HCO}_3{ }^{-}, \mathrm{HSO}_4{ }^{-}$and $\mathrm{NH}_3$ can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.
Answer:

Question 20.
Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid I Lewis base?
(a) $\mathrm{OH}^{-}$ions
(b) $\mathrm{F}^{-}$
(c) $\mathrm{H}^{+}$
(d) $\mathrm{BCI}_3$
Answer:
(a) $\mathrm{OH}^{-}$ions can donate an electron pair and act as Lewis base.
(b) $\mathrm{F}^{-}$ions can donate an electron pair and act as Lewis base.
(a) $\mathrm{H}^{+}$ions can accept an electron pair and act as Lewis base.
(b) $\mathrm{BCl}_3$ can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.
Question 21.
Predict the acidic, basic or neutral nature of the solutions of the following salts:
$\mathrm{NaCI}, \mathrm{KBr}, \mathrm{NaCN}, \mathrm{NH}_4 \mathrm{NO}_3, \mathrm{NaNO}_2, \mathrm{KF}$.
Answer:
$\mathrm{NaCN}, \mathrm{NaNO}_2, \mathrm{KF}$ solutions are basic, as they are salts of strong base, weak acid.
$\mathrm{NaCl}, \mathrm{KBr}$ solutions are neutral, as they are salts, of strong acid, strong base.
$\mathrm{NH}_4 \mathrm{NO}_3$ solution is acidic, as it is a salt of strong acid, weak base.
Question 22.
Ionic product of water at $310 \mathrm{~K}$ is $2.7 \times 10^{-14}$ What Is the $\mathrm{pH}$ of neutral water at this temperature?
Answer:
$
\begin{aligned}
{\left[\mathrm{H}^{+}\right] } & =\sqrt{\mathrm{K}_{\mathrm{w}}}=\sqrt{2.7 \times 10^{-14}}=1.643 \times 10^{-7} \mathrm{M} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=-\log \left(1.643 \times 10^{-7}\right)=7-0.2156=6.78
\end{aligned}
$

Question 23.
The aqueous solution of sugar does not conduct electricity whereas when sodium chloride is added to water, it conducts electricity. Justify this statement.
Answer:
1. Sugar is a non electrolyte and when it dissolves in water, there will be no ionisation takes place. If there is no free ions, it does not conduct electricity.
2. When sodium chloride is added to water, it is completely ionised to give $\mathrm{Na}$ ions and $\mathrm{Cl}^{-}$ions. Due to the presence of ions, they will be possibility of electhcal conductance. Because ions are carriers of electric current.
Question 24 .
A reaction betwen ammonia and boron trifluoride is given below
$
\stackrel{\mathrm{NH}_3}{ }+\mathrm{BF}_3 \rightarrow \mathrm{H}_3 \mathrm{~N}: \mathrm{BF}_3
$
Identify the acid and base in the reaction. Which theory explain it?
Answer:
$
\ddot{\mathrm{NH}_3}+\mathrm{BF}_3 \rightarrow \mathrm{H}_3 \mathrm{~N}: \mathrm{BF}_3
$
1. In the above reaction $\mathrm{BF}_3$ is an acid and $\mathrm{NH}_3$ is the base.
2. Lewis concept explain it as follows

3. A Lewis acid is an electron deficient molecule and capable of accepting a pair of electrons and a Lewis base is electron rich molecule and capable of donating a pair of electrons.
Question 25.
The salt of strong acid and strong base does not undergo hydrolysis. Explain.
Answer:
1. In this case, neither the cations nor the anions undergo hydrolysis. Therefore the solution remains neutral.
2. For example, in the aqueous solution of $\mathrm{NaCl}$, its ions $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions have no tendency to react with $\mathrm{H}^{+}$or $\mathrm{OH}^{-}$ions of water.
This is because the possible products of such reaction are $\mathrm{NaOH}$ and $\mathrm{HCI}$ which are completely dissociated. As a result, there is no change in the concentration of $\mathrm{W}$ and $\mathrm{OH}^{-}$ions and hence the solution continues to remain neutral.
3 Mark Questions ans Answers
Question 1.
Explain Lowry - Bronsted theory of acid and base.
Answer:
1. According to Lowry-Bronsted theory, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance.
2. An acid is a proton donor and a base is a proton acceptor.
3. When $\mathrm{HCI}$ is dissolved in $\mathrm{H}_2 \mathrm{O}, \mathrm{HCI}$ donates a proton to $\mathrm{H}_2 \mathrm{O}$. Thus $\mathrm{HCI}$ behaves as an acid and $\mathrm{H}_2 \mathrm{O}$ is a base.
$
\mathrm{HCI}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}
$
Question 2.
Explain the reaction of water with ammonia by proton theory.
Answer:
1. When ammonia dissolved in water, it accepts a proton from water. In this case, ammonia $\left(\mathrm{NH}_3\right)$ acts as a base and $\mathrm{H}_2 \mathrm{O}$ is acid.
2. The reaction is represented as

3. The species that remains after the donation of a proton is a base (Base ${ }_1$ ) and is called the conjugate base of Bronsted acid (Acid A $_1$. In other words, chemical species that differ only by a proton are called conjugate acid base pairs Conjugate acid - base pair

Question 3.
Explain about the strength of acids on the basis of $\mathrm{K}_{\mathrm{a}}$ value.
Answer:
1. $\mathrm{K}_{\mathrm{a}}$ is called the ionisation constant or dissociation constant of the acid. It measures the strength of an acid.
2. Acids such as $\mathrm{HCI}, \mathrm{HNO}_3$ are almost completely? onised and hence they have high $\mathrm{K}_{\mathrm{a}}$ value i.e., $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{HCI}$ at $25^{\circ} \mathrm{C}$ is $2 \times 10^6$.
3. Acids such as formic acid and acetic acid are partially ionised in solution and have low $\mathrm{K}_{\mathrm{a}}$ value. i.e., $\mathrm{K}_{\mathrm{a}}$ for acetic acid $1.8 \times 10^{-5}$ at $25^{\circ} \mathrm{C}$
4. Acids with $\mathrm{K}_{\mathrm{a}}$ value greater than ten are considered as strong acids and less than one considered as weak acids.
Question 4.
Write 3 formulas of strong acids, strong bases and weak acids.
Answer:
1. $\mathrm{HClO}_4, \mathrm{HCI}, \mathrm{H}_2 \mathrm{SO}_4$ - are strong acids
2. $\mathrm{NH}_2^{-}, \mathrm{O}^{2-}, \mathrm{H}^{-}-$are strong bases
3. $\mathrm{HNO}_2, \mathrm{HF}, \mathrm{CH}_3 \mathrm{COOH}$ are weak acids
Question 5 .
$\mathrm{pH}$ of a neutral solution is equal to 7 . Prove it.
Answer:
1. in neutral solutions, the concentration of $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$as well as $\left[\mathrm{OH}^{-}\right]$are equal to $1 \times 10^{-7} \mathrm{M}$ at $25^{\circ} \mathrm{C}$.
2. The $\mathrm{pH}$ of a neutral solution can be calculated by substituting this $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$çoncentration in the expression
$
\begin{aligned}
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\
& =-\log _{10}\left[1 \times 10^{-7}\right] \\
& =-(-7) \log \frac{1}{2}=+7(1)=7
\end{aligned}
$
3. $\mathrm{pH}=7$ for a neutral solution

