Text Book Back Questions and Answers - Chapter 8 - Ionic Equilibrium - 12th Chemistry Guide Samacheer Kalvi Solutions

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Ionic Equilibrium
Textual Evaluation Solved
Multiple Choice Questions
I. Choose the correct answer.
Question 1.
Concentration of the $\mathrm{Ag}^{+}$ions in a saturated solution of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2.24 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$ solubility product of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is ..............
(a) $2.42 \times 10^{-8} \mathrm{~mol}^3 \mathrm{~L}^{-3}$
(b) $2.66 \times 10^{-12} 12 \mathrm{~mol}^3 \mathrm{~L}^{-3}$
(c) $45 \times 10^{-11} \mathrm{~mol}^3 \mathrm{~L}^{-3}$
(d) $5.619 \times 10^{-12} \mathrm{~mol}^3 \mathrm{~L}^{-3}$
Answer:
(d) $5.619 \times 10^{-12} \mathrm{~mol}^3 \mathrm{~L}^{-3}$
$\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4 2 \mathrm{Ag}^{+}+\mathrm{C}_2 \mathrm{O}_4{ }^{2-}$
$\left[\mathrm{Ag}^{+}\right]=2.24 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
$\left[\mathrm{C}_2 \mathrm{O}_4{ }^{2-}\right]=\frac{2.24 \times 10^{-4}}{2} \mathrm{~mol} \mathrm{~L}^{-1}$
$=1.12 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{C}_2 \mathrm{O}_4^{2-}\right]$
$=\left(2.24 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\right)^2\left(1.12 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\right)$
$=5.619 \times 10^{-12} \mathrm{~mol}^3 \mathrm{~L}^{-3}$
Question 2.
Following solutions were prepared by mixing different volumes of $\mathrm{NaOH}$ of $\mathrm{HCl}$ different concentrations.
(i) $60 \mathrm{~mL} \frac{M}{10} \mathrm{HCI}+40 \mathrm{~mL} \frac{M}{10} \mathrm{NaOH}$
(ii) $55 \mathrm{~mL} \frac{M}{10} \mathrm{HCl}+45 \mathrm{~mL} \frac{M}{10} \mathrm{NaOH}$
(iii) $75 \mathrm{~mL} \frac{M}{5} \mathrm{HCI}+25 \mathrm{~mL} \frac{M}{5} \mathrm{MNaOH}$
(iv) $100 \mathrm{~mL} \frac{M}{10} \mathrm{HCI}+100 \mathrm{~mL} \frac{M}{10} \mathrm{NaOH}$

$\mathrm{pH}$ of which one of them wilt be equal to 1 ?
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
Answer:
(d) (iii) $75 \mathrm{~mL} \frac{M}{5} \mathrm{HCI}+25 \mathrm{~mL} \frac{M}{5} \mathrm{NaOH}$
No of moles of $\mathrm{HCl}=0.2 \times 75 \times 10^{-3}=15 \times 10^{-3}$
No of moles of $\mathrm{NaOH}=0.2 \times 25 \times 10^{-3}=5 \times 10^{-3}$
No of moles of $\mathrm{HCl}$ after mixing $=15 \times 10^{-3}-5 \times 10^{-3}$
$\therefore$ Concentration of $\mathrm{HCl}$
$\therefore$ Concentration of $\mathrm{HCl}=\frac{\text { No. of moles of } \mathrm{HCl}}{\text { Vol in litre }}=\frac{10 \times 10^{-3}}{100 \times 10^{-3}}=0.1 \mathrm{M}$
for (iii) solution, $\mathrm{pH}$ of $0.1 \mathrm{MHCI}=-\log _{10}(0.1)=1$.
Question 3.
The solubility of $\mathrm{BaSO}_4$ in water is $2.42 \times 10^{-3} \mathrm{gL}^{-1}$ at $298 \mathrm{~K}$. The value of its solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ will be
(Given molar mass of $\mathrm{BaSO}_4=233 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(a) $1.08 \times 10^{-14} \mathrm{~mol}^2 \mathrm{~L}^2$
(b) $1.08 \times 10^{-12} \mathrm{~mol}^2 \mathrm{~L}^2$
(c) $1.08 \times 10^{-10} \mathrm{~mol}^2 \mathrm{~L}^2$
(d) $1.08 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}$
Answer:
(c) $1.08 \times 10^{-10} \mathrm{~mol}^2 \mathrm{~L}^2$
$\mathrm{BaSO}_4 \rightleftharpoons \mathrm{Ba}^{2+}+\mathrm{SO}_4^{2-}$
$\mathrm{K}_{\mathrm{sp}}=(\mathrm{s})(\mathrm{s})$
$\mathrm{K}_{\mathrm{sp}}=(\mathrm{s})^2$
$=\left(2.42 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}\right)^2$
$=\left(\frac{2.42 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}}{233 \mathrm{~g} \mathrm{~mol}^{-1}}\right)^2$
$=\left(0.01038 \times 10^{-3}\right)^2$
$=\left(1.038 \times 10^{-5}\right)^2$
$=1.077 \times 10^{-10}$
$=1.08 \times 10^{-10} \mathrm{~mol}^2 \mathrm{~L}^{-2}$

Question 4.
$\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_2$ is 9 . The Solubility product $(\mathrm{K})$ of $\mathrm{Ca}(\mathrm{OH})_2$
(a) $0.5 \times 10^{-15}$
(b) $0.25 \times 10^{-10}$
(c) $0.125 \times 10^{-15}$
(d) $0.5 \times 10^{-10}$
Answer:
(a) $0.5 \times 10^{-15}$
$
\mathrm{Ca}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-}
$
Given that $\mathrm{pH}=9$
$
\begin{aligned}
& \mathrm{pOH}=14-9=5 \\
& {\left[\mathrm{p} 0 \mathrm{~K}=-1 \log _{10}\left[\mathrm{OH}^{-}\right]\right]} \\
& {\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{pOH}}} \\
& {\left[\mathrm{OH}^{-}\right]=10^{-5} \mathrm{M}} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\
& =\frac{10^{-5}}{2} \times\left(10^{-5}\right)^2=0.5 \times 10^{-15} \\
& =0.5
\end{aligned}
$
Question 5.
Conjugate base for bronsted acids $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{HF}$ are
(a) $\mathrm{OH}^{-}$and $\mathrm{H}_2 \mathrm{FH}^{+}$, respectively
(b) $\mathrm{H}_3 \mathrm{O}^{+}$and $\mathrm{F}^{-}$, respectively
(c) $\mathrm{OH}^{-}$and $\mathrm{F}^{-}$, respectively
(d) $\mathrm{H}_3 \mathrm{O}^{+}$and $\mathrm{H}_2 \mathrm{~F}^{+}$, respectively
Answer:
(c) $\mathrm{OH}^{-}$and $\mathrm{F}^{-}$, respectively
$
\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{OH}^{-}
$
acid 1 base 1 acid 2 base 2
$
\mathrm{HF}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{F}-
$
acid 1 base 1 acid 2 base 2
$\therefore$ Conjugate bases are $\mathrm{OH}^{-}$and $\mathrm{F}^{-}$respectively
Question 6.
Which will make basic buffer?
(a) $50 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}+25 \mathrm{~mL}$ of $01 \mathrm{M} \mathrm{CH} \mathrm{COOH}_3$
(b) $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}+100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$
(c) $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCI}+200 \mathrm{~mL}$ of $0.1 \mathrm{MNH}_4 \mathrm{OH}$

(d) $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCI}+100 \mathrm{~mL}$ of $\mathrm{O} .1 \mathrm{M} \mathrm{NaOH}$
Answer:
(c) $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCI}+200 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$
Basic buffer is the solution which has weak base and its salt
$
\mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O}+\mathrm{NH}_4 \mathrm{OH}
$
$200 \mathrm{ml} \quad 100 \mathrm{ml} \quad$ salt
$(100 \mathrm{ml}$ weak base)
Question 7.
Which of the following fluro - compounds is most likely to behave as a Lewis base?
(a) $\mathrm{BF}_3$
(b) $\mathrm{PF}_3$
(c) $\mathrm{CF}_4$
(d) $\mathrm{SiF}_4$
Answer:
(b) $\mathrm{PF}_3$
$\mathrm{BF}_3 \rightarrow$ electron deficient $\rightarrow$ Lewis acid
$\mathrm{PF}_3 \rightarrow$ electron rich $\rightarrow$ Lewis base
$\mathrm{CF}_4 \rightarrow$ neutral $\rightarrow$ neither lewis acid nor base
$\mathrm{SiF}_4 \rightarrow$ neutral $\rightarrow$ neither lewis acid nor base
Question 8.
Which of these is not likely to act as lewis base?
(a) $\mathrm{BF}_3$
(b) $\mathrm{PF}_3$
(c) $\mathrm{CO}$
(d) $\mathrm{F}$
Answer:
(a) $\mathrm{BF}_3$
$\mathrm{BF}_3 \rightarrow$ electron deficient $\rightarrow$ Lewis acid
$\mathrm{PF}_3 \rightarrow$ electron rich $\rightarrow$ Lewis base
$\mathrm{CO} \rightarrow$ having lone pair of electron $\rightarrow$ Lewis base
$\mathrm{F}^{-} \rightarrow$ unshared pair of electron $\rightarrow$ lewis base

Question 9.
What is the decreasing order of strength of bases?
$
\mathrm{OH}, \mathrm{NH}_2^{-}, \mathrm{H}-\mathrm{C}=\mathrm{C}^{-} \text {and } \mathrm{CH}_3-\mathrm{CH}_2^{-}
$
(a) $\mathrm{OH}^{-}>\mathrm{NH}_2^{-}>\mathrm{H}-\mathrm{C}=\mathrm{C}>\mathrm{CH}_3-\mathrm{CH}_2^{-}$
(b) $\mathrm{NH}_2^{-}>\mathrm{OH}>\mathrm{CH}_3-\mathrm{CH}_2^{-}>\mathrm{H}-\mathrm{C}=\mathrm{C}^{-}$
(c) $\mathrm{CH}_3-\mathrm{CH}_2^{-},>\mathrm{NH}_2^{-}>\mathrm{H}-\mathrm{C}=\mathrm{C}^{-}>\mathrm{OH}$
(d) $\mathrm{OH}^{-}>\mathrm{H}-\mathrm{C}=\mathrm{C}>\mathrm{CH}_3-\mathrm{CH}_2^{-}>\mathrm{NH}_2^{-}$
Answer:
(c) $\mathrm{CH}_3-\mathrm{CH}_2^{-},>\mathrm{NH}_2^{-}>\mathrm{H}-\mathrm{C}=\mathrm{C}^{-}>\mathrm{OH}$
Acid strength decreases in the order
$
\mathrm{HOH}>\mathrm{CH}=\mathrm{CH}>\mathrm{NH}_3>\mathrm{CH}_3 \mathrm{CH}_3
$
Its conjucate bases arc in the reverse order
$
\mathrm{CH}_3-\mathrm{CH}_2^{-}>\mathrm{NH}_2^{-}>\mathrm{H}-\mathrm{C}=\mathrm{C}>\mathrm{OH}^{-}
$
Question 10.
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
(a) acidic, acidic, basic
(b) basic, acidic, basic
(c) basic, neutral, basic
(d) none of these
Answer:
(b) basic, acidic, basic
$
\mathrm{HCOONa}+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{NaOH}+\mathrm{H}-\mathrm{COOH}
$
Basic in nature.
$
\begin{aligned}
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3 \mathrm{Cl}^{-}+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}_2+\mathrm{Cl}^{-} \\
& \mathrm{KCN}+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{KOH}+\mathrm{HCN} \\
&
\end{aligned}
$
basic strong base weak acid
Question 11.
The percentage of pyridine $\left(\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}\right)$ that forms pyridinium ion $\left(\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}\right)$ in a $0.10 \mathrm{M}$ aqueous pyridine solution $\left(\mathrm{Kb}\right.$ for $\left.\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}=1.7 \times 10^{-9}\right)$ is $\ldots \ldots \ldots \ldots \ldots$
(a) $0.006 \%$
(b) $0.013 \%$
(c) $0.77 \%$