Question 6.
Derive the relation between $\mathrm{pH}$ and $\mathrm{pOH}$
Answer:
$
\begin{aligned}
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}\right] \\
& \mathrm{pOH}=-\log _{10}[\mathrm{OH}]
\end{aligned}
$
Adding equations (1) and (2),
$
\begin{aligned}
& \mathrm{pH}+\mathrm{pOH}=\left(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right)+\left(-\log _{10}\left[\mathrm{OH}^{-}\right]\right)\right. \\
& =-\left[\left(\log _{10}\left[\mathrm{H}_3 \mathrm{O}\right]\right)+\left(\log _{10}\left[\mathrm{OH}^{-}\right)\right]\right. \\
& \mathrm{pH}+\mathrm{pOH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \\
& {\left[\mathrm{H}_3 \mathrm{O}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}} \\
& \mathrm{pH}+\mathrm{pOH}=-\log \mathrm{K}_{\mathrm{W}} \\
& \mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{W}} \\
& {\left[\mathrm{pK} \mathrm{W}_{\mathrm{W}}=-\log _{10 \mathrm{KW}}\right]}
\end{aligned}
$
At $25^{\circ} \mathrm{C}$, the ionic product of water $\mathrm{Kw}=1 \times 10^{-14}$.
$
\begin{aligned}
& \mathrm{pK}_{\mathrm{W}}=-\log _{10} 10^{-14}=14 \log _{10} 10=14 \\
& \mathrm{pK}_{\mathrm{W}}=14 \\
& \mathrm{pH}+\mathrm{pOH}=14 \text { at } 25^{\circ} \mathrm{C} \text {. }
\end{aligned}
$

Question 7.
When the dilution increases by 100 times, the dissociation increases by 10 times. Justify this statement.

Answer:
(i). Let us consideran acid with $\mathrm{K}_{\mathrm{a}}$ value $4 \times 10^4$. We are calculating the degree of dissociation of that acid at two different concentration $1 \times 10^{-2} \mathrm{M}$ and $1 \times 10^{-4} \mathrm{M}$ using Ostwalds dilution law $\alpha=\sqrt{\frac{K_a}{\mathrm{C}}}$
(ii) For $1 \times 10^{-2} \mathrm{M}$ acid, $\alpha=\sqrt{\frac{4 \times 10^{-4}}{1 \times 10^{-2}}}=\sqrt{4 \times 10^{-2}}=2 \times 10^{-1}=0.2$
(ii) For $1 \times 10^{-4} \mathrm{M}$ acid, $\alpha=\sqrt{\frac{4 \times 10^{-4}}{1 \times 10^{-4}}}=\sqrt{4}=2$
(iv) i.e., when the dilution increases by 100 times (concentration decreases from $1 \times 10^{-2} \mathrm{M}$ to $1 \times 10^{-4} \mathrm{M}$ ), the dissociation increases by 10 times.
(v) When dilution increases, the degree of dissociation of weak electrolyte also increase. (Ostwalsd's dilution law).

Question 8.
What Is buffer solution? Give an example for an acidic buffer and a basic buffer.

Answer:
1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.
2. This buffer solution resists drastic changes in its $\mathrm{pH}$ upon addition of a small quantities of acids (or) bases and this ability is called buffer action.
3. Acidic buffer solution, Solution containing acetic acid and sodium acetate. Basic buffer solution, Solution containing $\mathrm{NH}_4 \mathrm{O}$ and $\mathrm{NH}_4 \mathrm{Cl}$.
Question 9.
Define buffer capacity and buffer index.
Answer:
1. The buffering ability of a solution can be measured in terms of buffer capacity.
2. Buffer index ?, as a quantitative measure of the buffer capacity.
3. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its $\mathrm{pH}$ by unity.
4. $\beta=\frac{d B}{d(p H)}$. $\mathrm{dB}=$ number of gram equivalents of acid / base added to one litre of buffer solution. $\mathrm{d}(\mathrm{pH})$ $=$ The change in the $\mathrm{pH}$ after the addition of acid / base.
Question 10.
How is solubility product is used to decide the precipitation of ions?
Answer:
1. When the product of molar concentration of the constituent ions i.e., ionic product exceeds the solubility product then the compound gets precipitated.
2. When the
ionic Product $>\mathrm{K}_{\mathrm{sp}}$ precipitation will occur and the solution is super saturated. ionic Product $<\mathrm{K}_{\mathrm{sp}}$ no precipitation and the solution is unsaturated. ionic Product $=\mathrm{K}_{\text {sp }}$ equilibrium exist and the solution $i$ s saturated.
3. By this way, the solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contain the constituent ions are mixed.