(d) $1.6 \%$
Answer:
(b) $0.013 \%$
$
\begin{aligned}
& \mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{C}_5 \mathrm{H}_5 \stackrel{+}{\mathrm{NH}}+\mathrm{OH}^{-} \\
& \frac{\alpha^2 \mathrm{C}}{1-\alpha}=\mathrm{K}_{\mathrm{b}} \\
& \alpha^2 \mathrm{C} \approx \mathrm{K}_{\mathrm{b}} \\
& \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}}=\sqrt{\frac{1.7 \times 10^{-9}}{0.1}}=\sqrt{1.7} \times 10^{-4}
\end{aligned}
$
Percentage of dissociation
$
=\sqrt{1.7} \times 10^{-4} \times 100=1.3 \times 10^{-2}=0.013 \%
$
Question 12.
Equal volumes of three acid solutions of $\mathrm{pH} 1,2$ and 3 are mixed in a vessel. What will be the $\mathrm{H}^{+}$ion concentration in the mixture?
(a) $37 \times 10^{-2}$
(b) $10^{-6}$
(c) 0.111
(d) none of these
Answer:
$
\begin{aligned}
& \text { (a) } 3.7 \times 10^{-2} \\
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\
& {\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}}
\end{aligned}
$
Let the volume be $\mathrm{x} \mathrm{mL}$
$
\begin{aligned}
& \mathrm{V}_1 \mathrm{M}_1+\mathrm{V}_2 \mathrm{M}_2+\mathrm{V}_3 \mathrm{M}_3=\mathrm{VM} \\
& \mathrm{x} \mathrm{mL} \text { of } 10^{-1} \mathrm{M}+\mathrm{x} \mathrm{mL} \text { of } 10^{-2} \mathrm{M}+\mathrm{x} \mathrm{mL} \text { of } 10^{-3} \mathrm{M} \\
& =3 \mathrm{x} \mathrm{mL} \text { of }\left[\mathrm{H}^{+}\right] \\
& =3 \mathrm{xmL} \text { of }\left[\mathrm{H}^{+}\right] \\
& {\left[\mathrm{H}^{+}\right]=} \\
& \therefore\left[\mathrm{H}^{+}\right]=\frac{x[0.1+0.01+0.001]}{3 x}=\frac{0.1+0.01+0.001}{3}=\frac{0.111}{3} \\
& =0.037=3.7 \times 10^{-2}
\end{aligned}
$

Question 13.
The solubility of $\mathrm{AgCl}$ (s) with solubility product $1.6 \times 10^{-10}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ solution would be
(a) $1.26 \times 10^{-5} \mathrm{M}$
(b) $1.6 \times 10^{-9} \mathrm{M}$
(c) $1.6 \times 10^{-11} \mathrm{M}$
(d) Zero
Answer:
(b) $1.6 \times 10^{-9} \mathrm{M}$
$\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$
$\mathrm{K}_{\mathrm{sp}}=1.6 \times 10^{-10}$
$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$\mathrm{K}=(\mathrm{s})(\mathrm{s}+0.1)$
$0.1 \gg \mathrm{s}$
$\therefore \mathrm{s}+0.1 \simeq 0.1$
$\therefore s=\frac{1.6 \times 10^{-10}}{0.1}=1.6 \times 10^{-9}$
Question 14.
If the solubility product of lead iodide is $3.2 \times 10^{-8}$, its solubility will be
(a) $2 \times 10^{-3} \mathrm{M}$
(b) $4 \times 10^{-4} \mathrm{M}$
(c) $1.6 \times 10^{-5} \mathrm{M}$
(d) $1.8 \times 10^{-5} \mathrm{M}$
Answer:
(a) $2 \times 10^{-3} \mathrm{M}$
$\mathrm{PbI}_2(\mathrm{~s}) \rightarrow \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})$
$\mathrm{K}_{\mathrm{sp}}=(\mathrm{s})(2 \mathrm{~s})^2$
$3.2 \times 10^{-8}=4 \mathrm{~s}^3$
$s=\left(\frac{3.2 \times 10^{-8}}{4}\right)^{1 / 3}=\left(8 \times 10^{-9}\right)^{1 / 3}=2 \times 10^{-3} \mathrm{M}$

Question 15 .
Using Gibb's free energy change, $\Delta \mathrm{G}^0=57.34 \mathrm{KJ} \mathrm{mol}^{-1}$, for the reaction, $\mathrm{X}_2 \mathrm{Y}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{X}^{+}+\mathrm{Y}^{2-}(\mathrm{aq})$, calculate the solubility product of $\mathrm{X}_2 \mathrm{Y}$ in water at $300 \mathrm{~K}\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{Mol}^{-1}\right)$
(a) $10^{-10}$
(b) $10^{-12}$
(c) $10^{-14}$
(d) can not be calculated from the given data
Answer:
(a) $10^{-10}$
$57.34 \mathrm{KJ} \mathrm{mol}^{-1}=-2.303 \times 8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \log \mathrm{K}_{\mathrm{sp}}$
$\log \mathrm{K}_{\mathrm{sp}}=\frac{-57.34 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}$
$\log _{10} \mathrm{~K}_{\mathrm{sp}}=-10$
$\therefore \mathrm{K}_{\mathrm{sp}}=10^{-10}$
$\Delta \mathrm{G}^0=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}}$
$\mathrm{X}_2 \mathrm{Y}_{(\mathrm{s})} \rightleftharpoons 2 \mathrm{X}^{+}(\mathrm{aq})+\mathrm{Y}^{2-}(\mathrm{aq})$
$\mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{X}^{+}\right]^2\left[\mathrm{Y}^{2-}\right]}{\left[\mathrm{X}_2 \mathrm{Y}\right]}$
$\mathrm{K}_{\mathrm{eq}}=\left[\mathrm{x}^{+}\right]^2\left[\mathrm{Y}^{2-}\right]\left(\mathrm{X}_2 \mathrm{Y}_{(\mathrm{s})}=1\right)$
$\mathrm{K}_{\text {eq }}=\mathrm{K}$