Question 11.
Derive the value of solubility product from molar solubility.
Answer:
1. Solubility can be calculated from molar solubility.i.e., the maximum number of moles of the solute that can be dissolved in one litre of the solution.
2. For a solute $\mathrm{X}_{\mathrm{m}} \mathrm{Y}_{\mathrm{n}}$
$
\mathrm{X}_{\mathrm{m}} \mathrm{Y}_{\mathrm{n}(\mathrm{s})} \rightleftharpoons \mathrm{mX}_{(\mathrm{aq})}^{\mathrm{n}+}+\mathrm{n}^{\mathrm{m}-}(\mathrm{aq})
$
3. From the above stoichiometrically balanced equation, it is clear that $\mathrm{I}$ mole of $\mathrm{X}_{\mathrm{m}} \mathrm{Y}_{\mathrm{n}(\mathrm{s})}$ dissociated to furnish ' $m$ ' moles of $\mathrm{x}$ and ' $n$ ' moles of Y. If's' is the molar solubility of $\mathrm{X}_{\mathrm{m}} \mathrm{Y}_{\mathrm{n}}$ then
Answer:
$
\begin{aligned}
& {\left[\mathrm{X}^{\mathrm{n}+}\right]=\mathrm{ms} \text { and }\left[\mathrm{Y}^{\mathrm{m}-}\right]=\mathrm{ns}} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{X}^{\mathrm{n}+}\right]^{\mathrm{m}}\left[\mathrm{Y}^{\mathrm{m}-}\right]^{\mathrm{n}} \\
& \mathrm{K}_{\mathrm{sp}}=(\mathrm{ms})^{\mathrm{m}}(\mathrm{ns})^{\mathrm{n}} \\
& \mathrm{K}_{\mathrm{sp}}=(\mathrm{m})^{\mathrm{m}}(\mathrm{n})^{\mathrm{n}}(\mathrm{s})^{\mathrm{m}+\mathrm{n}}
\end{aligned}
$
Question 12.
The concentration of hydrogen ions $; \mathrm{n}$ a sample of soft drink is $3.8 \times 10^{-3} \mathrm{~m}$. What is the. $\mathrm{pH}$ value? Whether the soft drink is acidic (or) basic?
Answer:
$
\begin{aligned}
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\
& =-\log _{10}\left[3.8 \times 10^{-}\right] \\
& =-\log 3.8+3 \\
& =3-0.5798=2.4202 \\
& \mathrm{pH}=2.42
\end{aligned}
$
When $\mathrm{pH}<7$, the soft drink is acidic.
Question 13.
The $\mathrm{pH}$ of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
$
\begin{aligned}
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\
& =-\log _{10}=-\mathrm{pH}=-3.76 \\
& =\overline{4} .24 \\
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\operatorname{antilog} \overline{4} .24} \\
& =1.738 \times 10^{-4} \\
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=1.74 . \times 10^{-4} \mathrm{M}}
\end{aligned}
$

Question 14.
The ionisation constant of $\mathrm{HF}, \mathrm{HCOOH}, \mathrm{HCN}$ at $298 \mathrm{~K}$ are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionisation constant of the corresponding conjugate base.
Answer:
1. $\mathrm{HF}$, conjugate base is $\mathrm{F}$
$
\mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}} / \mathrm{K}_{\mathrm{a}}=\frac{1 \times 10^{-4}}{6.8 \times 10^{-4}}=1.47 \times 10^{-11}=1.5 \times 10^{-11}
$
2. for $\mathrm{HCOO}^{-}$
$
\mathrm{K}_{\mathrm{b}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}=5.6 \times 10^{-11}
$
3. for $\mathrm{CN}^{-}$
$
\mathrm{K}_{\mathrm{b}}=\frac{1 \times 10^{-14}}{4.8 \times 10^{-4}}=2.8 \times 10^{-6}
$
Question 15 .
The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of cyanic acid (HCNO) is 2.34 . Calculate the ionization constant of the acid and its degree of ionization in the solution.
$
\begin{aligned}
& \mathrm{HCNO} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CNO}^{-} \\
& \mathrm{pH}=2.34 \text { means }-\log \left[\mathrm{H}^{+}\right]=2.34 \text { or } \log \left[\mathrm{H}^{+}\right]=-2.34=3.86 \\
& \text { or } \\
& {\left[\mathrm{H}^{+}\right]=\text {Antilog } 3.86=4.57 \times 10^{-3} \mathrm{M}} \\
& {\left[\mathrm{CNO}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.57 \times 10^{-3} \mathrm{M}} \\
& \mathrm{K}_{\mathrm{a}}=\frac{\left(4.57 \times 10^{-3}\right)\left(4.57 \times 10^{-3}\right)}{0.1}=2.09 \times 10^{-4} \\
& \alpha=\sqrt{\mathrm{K}_{\mathrm{a}} / \mathrm{C}}=\sqrt{2.09 \times 10^{-4} / 0.1}=0.0457
\end{aligned}
$

Question 16.
The Ionization constant of nitrous acid is $4.5 \times 10^{-4}$. Calculate the $\mathrm{pH}$ of $0.04 \mathrm{M}$ sodium nitrite solution and also its degree of hydrolysis.
Answer:
Sodium mtrite is a salt of weak acid, strong base. Hence,
$
\begin{aligned}
& \mathrm{K}_{\mathrm{h}}=2.22 \times 10_{-11} \mathrm{~K}_{\mathrm{W}} / \mathrm{K}_{\mathrm{b}}=10^{-14} /\left(4.5 \times 10^{-4}\right) \\
& \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}} / \mathrm{c}}=\sqrt{2.22 \times 10^{-11} / 0.04}=\sqrt{5.5 \times 10^{-11}}=2.36 \times 10^{-5} \\
& \mathrm{NO}_2^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{HNO}_2+\mathrm{OH}^{-} \\
& \text {Initial } \quad \mathrm{c} \quad \mathrm{ch} \\
& \text { After hydrolysis } \quad \mathrm{c}-\mathrm{ch} \quad \mathrm{ch} \\
& {\left[\mathrm{OH}^{-}\right]=\mathrm{ch}=0.04 \times 2.36 \times 10^{-5}=944 \times 10^{-7}} \\
& \text { pOH }=-\log \left(9.44 \times 10^{-7}\right)=7-0.9750=6.03 \\
& \mathrm{pH}=14-\mathrm{pOH}=14-6.03=7.97
\end{aligned}
$
Question 17
What is the minimum volume of water required to dissolve $1 \mathrm{~g}$ of calcium sulphate at $298 \mathrm{~K}$. For calcium sulphate, $\mathrm{K}_{\mathrm{sp}}=9.1 \times 10^{-6}$
Answer:
$
\mathrm{CaSO}_4(\mathrm{~s}) \mathrm{Ca}^2(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq})
$
If ' $\mathrm{s}$ ' is the solubility of $\mathrm{CaSO}_4$ in moles $\mathrm{L}^{-}$, then $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \mathrm{x}\left[\mathrm{SO}_4^{2-}\right]=\mathrm{s}^2$
or
$
\begin{aligned}
& \mathrm{s}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=\sqrt{9.1 \times 10^{-6}}=3.02 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\
& =3.02 \times 10^{-3} \times 136 \mathrm{~g} \mathrm{~L}^{-1}=0.411 \mathrm{gL}^{-1} \\
& =3.02 \times 10^{-3} \times 136 \mathrm{gL}^{-1}=0.411 \mathrm{gL}^{-1} \\
& \left(\text { Molar mass of } \mathrm{CaSO}_4=136 \mathrm{~g} \mathrm{~mol}^{-1}\right)
\end{aligned}
$
Thus, for dissolving $0.441 \mathrm{~g}$, water required $=\mathrm{I} \mathrm{L}$
For dissolving $1 \mathrm{~g}$, water required $=\frac{1}{0.411} \mathrm{~L}=2.43 \mathrm{~L}$