Question 16.
$\mathrm{MY}$ and $\mathrm{NY}_3$, are insoluble salts and have the same $\mathrm{K}_{\mathrm{sp}}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true with regard to $\mathrm{MY}$ and $\mathrm{NY}_3$ ?
(a) The salts $\mathrm{MY}$ and $\mathrm{NY}_3$ are more soluble in $\mathrm{O} .5 \mathrm{M} \mathrm{KY}$ than in pure water
(b) The addition of the salt of $\mathrm{KY}$ to the suspension of $\mathrm{MY}$ and $\mathrm{NY}_3$ will have no effect on
(c) The molar solubities of $\mathrm{MY}$ and $\mathrm{NY}_3$ in water are identical
(d) The molar solubility of $\mathrm{MY}$ in water is less than that of $\mathrm{NY}_3$
Answer:
(d) The molar solubility of $\mathrm{MY}$ in water is less than that of $\mathrm{NY}_3$
Addition of salt $\mathrm{KY}$ (having a common ion $\mathrm{Y}$ ) decreases the solubility of $\mathrm{MY}$ and $\mathrm{NY}_3$ due to common ion effect. Option (a) and (b) are wrong.
For salt $\mathrm{MY}, \mathrm{MY} \rightleftharpoons \mathrm{M}^{+}+\mathrm{Y}^{-}$
$\mathrm{K}_{\mathrm{sp}}=$ (s) (s)
$6.2 \times 10^{-13}=\mathrm{s}^2$

$
\begin{aligned}
& \therefore \mathrm{s}=\sqrt{6.2 \times 10^{-13}} \simeq 10^{-7} \\
& \text { for salt } \mathrm{NY}_3, \mathrm{NY}_3 \rightleftharpoons \mathrm{N}^{3+}+3 \mathrm{Y}^{-} \\
& \mathrm{K}_{\mathrm{sp}}=(\mathrm{s})(3 \mathrm{~s})^3 \\
& \mathrm{~K}_{\mathrm{sp}}=27 \mathrm{~s}^4 \\
& \mathrm{~s}=\left(\frac{6.2 \times 10^{-13}}{27}\right)^{1 / 4} \\
& \mathrm{~s} \simeq 10^{-4}
\end{aligned}
$
The molar solubility of $\mathrm{MY}$ in water is less than of $\mathrm{NY}_3$
Question 17.
What is the $\mathrm{pH}$ of the resulting solution when equal volumes of $0.1 \mathrm{M} \mathrm{NaOH}$ and $0.01 \mathrm{M} \mathrm{HCl}$ are mixed?
(a) 2.0
(b) 3
(c) 7.0
(d) 12.65
Answer:
(d) 12.65
$\mathrm{x} \mathrm{ml}$ of $0.1 \mathrm{~m} \mathrm{NaOH}+\mathrm{x} \mathrm{ml}$ of $0.01 \mathrm{M} \mathrm{HCI}$
No. of moles of $\mathrm{NaOH}=0.1 \times \times \times 10^{-3}=0.1 \times 10^{-3}$
No. of moles of $\mathrm{HCl}=0.01 \times 1 \times 10^{-3}=0.01 \times 10^{-3}$
No. of moles of $\mathrm{NaOH}$ after mixing $=0.1 \times 10^{-3}-0.01 \times 10^{-3}$
$=0.09 \times 10^{-3}$
Concentration of $\mathrm{NaOH}=$
$
\begin{aligned}
& \mathrm{NaOH}=\frac{0.09 x \times 10^{-3}}{2 x \times 10^{-3}}=0.045 \\
& {\left[\mathrm{OH}^{-}\right]=0.045} \\
& \mathrm{p}^{\mathrm{OH}}=-\log \left(4.5 \times 10^{-2}\right) \\
& =2-\log 4.5 \\
& =2-0.65=1.35 \\
& \mathrm{pH}=14-1.35=12.65
\end{aligned}
$

Question 18.
The dissociation constant of a weak acid is $1 \times 10^{-3}$. In order to prepare a buffer solution with a $\mathrm{pH}=4$, the [Acid] / [Salt] ratio should be
(a) $4: 3$
(b) $3: 4$
(c) $10: 1$
(d) $1: 10$
Answer:
(d) $1: 10$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=1 \times 10^{-3} ; \mathrm{pH}=4 \\
& \frac{[\text { Salt }]}{[\text { Acid }]}=? \\
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 4=-\log _{10}\left(1 \times 10^{-3}\right)+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 4=3+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& 1=\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]} \Rightarrow \frac{[\text { Salt }]}{[\text { Acid }]}=10^1 \\
& \text { i.e., } \frac{[\text { acid }]}{[\text { salt }]}=\frac{1}{10} \Rightarrow 1: 10
\end{aligned}
$

Question 19.
The $\mathrm{pH}$ of $10^{-5} \mathrm{M} \mathrm{KOH}$ solution will be
(a) 9
(b) 5
(c) 19
(d) none of these
Answer:
(a) 9
$\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-}$
acid 1 base $1 \quad$ acid 2 base 2
$
\begin{aligned}
& {\left[\mathrm{OH}^{-}\right]=10^{-5} \mathrm{M} .} \\
& \mathrm{pH}=14-\mathrm{pOH} \\
& \mathrm{pH}=14-\left(-\log \left[\mathrm{OH}^{-}\right]\right) \\
& =14+\log \left[\mathrm{OH}^{-}\right]=14+\log 10^{-5} \\
& =14-5=9
\end{aligned}
$
Question 20 .
$\mathrm{H}_2 \mathrm{PO}_4^{-}$the conjugate base of
(a) $\mathrm{PO}_4$
(b) $\mathrm{P}_2 \mathrm{O}_5$
(c) $\mathrm{H}_3 \mathrm{PO}_4$
(d) $\mathrm{HPO}_4^{2-}$
Answer:
(c) $\mathrm{H}_3 \mathrm{PO}_4$
$\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-}$
acid 1 base 1 acid 2 base 2
$\mathrm{H}_2 \mathrm{PO}_4$ is the conjugate base of $\mathrm{H}_3 \mathrm{PO}_4$
Question 21.
Which of the following can act as lowery - Bronsted acid well as base?
(a) $\mathrm{HCl}$
(b) $\mathrm{SO}_4{ }^{2-}$
(c) $\mathrm{HPO}_4^{2-}$
(d) $\mathrm{Br}^{-}$

Answer:
(c) $\mathrm{HPO}_4{ }^{2-}$
$\mathrm{HPO}_4{ }^{2-}$ can have the ability to accept a proton to form $\mathrm{H}_2 \mathrm{PO}_4$.
It can also have the ability to donate a proton to form $\mathrm{PO}_4{ }^{-3}$.
Question 22.
The $\mathrm{pH}$ of an aqueous solution is Zero. The solution is
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(b) strongly acidic
$
\begin{aligned}
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\
& {\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}} \\
& =10^0=1 \\
& {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}}
\end{aligned}
$
The, solution is strongly acidic
Question 23.
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by
(a) $\left[\mathrm{H}^{+}\right]=\frac{\left.\mathrm{K}_{\mathrm{a}} \text { [acid }\right]}{\text { [salt }]}$
(b) $\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}}[$ salt $]$
(c) $\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}}[\mathrm{acid}]$
(d) $\left[\mathrm{H}^{+}\right]=\frac{\mathrm{K}_{\mathrm{a}}[\text { salt }]}{[\mathrm{acid}]}$
Answer:
According to Henderson equation
(a) $\left[\mathrm{H}^{+}\right]=\frac{\left.\mathrm{K}_{\mathrm{a}} \text { acid }\right]}{[\text { salt }]}$

According to Henderson equation
$
\begin{aligned}
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& \text { i.e. }-\log \left[\mathrm{H}^{+}\right]=-\log \mathrm{K}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& -\log \left[\mathrm{H}^{+}\right]=\log \frac{[\text { salt }]}{[\mathrm{acid}]} \times \frac{1}{\mathrm{~K}_{\mathrm{a}}} \\
& \log \frac{1}{\left[\mathrm{H}^{+}\right]}=\log \frac{[\text { salt }]}{[\text { acid }]} \times \frac{1}{\mathrm{~K}_{\mathrm{a}}} \\
& \therefore\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \frac{[\text { acid }]}{[\text { salt }]}
\end{aligned}
$
Question 24.
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?
(a) $\mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{C}}}$
(b) $\mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{b}}}}$
(c) $\mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}}}$
(d) $\mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{h}}}}$
Answer:
(c) $\mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}}}$
Question 25
Dissociation constant of $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$ the hydrolysis constant of $\mathrm{NH}_4 \mathrm{Cl}$ would be
(a) $1.8 \times 10^{-19}$
(b) $5.55 \times 10^{-10}$
(c) $5.55 \times 10^{-5}$
(d) $1.80 \times 10^{-5}$
Answer:
(b) $5.55 \times 10^{10}$

$
\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}=0.55 \times 10^{-9}=5.5 \times 10^{-10}
$
II. Answer the following questions.
Question 1.
What are lewis acids and bases? Give two example for each.
Answer:
1. Lewis acids:
- Lewis acid is a species that accepts an electron pair.
- Lewis acid is a positive ion (or) an electron deficient molecule.
- Example, $\mathrm{Fe}^{2+}, \mathrm{CO}_2, \mathrm{BF}_3, \mathrm{SiF}_4$ etc $\ldots$
2. Lewis bases:
- Lewis base is a species that donates an electron pair.
- Lewis base is an anion (or) neutral molecule with atleast one lone pair of electrons.
- Example, $\mathrm{NH}_3, \mathrm{~F}^{-}, \mathrm{CH}_2=\mathrm{CH}_2, \mathrm{CaO}$ etc....
Question 2.
Discuss the Lowry - Bronsted concept of acids and bases.
Answer:
According to Lowry - Bronsted concept, 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. When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, $\mathrm{HCl}$ behaves as an acid and $\mathrm{H}_2 \mathrm{O}$ is base. The proton transfer from the acid to base can be represented as
$
\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}
$

When ammonia is 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. The reaction is represented as
$
\mathrm{H}_2 \mathrm{O}+\mathrm{NH}_3 \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}
$
Let us consider the reverse reaction in the following equilibrium
$\mathrm{H}_3 \mathrm{O}^{+}$donates a proton to $\mathrm{Cl}^{-}$to form $\mathrm{HCI}$ i.e., the products also behave as acid and base. In general, Lowry - Bronsted (acid - base) reaction is represented as
$\mathrm{Acid}_1+$ Base $_2 \rightleftharpoons \mathrm{Acid}_2+$ Base $_1$ the Bronsted acid $\left(\mathrm{Acid}_1\right.$ ). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs. Conjugate acid - base pair

$\mathrm{HCl}$ and $\mathrm{Cl}^{-}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_3 \mathrm{O}$ are two conjugate acid - base pairs. i.e., $\mathrm{Cl}_{-}$is the conjugate base of the acid $\mathrm{HCl}$ (or) $\mathrm{HCl}$ is conjugate acid of $\mathrm{Cl}^{-}$Similarly $\mathrm{H}_3 \mathrm{O}$ is the conjugate acid of $\mathrm{H}_2 \mathrm{O}$. Limitations of Lowry Bronsted theory. Substances like $\mathrm{BF}_3, \mathrm{AICl}_3$ etc., that do not donate protons are known to behave as acids.
Question 3.
Indentify the conjugate acid base pair for the following reaction in aqueous solution.
1. $\mathrm{HS}^{-}(\mathrm{aq})+\mathrm{HF} \rightleftharpoons \mathrm{F}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{~S}(\mathrm{aq})$
2. $\mathrm{HPO}_4{ }^{2-}+\mathrm{SO}_3{ }^{2-} \rightleftharpoons \mathrm{PO}_4{ }^{3-}+\mathrm{HSO}_3{ }^{-}$
3. $\mathrm{NH}_4^{+}+\mathrm{CO}_3^{2-} \rightleftharpoons \mathrm{NH}_3+\mathrm{HCO}_3^{-}$
Answer:

1.