Question 18.
1. Point out the differences between ionic product and solubility product.
2. The solubllity of $\mathrm{AgCI}$ in water at $298 \mathrm{~K}$ is $1.06 \times 10^{-5}$ mole per litre. Calculate is solubility product at this temperature.
Answer:
1. Ionic product
1. It is applicable to all types of solutions.
2. Its value changes with the change in con centration of the ions.
Solubility product
1. It is applicable to the saturated solutions.
2. It has a definite value for an electrolyte at a constant temperature.
2. The solubility equilibrium in the saturated solution is $\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$
The solubility of $\mathrm{AgCl}$ is $1.06 \times 10^{-5}$ mole per litre.
$
\begin{aligned}
& {\left[\mathrm{Ag}^{+}(\mathrm{aq})\right]=1.06 \times 10^{-5} \mathrm{~mol} \mathrm{~L}} \\
& {\left[\mathrm{Cl}^{-}(\mathrm{aq})\right]=1.06 \times 10^{-5} \mathrm{~mol} \mathrm{~L}} \\
& \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}(\mathrm{aq})\right]\left[\mathrm{Cl}^{-}(\mathrm{aq})\right] \\
& =\left(1.06 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\right) \times\left(1.06 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\right) \\
& =1.12 \times 10^{-2} \mathrm{moI}^2 \mathrm{~L}^{-2}
\end{aligned}
$
Question 19.
The value of $\mathrm{K}$ of two sparingly soluble salts $\mathrm{Ni}(\mathrm{OH})_2$ and $\mathrm{AgCN}$ are $2.0 \times 10^{-15}$ and $6 \times 10^{-17}$ respectively. Which salt is more soluble? Explain.
Answer:

$
\begin{aligned}
\mathrm{AgCN} & \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{CN}^{-} \\
\mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right]=6 \times 10^{-17} \\
\mathrm{Ni}(\mathrm{OH})_2 & \rightleftharpoons \mathrm{Ni}^{2+}+2 \mathrm{OH}^{-} \\
\mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=2 \times 10^{-15} \\
\text { Let }\left[\mathrm{Ag}^{+}\right] & =\mathrm{S}_1, \text { then }\left[\mathrm{CN}^{-}\right]=\mathrm{S}_1 \\
\text { Let }\left[\mathrm{Ni}^{2+}\right] & =\mathrm{S}_2, \text { then }\left[\mathrm{OH}^{-}\right]=2 \mathrm{~S}_2 \\
\mathrm{~S}_1^2 & =6 \times 10^{-17}, \mathrm{~S}_1=7.8 \times 10^{-9} \\
\left(\mathrm{~S}_2\right)\left(2 \mathrm{~S}_2\right)^2 & =10^{-15}, \mathrm{~S}_2=0.58 \times 10^{-4}
\end{aligned}
$
$\mathrm{Ni}(\mathrm{OH})_2$ is more soluble than $\mathrm{AgCN}$.
Question 20.
If $0.561 \mathrm{~g} \mathrm{KOH}$ is dossolved in water to give $200 \mathrm{~mL}$ of solution at $298 \mathrm{~K}$, calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its $\mathrm{pH}$ ?
Answer:
$
\begin{aligned}
{[\mathrm{KOH}] } & =\frac{0.561}{56} \times \frac{1000}{200} \mathrm{M}=0.050 \mathrm{M} \\
\text { As } \mathrm{KOH} & \rightarrow \mathrm{K}^{+}+\mathrm{OH}^{-}, \quad \therefore\left[\mathrm{K}^{+}\right]=\left[\mathrm{OH}^{-}\right]=0.05 \mathrm{M} \\
{\left[\mathrm{H}^{+}\right] } & =\mathrm{K}_{\mathrm{w}} /\left[\mathrm{OH}^{-}\right]=10^{-14} / 0.05=10^{-14} /\left(5 \times 10^{-2}\right)=2.0 \times 10^{-13} \mathrm{M} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=-\log \left(2.0 \times 10^{-13}\right)=13-0.3010=12.699
\end{aligned}
$

5 Marks Questions and Answers
Question 1.
Differentiate Lewis acids and Lewis bases.
Answer:
Lewis Acids
1. Lewis acids are substances that can accept one or more lone pair of electrons.
2. All metal ions (or) atoms can act as Lewis acids. Examples: $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Cu}^{2+}, \mathrm{Cr}^{3+}$
3. Molecules that contain a polar double bond can act as Lewis acids. Examples: $\mathrm{SO}_2, \mathrm{CO}_2, \mathrm{SO}_3$
4. Molecules in which the central atom can expand its act due to the availability of empty d-orbitais can act as Lewis acid. Example: $\mathrm{SiF}_4, \mathrm{SF}_4, \mathrm{FeCI}_3$
5. Carbonium ion $\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}$can act as Lewis acid
6. Electron deficient molecules such as $\mathrm{BF}_3, \mathrm{AlCl}_3, \mathrm{BeF}_2$ act as Lewis acid (electron pair acceptors)
Lewis Bases
1. Lewis bases are substances that can donate one or more lone pair of electrons.
2. All anions can act as Lewis bases. Examples: $\mathrm{F}^{-}, \mathrm{Cl}^{-}, \mathrm{CN}^{-}, \mathrm{SO}_4{ }^{2-}$
3. Molecules that contain carbon-carbon multiple bond. Example: $\mathrm{CH}_2=\mathrm{CH}_2, \mathrm{CH}=\mathrm{CH}$
4. All metal oxides can act as Lewis bases. Examples : $\mathrm{CaO}, \mathrm{MgO}, \mathrm{Na}_2 \mathrm{O}$
5. $\mathrm{CH}_2^{-}$carbanion cari act as Lewis acid
6. Electron rich molecules such as $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{ROH}, \mathrm{R}-\mathrm{O}-\mathrm{R}, \mathrm{R}-\mathrm{NH}_2$ act as Lewis base (Electron pair donors)

Question 2.
Explain about the ionisation of weak acid and how $\mathrm{K}_2$ is derived?
Answer:
1. Weak acids are partially dissociated in water and there is an equilibrium between the undissociated acid and its dissociated ions.
2. Consider the ionisation of weak monobasic acid $\mathrm{HA}$ in water
$\mathrm{HA}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-}$
3. Applying law of chemical equilibrium, the equilibrium constant $\mathrm{K}_{\mathrm{c}}$ is given by the expression
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]\left[\mathrm{H}_2 \mathrm{O}\right]}
$
4. In dilute solutions, water is present in large excess, hence its concentration may be taken as constant say $\mathrm{K}$. Further $\mathrm{H}_3 \mathrm{O}^{+}$indicates hydrated hydrogen ions, for simplicity, it may be replaced by $\mathrm{H}^{+}$. So the equation (2) becomes
$
\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}] \times \mathrm{K}}
$
5. The product of two constants $\mathrm{K}$ and $\mathrm{K}$ gives another constant. Let is be $\mathrm{K}_2$
$
\therefore \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}
$
The constant $\mathrm{K}_{\mathrm{a}}$ is called dissociation constant of weak acid.