- $\mathrm{HF}$ and $\mathrm{F}^{-}, \mathrm{HS}^{-}$and $\mathrm{H}_2 \mathrm{~S}$ are two conjugate acid - base pairs.
- $\mathrm{F}^{-}$is the conjugate base of the acid $\mathrm{HF}$ (or) $\mathrm{HF}$ is the conjugate acid of $\mathrm{F}^{-}$
- $\mathrm{H}_2 \mathrm{~S}$ is the conjugate acid of $\mathrm{HS}^{-}$(or) $\mathrm{HS}^{-}$is the conjugate base of $\mathrm{H}_2 \mathrm{~S}$.

- $\mathrm{HPO}_4{ }^{2-}$ and $\mathrm{PO}_4{ }^{3-}, \mathrm{SO}_3{ }^{2-}$ and $\mathrm{HSO}_3{ }^{-}$are two conjugate acid - base pairs. $\mathrm{PO}_4{ }^{3-}$ is the conjugate base of the acid $\mathrm{HPO}_4{ }^{2-}$ (or) $\mathrm{HPO}_4{ }^{2-}$ is the conjugate acid of $\mathrm{PO}_4$. - $\mathrm{HSO}_3{ }^{-}$is the conjugate acid of $\mathrm{SO}_3{ }^{2-}$ (or) $\mathrm{SO}_3{ }^{2-}$ is the conjugate base of $\mathrm{HSO}_3{ }^{-}$.

- $\mathrm{NH}^{+}$and $\mathrm{NH} 3, \mathrm{CO}^{2-}$ and $\mathrm{HCO}^{-}$are two conjugate acid - base pairs.
- $\mathrm{HCO}_3{ }^{-}$is the conjugate of acid $\mathrm{CO}_3{ }^{2-}$ (or) $\mathrm{CO}_3{ }^{2-}$ is the conjugate bases of $\mathrm{HCO}_3{ }^{-}$.
- $\mathrm{NH}_3$ is the conjugate base of $\mathrm{NH}_4^{+}$(or) $\mathrm{NH}_4^{+}$is the conjugate acid of $\mathrm{NH}_3$.
Question 4.
Account for the acidic nature of $\mathrm{HCIO} 4$. In terms of Bronsted - Lowry theory, identify its conjugate base.

Answer:
$
\mathrm{HClO}_4 \rightleftharpoons \mathrm{H}^{+}+\mathrm{ClO}_4^{-}
$
1. According to Lowry - Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.
2. Let us consider the stabilities of the conjugate bases $\mathrm{ClO}_4{ }^{-}, \mathrm{ClO}_3{ }^{-}, \mathrm{CIO}_2{ }^{-}$and $\mathrm{ClO}^{-}$formed from these acid $\mathrm{HClO}_4, \mathrm{HClO}_3, \mathrm{HCIO}_2, \mathrm{HOCI}$ respectively.

These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.
3. The charge stabilization mercases in the order, $\mathrm{ClO}^{-}<\mathrm{ClO}_2^{-}<\mathrm{ClO}_3^{-}<\mathrm{ClO}_4^{-}$.
This means $\mathrm{ClO}_4^{-}$will have maximum stability and therefore will have minimum attraction for W. Thus $\mathrm{CIO}_4^{-}$will be weakest base and its conjugate acid $\mathrm{HCIO}_4$ is the strongest acid.
4. $\mathrm{CIO}_4^{-}$is the conjugate base of the acid $\mathrm{HClO}_4$.
Question 5.
When aqueous ammonia is added to $\mathrm{CuSO}_4$ solution, the solution turns deep blue due to the formation of tetrammine copper (II) complex, $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}(\mathrm{aq})+4 \mathrm{NH}_3(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ (aq), among $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ which is stronger Lewis base.
Answer:
Copper (II) sulphate solution, for example contains the blue hexaaqua copper (II) complex ion. In the first stage of the reaction, the ammonia acts as a Bronsted - Lowry base. With a small amount of ammonia solution, hydrogen ions are pulled off two water molecules in the hexaaqua ion. This produces a neutral complex, one carrying no charge.

If you remove two positively charged hydrogen ions from a $2+$ ion, then obviously there isn't going to be any charge left on the ion. Because of the lack of charge, the neutral complex isn't soluble in water and so you get a pale blue precipitate. $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)^6\right]^{2+}+2 \mathrm{NH}_3\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{OH}\right]+2 \mathrm{NH}_4^{+}$

This precipitate is often written as $\mathrm{Cu}(\mathrm{OH})_2$ and called copper (II) hydroxide. The reaction is reversible because ammonia is only a weak base. That precipitate dissolves if you add an excess of ammonia solution, giving a deep blue solution. The ammonia replaces four of the water molecules around the copper to give tetramminc diaqua copper (II) ions. The ammonia uses its lone pair to form a coordinate covalent bond with the copper. It is acting as an electron pair donor a Lewis base.
$
\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+4 \mathrm{NH}_3 \rightarrow\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2\right]^{2+}+\mathrm{H}_2 \mathrm{O}
$
Pale blue
Deep blue solution
Question 6.
The concentration of hydroxide ion in a water sample is found to be $2.5 \times 10^{-6} \mathrm{M}$. Identify the nature of the solution.
Answer:
The concentration of $\mathrm{OH}$ ion in a water sample is found to be $2.5 \times 10 \mathrm{M}$ $\mathrm{pOH}=-\log _{10}\left[\mathrm{OH}^{-}\right]$
$
\begin{aligned}
& \text { pOH }=-\log _{10}\left[2.5 \times 10^{-6}\right] \\
& =-\log _{10}[2.5]-\log _{10}\left[10^{-6}\right] \\
& =-0.3979-(-6) \\
& =-0.3979+6 \\
& \text { pOH }=5.6
\end{aligned}
$
We know that,
$
\begin{aligned}
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{pH}+5.6=14 \\
& \mathrm{pH}=14-5.6 \\
& \mathrm{pH}=8.4
\end{aligned}
$
$\mathrm{pH}=8.4$, shows the nature of the solution is basic.