Question 3.
Explain Buffer action with suitable example.
Answer:
Buffer action:
1. Let us consider buffer solution containing $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COO} \mathrm{Na}$. The dissociation the buffer components occur as below.
$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_3-\mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}_{(\mathrm{aq})}^{+} \\
& \mathrm{CH}_3 \mathrm{COONa}(\mathrm{s}) \stackrel{\mathrm{H}_2 \mathrm{O}(\mathrm{l})}{\longrightarrow} \mathrm{CH}_3-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Na}^{+}(\mathrm{aq})
\end{aligned}
$
2. If an acid is added to this mixture, it will be consumed by the conjugate base $\mathrm{CH}_3 \mathrm{COO}^{-}$to form undissociated weak acid. i.e., the increase in the concentration of $\mathrm{H}^{+}$does not reduce the $\mathrm{pH}$ significantly.
$
\mathrm{CH}_3 \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow \mathrm{CH}_3 \mathrm{COOH}_{(\mathrm{aq})}
$
3. If a base is added, it will be neutralised by $\mathrm{H} 3 \mathrm{O}$ and the acetic acid is dissociated to maintain the equilibrium. Hence the $\mathrm{pH}$ is not altered.

4. On the addition of an acid (or) base to a buffer solution, there will be no change in its $\mathrm{pH}$ value. Because the buffer solution should contain both acidic as well as basic components so as to neutralise the effect of added acid (or) base at the same time, these components should not consume each other.
Question 4.
Prove the buffer action of acetic acid and sodium acetate by the addition of $0.01 \mathrm{~mol}$ of solid sodium hydroxide.
Answer:
1. Consider one litre of buffer solution containing $0.8 \mathrm{~m} \mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{O} .8 \mathrm{~m} \mathrm{CH}_3 \mathrm{COONa}$. Assume that the volume change due to the addition of $0.01 \mathrm{~mol}$ of solid $\mathrm{NaOH}$ is negligible. $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_3 \mathrm{COOH}$ is $1.8 \mathrm{x}$ $10^{-5}$
2.

3. The dissociation constant for $\mathrm{CH}_3 \mathrm{COOH}$ is given by
$
\begin{aligned}
\mathrm{K}_{\mathrm{a}} & =\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\
{\left[\mathrm{H}^{+}\right] } & =\frac{\mathrm{K}_{\mathrm{a}} \cdot\left[\mathrm{CH}_3 \mathrm{COOH}\right]}{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}
\end{aligned}
$
The above expression shows that the concentration of $\mathrm{H}^{+}$is directly proportional to $\frac{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}$
degree of dissociation of $\mathrm{CH}_3 \mathrm{COOH}=\alpha$

4.
$
\begin{aligned}
& {\left[\mathrm{CH}_3 \mathrm{COOH}\right]=0.8-\alpha \text { and }\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=\alpha+0.8} \\
& \therefore\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \cdot \frac{[0.8-\alpha]}{[0.8+\alpha]} \\
& \text { If } \alpha<<0.8,0.8-\alpha \simeq 0.8 \\
& \quad 0.8+\alpha \simeq 0.8 \\
& \therefore\left[\mathrm{H}^{+}\right]=\frac{\mathrm{K}_{\mathrm{a}} \cdot[0.8]}{[0.8]}=\mathrm{K}_{\mathrm{a}} \\
& {\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}}}
\end{aligned}
$
5. Given that $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_3 \mathrm{COOH}$ is $1.8 \times 10^{-5}$
$
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=1.8 \times 10^{-5}} \\
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& =-\log \left[1.8 \times 10^{-5}\right] \\
& =5-\log 1.8 \\
& =5-0.26 \\
& \mathrm{pH}=4.74
\end{aligned}
$

6. After adding $0.01 \mathrm{moI} \mathrm{NaOH}$ to I litre of buffer. Given that volume change due to the addition of $\mathrm{NaOH}$ is negligible. $\left[\mathrm{OH}^{-}\right]=0.01 \mathrm{M}$. The consumption of $\mathrm{OH}^{-}$are expressed by the following equation.
$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH}+\mathrm{OH}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{CH}_3 \mathrm{COO}_{(\text {(aq) }}^{-}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \\
& \therefore\left[\mathrm{CH}_3 \mathrm{COOH}\right]=0.8-\alpha-0.01=0.79-\alpha \\
& {\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=\alpha+0.8+0.01} \\
& =0.81+\alpha \quad \alpha<<0.8 \\
& 0.79-\alpha \simeq 0.79 \text { and } 0.81+\alpha \simeq 0.81 \\
& \therefore\left[\mathrm{H}^{+}\right]=1.8 \times 10^{-5} \times \frac{0.79}{0.81} \\
& {\left[\mathrm{H}^{+}\right]=1.76 \times 10^{-5}} \\
& \therefore \mathrm{pH}=-\log \left(1.76 \times 10^{-5}\right) \\
& =5-\log 1.76 \\
& =5-0.25 \\
& \mathrm{pH}=4.75 \\
&
\end{aligned}
$
7. The addition of a strong base ( $0.01 \mathrm{M} \mathrm{NaOH})$ increased the $\mathrm{pH}$ only slightly i.e., from 4.74 to 4.75 . So the buffer action is verified.

Question 5.
DerIve Henderson - Hasselbalch equation
Answer:
1. The concentration of hydronium ion in acidic buffer solution depends on the ratio of concentration of the weak acid to the concentration of its conjugate base present in the solution. i.e.,
$
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\mathrm{K}_{\mathrm{a}} \cdot \frac{[\text { acid }]_{\mathrm{aq}}}{[\text { base }]_{\mathrm{aq}}}
$
2. The weak acid is dissociated only to a small extent. Moreover due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.
$
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\mathrm{K}_{\mathrm{a}} \cdot \frac{[\mathrm{acid}]}{[\text { salt }]}
$
3. [Acid] and [Salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution.
4. Taking logarithm on both sides of the equation
$\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\log \mathrm{K}_{\mathrm{a}}+\log \frac{\text { [acid] }}{\text { [salt] }}$
5. reverse the sign on both sides
$
-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log \mathrm{K}_{\mathrm{a}}-\log \frac{\text { [acid] }}{\text { [salt }]}
$
We know that,
$
\begin{aligned}
\mathrm{pH} & =-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right] \text {and } \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}} \\
\therefore \mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}-\log \frac{\text { [acid] }}{\text { [salt] }} \\
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [acid] }}{\text { [salt }]}
\end{aligned}
$
Similarly for basic buffer,
$
\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [acid] }}{\text { [base] }}
$
Eqaution (6) \& (7) are called Henderson - Hasselbalch equations.

Question 6.
Explain about the hydrolysis of salt of strong acid and a strong base with a suitable example.
Answer:
1. Let us consider the neutralisation reaction between $\mathrm{NaOH}$ and $\mathrm{HNO}_3$ to give $\mathrm{NaNO}_3$ and water.
$
\mathrm{NaOH}_{(\mathrm{aq})}+\mathrm{HNO}_{3(\mathrm{aq})} \rightarrow \mathrm{NaNO}_{3(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}_{(1)}
$
2. The salt $\mathrm{NaNO}_3$ completely dissociates in water to produce $\mathrm{Na}^{+}$and $\mathrm{NO}_3$ ions
$
\mathrm{NaNO}_{3(\mathrm{aq})} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{3(\mathrm{aq})}^{-}
$
3. Water dissociates to a small extent as
$
\mathrm{H}_2 \mathrm{O}_{(1)} \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}^{-}(\mathrm{aq})
$
Since $\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]$, water is neutral.
4. $\mathrm{NO}_3$ ion is the conjugate base of strong acid $\mathrm{HNO}_3$ and hence it has no tendency to react withH ${ }^{+}$,
5. Similarly $\mathrm{Na}$ is the conjugate acid of the strong base $\mathrm{NaOH}$ and it has no tendency to react with $\mathrm{OH}$
6. It means that there is no hydrolysis. In such cases $\left[\mathrm{H}^{+}\right]\left(\mathrm{OH}^{-}\right), \mathrm{pH}$ is maintained and there fore the solution is neutral.

Question 7.
Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of $\mathrm{K}_{\mathrm{h}}$ for that reaction.

Answer:
1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water. $\mathrm{NaOH}_{(\mathrm{aq})}+\mathrm{CH}_3 \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_3 \mathrm{COONa}_{(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}_{(1)}$
2. In aqueous solution, $\mathrm{CH}_3 \mathrm{COONa}$ is completely dissociated as follows.
$\mathrm{CH}_3 \mathrm{COONa}_{(\mathrm{aq})} \mathrm{CH}_3 \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Na}^{+}{ }_{(\mathrm{aq})}$
3. $\mathrm{CH}_3 \mathrm{COO}$ is a conjugate base of the weak acid $\mathrm{CH}_3 \mathrm{COOH}$ and it has a tendency to react with $\mathrm{H}^{+}$from water to produce unionised acid. But there is no such tendency for $\mathrm{Na}^{+}$to react with $\mathrm{OH}^{-}$
4. $\mathrm{CH}_3 \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_2 \mathrm{O}(1) \mathrm{CH}_3 \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{OH}_3^{-}$and therefore $\left[\mathrm{OH}^{-}\right]>\left[\mathrm{H}^{+}\right]$, in such cases, the solution is basic due to the hydrolysis and $\mathrm{pH}$ is greater than 7 .
5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:
$
\begin{aligned}
\mathrm{K}_{\mathrm{h}} & =\frac{\left[\mathrm{CH}_3 \mathrm{COOH}\left[\mathrm{OH}^{-}\right]\right.}{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}_2 \mathrm{O}\right]} \\
\mathrm{K}_{\mathrm{h}} & =\frac{\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}
\end{aligned}
$
$
\mathrm{CH}_3 \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+}
$
$
\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}
$
Equation (1) $\mathrm{x}(2)$
$\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{a}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]$
$\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{W}}$
$\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{a}}=\mathrm{K}_{\mathrm{W}}$
$\mathrm{K}_{\mathrm{h}}$ value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald's dilution law $\mathrm{K}_{\mathrm{h}}=\mathrm{h}^2 \mathrm{C}$ and $\left[\mathrm{OH}^{-}\right]=$
$
\mathrm{K}_{\mathrm{h}}=\mathrm{h}^2 \mathrm{C} \text { and }\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{c}} \text {. }
$

Question 8.
DerIve the value of $\mathrm{pH}$ of salt solution in terms of $\mathrm{K}_{\mathrm{a}}$ and concentration of electrolyte.
Answer:
$
\begin{aligned}
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{pH}=14-\mathrm{pOH} \\
& =14-\left[-\log _{10}\left[\mathrm{OH}^{-}\right]\right. \\
& =14+\log \left[\mathrm{OH}^{-}\right] \\
& \therefore \mathrm{pH}=14+\log \left[\mathrm{K}_{\mathrm{h}} \mathrm{c}\right]^{1 / 2} \quad \because \mathrm{OH}^{-}=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{c}} \\
& \mathrm{pH}=14+\log \left[\frac{\mathrm{K}_{\mathrm{w}} \mathrm{c}}{\mathrm{K}_{\mathrm{a}}}\right]^{1 / 2} \\
& \mathrm{pH}=14+\left(1 / 2 \log \mathrm{K}_{\mathrm{w}}+1 / 2 \log \mathrm{c}-1 / 2 \log \mathrm{K}_{\mathrm{a}}\right) \because \mathrm{K}_{\mathrm{w}}=10^{-14} \\
& \mathrm{pH}=14-7+1 / 2 \log \mathrm{c}-1 / 2 \log \mathrm{K}_{\mathrm{a}} \\
& \because 1 / 2 \log \mathrm{K}_{\mathrm{w}}=1 / 2 \log 10^{-14}=-14 / 2=-7 \\
& \mathrm{pH}=14-7+1 / 2 \log \mathrm{c}+1 / 2 \mathrm{pK}_{\mathrm{a}} \quad \because-\log \mathrm{K}_{\mathrm{a}}=\mathrm{pK}_{\mathrm{a}} \\
& \therefore \mathrm{pH}=7+1 / 2 \mathrm{pK}_{\mathrm{a}}+1 / 2 \log \mathrm{c} \\
&
\end{aligned}
$

Question 9.
Explain about the hydrolysis of salt of strong acid and weak base. Derive $\mathrm{K}_{\mathrm{h}}$ and $\mathrm{pH}$ for that solution.
Answer:
1. Consider a reaction between strong acid $\mathrm{HCl}$ and a weak base $\mathrm{NH}_4 \mathrm{OH}$ to produce a salt $\mathrm{NH}_4 \mathrm{CI}$ and water
$
\begin{aligned}
& \mathrm{HCl}_{(\mathrm{aq})}+\mathrm{NH}_4 \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_4 \mathrm{Cl}_{(\mathrm{aq})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \\
& \mathrm{NH}_4 \mathrm{Cl}_{(\mathrm{aq})} \longrightarrow \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \\
&
\end{aligned}
$
2. $\mathrm{NH}_4$ is a strong conjugate acid of the weak base $\mathrm{NH}_4 \mathrm{OH}$ and it has a tendency to react with $\mathrm{OH}$ - from water to produce unionised $\mathrm{NH}_4$ as below,
$
\mathrm{NH}_4^{+}{ }_{(\mathrm{aq})}^{+}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightleftharpoons \mathrm{NH}_4 \mathrm{OH}_{(\mathrm{aq})}+\mathrm{H}_{(\mathrm{aq})}^{+}
$
3. There is no such tendency shown by $\mathrm{Cl}^{-}$and therefore $\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]$the solution is acidic and the $\mathrm{pH}$ is less than 7 .
4. In the salt hydrolysis of strong base and weak acid, we have to derive a relationship between $\mathrm{K}_{\mathrm{h}}$ and $\mathrm{K}_{\mathrm{b}}$ as
$
\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{W}}
$

5.
$
\begin{gathered}
\mathrm{K}_{\mathrm{h}}=\mathrm{h}^2 \mathrm{C} \text { and }\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \\
\therefore\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}} \\
\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
=-\log \left[\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}\right]^{1 / 2} \\
=-1 / 2 \log \mathrm{K}_{\mathrm{w}}-1 / 2 \log \mathrm{C}+1 / 2 \log \mathrm{K}_{\mathrm{b}} \\
\mathrm{pH}=7-1 / 2 \mathrm{pK}_{\mathrm{b}}-1 / 2 \log \mathrm{C}
\end{gathered}
$

Question 10 .
Discuss about the hydrolysis of salt of weak acid and weak base and derive $\mathrm{pH}$ value for the solution.

Answer:
1. Consider the hydrolysis of ammonium acetate
$
\mathrm{CH}_3 \mathrm{COONH}_{4(\mathrm{aq})} \rightarrow \mathrm{CH}_3 \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{NH}_4^{+}(\mathrm{aq})
$
2. In this case both the cation $\left(\mathrm{NH}_4^{+}\right)$and $\left(\mathrm{CH}_3 \mathrm{COO}^{-}\right)$anion have the tendency to react with water.
$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}+\mathrm{OH}^{-} \\
& \mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{NH}_4 \mathrm{OH}+\mathrm{H}_3
\end{aligned}
$
3. The nature of the solution depends on the strength of acid (or) base i.e., if $\mathrm{K}_{\mathrm{a}}>\mathrm{K}_{\mathrm{b}}$, then the solution is acidic and $\mathrm{pH}<7$, if $\mathrm{K}_{\mathrm{a}}<\mathrm{K}_{\mathrm{b}}$ then the solution is basic and $\mathrm{pH}>7$. If $\mathrm{K}_{\mathrm{a}}=\mathrm{K}_{\mathrm{b}}$, then the solution is neutral.
4. The relation between the dissociation constant $\mathrm{K}_{\mathrm{a}}, \mathrm{K}_{\mathrm{b}}$ and hydrolysis constant is given by the following expression.
$
\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}} \cdot \mathrm{K}_{\mathrm{h}}=\mathrm{K}_{\mathrm{w}}
$
5. $\mathrm{pH}$ of the solution
$
\mathrm{pH}=7+\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}
$
Question 11.
It has been found that the $\mathrm{pH}$ of a $0.01 \mathrm{M}$ solution of an organic acid Is 4.15 . Calculate the concentration of the anion, the Ionization constant of the acid and its $\mathrm{pK}_{\mathrm{a}}$.
Answer:
$
\begin{aligned}
& \mathrm{HA} \rightleftharpoons \mathrm{H}^{+} \\
& \mathrm{pH}=\log \left[\mathrm{H}^{+}\right] \text {or } \log \left[\mathrm{H}^{+}\right]=-4.15=5.85 \\
& {\left[\mathrm{H}^{+}\right]=7.08 \times 10^{-5} \mathrm{M}=7.08 \times 10^{-5} \mathrm{M}} \\
& {\left[\mathrm{A}^{-}\right]=\left[\mathrm{H}^{+}\right]=7.08 \times 10^{-5} \mathrm{M}} \\
& \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{\left(7.08 \times 10^{-5}\right)\left(7.08 \times 10^{-5}\right)}{10^{-2}}=5.0 \times 10^{-7} \\
& \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}=-\log \left(5.0 \times 10^{-7}\right)=7-0.699=6.301 \\
&
\end{aligned}
$

Question 12 .
Assuming complete dissociation, calculate the $\mathrm{pH}$ of the following solutions.
(i) $0.003 \mathrm{M} \mathrm{HCl}$
(ii) $0.005 \mathrm{M} \mathrm{NaOH}$
(iii) $0.002 \mathrm{M} \mathrm{HBr}$
(iv) $0.002 \mathrm{M} \mathrm{KOH}$
Answer:
(i) $\mathrm{HCl}+\mathrm{aq} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}$
$
\therefore\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=3 \times 10^{-3} \mathrm{M}, \mathrm{pH}=-\log \left(3 \times 10^{-3}\right)=2.52
$
(ii) $\mathrm{NaOH}+\mathrm{aq} \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}$
$
\begin{aligned}
\therefore\left[\mathrm{OH}^{+}\right] & =5 \times 10^{-3} \mathrm{M},\left[\mathrm{H}^{+}\right]=10^{-14} /\left(5 \times 10^{-3}\right)=2 \times 10^{-12} \mathrm{M} \\
\mathrm{pH} & =-\log \left(2 \times 10^{-12}\right)=11.70
\end{aligned}
$
(iii) $\mathrm{HBr}+\mathrm{aq} \longrightarrow \mathrm{H}^{+}+\mathrm{Br}^{-}$
$
\therefore\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M}, \mathrm{pH}=-\log \left(2 \times 10^{-3}\right)=2.70
$
(iv) $\mathrm{KOH}+\mathrm{aq} \longrightarrow \mathrm{K}^{+}+\mathrm{OH}^{-}$
$
\begin{aligned}
\therefore\left[\mathrm{OH}^{+}\right] & =2 \times 10^{-3} \mathrm{M},\left[\mathrm{H}^{+}\right]=10^{-14} /\left(2 \times 10^{-3}\right)=5 \times 10^{-12} \\
\mathrm{pH} & =-\log \left(5 \times 10^{-12}\right)=11.30
\end{aligned}
$
Question 13.
What is the $\mathrm{pH}$ of $0.001 \mathrm{M}$ aniline solution? The ionisation constant of aniline is $4.27 \times 10^{-10}$. Calculate degree of ionization of aniline in the solution. Also calculate the ionisation constant of the conjugate acid of anile.
Answer:
1. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3+\mathrm{OH}^{-}$
$
\begin{aligned}
\mathrm{K}_{\mathrm{a}} & =\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\right]}=\frac{\left[\mathrm{OH}^{-}\right]^2}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\right]} \\
{\left[\mathrm{OH}^{-}\right] } & =\sqrt{\mathrm{K}_{\mathrm{a}}\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\right]}=\sqrt{\left(4.27 \times 10^{-10}\right)\left(10^{-3}\right)}=6.534 \times 10^{-7} \mathrm{M} \\
\mathrm{pOH} & =-\log \left(6.534 \times 10^{-7}\right)=7-0.8152=6.18 \\
\therefore \mathrm{pH} & =14-6.18=7.82
\end{aligned}
$

2. Also
Also $\quad \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}+\mathrm{OH}^{-}$
Initial
At. eqm. $\mathrm{C}-\mathrm{C} \alpha \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha$
$
\begin{aligned}
\mathrm{K}_{\mathrm{b}} & =\frac{\mathrm{C} \alpha \cdot \mathrm{C} \alpha}{\mathrm{C}(1-\alpha)}=\frac{C \alpha^2}{1-\alpha} \simeq \mathrm{C} \alpha^2 \\
\therefore \alpha & =\sqrt{\mathrm{K}_{\mathrm{b}} / \mathrm{C}}=\sqrt{4.27 \times 10^{-10} / 10^{-3}}=6.53 \times 10^{-4}
\end{aligned}
$
3. $\mathrm{pK}_{\mathrm{a}}+\mathrm{pK}_{\mathrm{b}}=14$ (for a pair of conjugate acid and base)
$
\begin{aligned}
& \mathrm{pK}_{\mathrm{b}}=-\log \left(4.27 \times 10^{-10}\right)=10-0.62=9.38 \\
& \mathrm{pK}_{\mathrm{a}}=14-9.38=4.62 \\
& \text { i.e., }-\log \mathrm{K}_{\mathrm{a}} 4.62 \text { or } \log \mathrm{K}_{\mathrm{a}}=-4.62= \\
& \text { i.e., }-\log \mathrm{K}_{\mathrm{a}}=4.62 \text { or } \log \mathrm{K}_{\mathrm{a}}=-4.62=\overline{5} .38 \\
& \mathrm{~K}_{\mathrm{a}}=\text { Antrilog } \overline{5} .38=2.399 \times 10^{-5} \simeq 2.4 \times 10^{-5} \\
&
\end{aligned}
$

Question 14.
Calculate the degree of ionization of $0.05 \mathrm{M}$ acetic acid If its $\mathrm{pK}_{\mathrm{a}}$ value is 4.74 . How is the degree of dissociation affected when its solution also contains
1. $0.01 \mathrm{M}$
2. $0.1 \mathrm{M} \mathrm{HCI}$
Answer:
$\mathrm{PK}_{\mathrm{a}}=$ i.e., $-\log \mathrm{K}_{\mathrm{a}}=4.74$
or $\log \mathrm{K}_{\mathrm{a}}=4.74=5.26$
$\mathrm{K}_{\mathrm{a}}=1.82 \times 10^{-5}$
$\alpha=\sqrt{\mathrm{K}_{\mathrm{a}} / \mathrm{C}}=\sqrt{\left(1.82 \times 10^{-5}\right) /\left(5 \times 10^{-2}\right)}=1.908 \times 10^{-2}$
In presence of $\mathrm{HCI}$, due to high concentration of $\mathrm{H}^{+}$ion, dissociation equilibrium will shift backward, Le., dissociation of acetic acid will decrease.
1. In presence of $0.01 \mathrm{M} \mathrm{HCI}$, if $x$ is the amount dissociated, then

2. In the presence of $0.1 \mathrm{M} \mathrm{HCl}$, if $y$ is the amount of acetic acid dissociated, then at equilibrium
$
\begin{aligned}
& {\left[\mathrm{CH}_3 \mathrm{COOH}\right]=0.05-\mathrm{y} \simeq 0.05 \mathrm{M}} \\
& {\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=y\left[\mathrm{H}^{+}\right]=0.1 \mathrm{M}+y \simeq 0.1 \mathrm{M}} \\
& \mathrm{K}_{\mathrm{a}}=\frac{y(0.1)}{0.05} \text { or } \frac{y}{0.05}=\frac{\mathrm{K}_{\mathrm{a}}}{0.1}=\frac{1.82 \times 10^{-5}}{10^{-1}}=1.82 \times 10^{-4} \\
& \alpha=1.82 \times 10^{-4}
\end{aligned}
$

Question 15.
The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05 \mathrm{M}$ solution. Calculate the concentration of acetate ions in the solution and its $\mathrm{pH}$.
Answer:
$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \\
& \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_2 \mathrm{COOH}\right]}=\frac{\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\
& \text { (or) }\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a}}\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\sqrt{\left(1.74 \times 10^{-5}\right)\left(5 \times 10^{-2}\right)}=9.33 \times 10^{-4} \mathrm{M} \\
& {\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=\left[\mathrm{H}^{+}\right]=9.33 \times 10^{-4} \mathrm{M}} \\
& \mathrm{pH}=-\log \left(9.33 \times 10^{-4}\right)=4-0.9699=4-0.97=3.03
\end{aligned}
$
Common Errors
1. Acid and Base - Definiton
2. $\mathrm{pH}$ value
3. Buffer mixture
4. Conjugate Acid base pair
Rectifications
1.

2. $\mathrm{pH}$ neutral
3. $\mathrm{pH}$ less than 7 - Acid
4. $\mathrm{pH}$ more than 7 -Base
5. Always either weak acid and its salt (or) weak base and its salt.
6. They differ by $\mathrm{H}^{+}$. For e.g., $\mathrm{CH}_3 \mathrm{COOH}$. Its conjugate base is $\mathrm{CH}_3 \mathrm{COO}^{-} \cdot \mathrm{H}_2 \mathrm{O}-$ Acid and its conjugate base is $\mathrm{OH}$