Question 7.
A lab assistant prepared a solution by adding a calculated quantity of $\mathrm{HCl}$ gas $25^{\circ} \mathrm{C}$ to get a solution with $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=4 \times 10^5 \mathrm{M}$. Is the solution neutral (or) acidic (or) basic.
Answer:
$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=4 \times \mathrm{M}} \\
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\
& \mathrm{pH}=-\log _{10}\left[4 \times 10^5\right] \\
& \mathrm{pH}=-\log _{10}[4]-\log _{10}\left[10^{-5}\right] \\
& \mathrm{pH}=-0.6020-(-5)=-0.6020+5 \\
& \mathrm{pH}=4.398
\end{aligned}
$
Therefore, the solution is acidic.
Question 8.
Calculate the $\mathrm{pH}$ of $0.04 \mathrm{MHNO}_3$ Solution.
Answer:
Concentration of $\mathrm{HNO}_3=0.04 \mathrm{M}$
$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=0.04 \mathrm{~mol} \mathrm{dm}^{-3}} \\
& \mathrm{pH}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\
& =-\log (0.04) \\
& =-\log \left(4 \times 10^{-2}\right) \\
& =2-\log 4=2-0.6021 \\
& =1.3979=1.40
\end{aligned}
$

Question 9.
Define solubility product.
Answer:
Solubility product:
It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric coefficient in a balanced equilibrium equation.

Question 10 .
Define ionic product of water. Give its value at room temperature.
Answer:
1. The product of concentration of $\mathrm{H}^{+}$and $\mathrm{OH}$ ions in water at a particular temperature is known as ionic product.
2. The ionic product of water at room temperature $\left(25^{\circ} \mathrm{C}\right)$ is,
$
\begin{aligned}
& \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{+}\right](\text {or }) \\
& \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{+}\right] \\
& \mathrm{K}_{\mathrm{W}}=\left(1 \times 10^{-7}\right)\left(1 \times 10^{-7}\right) \\
& \mathrm{K}_{\mathrm{W}}=1 \times 10^{-14} \mathrm{~mol}^2 \mathrm{dm}^{-6}
\end{aligned}
$
Question 11.
Explain common ion effect with an example.
Answer:
Common ion Effect:
When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists.
$
\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})
$
However, the added salt, sodium acetate, completely dissociates to produce $\mathrm{Na}^{+}$and $\mathrm{CH}_3 \mathrm{COO}_{\text {ion. }}$.
$\mathrm{CH}_3 \mathrm{COONa}(\mathrm{aq}) \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{CH}_3 \mathrm{COO}(\mathrm{aq})$ Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.

We know from Le chatelier's priñciple that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess $\mathrm{CH}_3 \mathrm{COO}^{-}$ions combines with $\mathrm{H}$ ions to produce much more unionized $\mathrm{CH}_3 \mathrm{COOH}$ i.e.,
the equilibrium will shift towards the left. In other words, the dissociation of $\mathrm{CH}_3 \mathrm{COOH}_{\text {is suppressed. }}$. Thus, the dissociation of a weak acid $\left(\mathrm{CH}_3 \mathrm{COOH}\right)$ is suppressed in the presence of a salt $\left(\mathrm{CH}_3 \mathrm{COONa}\right)$ containing an ion common to the weak electrolyte. It is called the common ion effect.

Question 12.
Derive an expression for Ostwald's dilution law.
Answer:
Ostwald's dflution law:
It relates the dissociation constant of the weak acid $\left(\mathrm{K}_{\mathrm{a}}\right)$ with its degree of dissociation $(\alpha)$ and the concentration (c). Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as,
$
\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}
$
The dissociation constant of acetic acid is,
$
\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}
$

Substituting the equilibrium concentration in equation
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\frac{(\alpha \mathrm{C})(\alpha \mathrm{C})}{(1-\alpha) \mathrm{C}} \Rightarrow \mathrm{K}_{\mathrm{a}}=\frac{\alpha^2 \mathrm{C}^2}{(1-\alpha) \mathrm{C}} \\
& \mathrm{K}_{\mathrm{a}}=\frac{\alpha^2 \mathrm{C}}{(1-\alpha)}
\end{aligned}
$
We know that weak acid dissociates only to a very small extent compared to one, a is so small. equation (1) becomes,
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{C} \\
& \alpha^2=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}} \Rightarrow \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}
\end{aligned}
$
Similarly, for a weak base,
$
\begin{gathered}
\mathrm{K}_{\mathrm{b}}=\alpha^2 \mathrm{C} \\
\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}}
\end{gathered}
$
The concentration of $\mathrm{H}$ can be calculated using the $\mathrm{K}_{\mathrm{a}}$ value as below,
$
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\alpha \mathrm{C}} \\
& \alpha=\frac{\left[\mathrm{H}^{+}\right]}{\mathrm{C}}
\end{aligned}
$
Substituting a value in equation (2),
$
\begin{aligned}
& \frac{\left[\mathrm{H}^{+}\right]}{\mathrm{C}}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}} \Rightarrow\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}} \cdot \mathrm{C} \\
& {\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}^2}{\mathrm{C}}} \Rightarrow\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}}
\end{aligned}
$
For weak base
$
\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{b}} \cdot \mathrm{C}}
$
Question 13.
Define $\mathrm{pH}$.
Answer:
$\mathrm{pH}$ of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium

ions present in the solution.
$
\mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}\right] \text { (or) } \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]
$
Question 14.
Calculate the $\mathrm{pH}$ of $1.5 \times 10^{-3} \mathrm{M}$ solution of $\mathrm{Ba}(\mathrm{OH})_2$
Answer:
$
\begin{aligned}
& \mathrm{Ba}(\mathrm{OH})_2 \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-} \\
& 1.5 \times 10^{-3} \mathrm{M} \quad 2 \times 1.5 \times 10^{-3} \mathrm{M} \\
& {\left[\mathrm{OH}^{-}\right]=3 \times 10^3 \mathrm{M} .} \\
& {[\mathrm{pH}+\mathrm{pOH}=14]} \\
& \mathrm{pH}=14-\mathrm{pOH} \\
& \mathrm{pH}=14-\left(-\log \left[\mathrm{OH}^{-}\right]\right) \\
& =14+\log \left[\mathrm{OH}^{-}\right] \\
& =14+\log \left(3 \times 10^{-3}\right) \\
& =14+\log 3+\log 10^{-3} \\
& =11+0.4771 \\
& \mathrm{pH}=11.48
\end{aligned}
$
Question 15 .
$50 \mathrm{ml}$ of $0.05 \mathrm{M} \mathrm{HNO}_3$ is added to $50 \mathrm{ml}$ of $0.025 \mathrm{M} \mathrm{KOH}$. Calculate the $\mathrm{pH}$ of the resultant solution.
Solution.
Number of moles of $\mathrm{HNO}_3=0.05 \times 50 \times 1.5 \times 10^{-3}$
Number of moles of $\mathrm{KOH}=0.025 \times 50 \times 10^{-3}=1.25 \times 10^{-3}$
Number of moles of $\mathrm{HNO}_3$ after mixing $=2.5 \times 10^{-3}-1.5 \times 10^{-3}$
$
=1.25 \times 10^{-3}
$
$\therefore$ concentration of $\mathrm{HNO}_3=\frac{\text { Number of moles of } \mathrm{HNO}_3}{\text { Volume in litre }}$
After mixing, total volume $=100 \mathrm{ml}=100 \times 10^{-3} \mathrm{~L}$
$
\begin{aligned}
& \therefore\left[\mathrm{H}^{+}\right]=\frac{1.25 \times 10^{-3} \text { moles }}{100 \times 10^{-3} \mathrm{~L}}=1.25 \times 10^{-2} \text { moles } \mathrm{L}^{-1} \\
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& \mathrm{pH}=-\log \left(1.25 \times 10^{-2}\right)=2-0.0969 \\
& =1.9031
\end{aligned}
$
Question 16.
The $\mathrm{K}_{\mathrm{a}}$ value for $\mathrm{HCN}$ is $10^{-9}$. What is the $\mathrm{pH}$ of $0.4 \mathrm{M} \mathrm{HCN}$ solution?
Answer:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=10^{-9} \\
& \mathrm{c}=\mathrm{O} .4 \mathrm{M} \\
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& {\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a}} \times \mathrm{c}}} \\
& {\left[\mathrm{H}^{+}\right]=\sqrt{10^{-9} \times 0.4}} \\
& {\left[\mathrm{H}^{+}\right]=2 \times 10^{-5}}
\end{aligned}
$
$
\begin{aligned}
& \therefore \mathrm{pH}=-\log \left(2 \times 10^{-5}\right) \\
& =-\log 2-\log \left(10^{-5}\right) \\
& =-0.3010+5 \\
& \mathrm{pH}=4.699
\end{aligned}
$
Question 17.
Calculate the extent of hydrolysis and the $\mathrm{pH}$ of $0.1 \mathrm{M}$ ammonium acetate Given that.
$
\mathrm{K}_{\mathrm{a}}=\mathrm{K}_{\mathrm{b}}=1.8 \times 10^{-5}
$
Solution.
$
\begin{aligned}
& \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}}}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}}=\sqrt{\frac{1}{1.8} \times 10^{-4}}=\sqrt{0.5555 \times 10^{-4}} \\
& =0.7453 \times 10^{-2} \\
& \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{W}}+\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \mathrm{pK}_{\mathrm{b}} \\
& \text { Given that } \mathrm{K}_{\mathrm{a}}=\mathrm{K}_{\mathrm{b}}=1.8 \times 10^{-5} \\
& \text { if } \mathrm{K}_{\mathrm{a}}=\mathrm{K}_{\mathrm{b}} \text {, then. } \mathrm{pK}_{\mathrm{a}}=\mathrm{pK}_{\mathrm{b}} \\
& \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{W}}=\frac{1}{2}(14)=7
\end{aligned}
$
Question 18 .
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.
Answer:
Let us consider the reactions between a strong acid, $\mathrm{HCl}$, and a weak base, $\mathrm{NH}_4 \mathrm{OH}$, to produce a salt, $\mathrm{NH}_4 \mathrm{Cl}$, 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{I}) \\
& \mathrm{NH}_4 \mathrm{CI}(\mathrm{aq}) \rightarrow \mathrm{NH}_4^{+}+\mathrm{Cl}^{-}(\mathrm{aq})
\end{aligned}
$
$\mathrm{NH}_4{ }^{+}$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 $\mathrm{NH}_4 \mathrm{OH}$ shown below.
$
\mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(1) \rightleftharpoons \mathrm{NH}_4 \mathrm{OH}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})
$

There is no such tendency shown by $\mathrm{Ct}$ and therefore $\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]$the solution is acidic and the $\mathrm{pH}$ is less than 7.
As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the $\mathrm{K}_{\mathrm{a}}$ and $\mathrm{K}_{\mathrm{b}}$ as
$\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}$
Let us calculate the $\mathrm{Kb}$ value in terms of degree of hydrolysis (h) and the concentration of salt $\mathrm{K}_{\mathrm{h}}=\mathrm{h}^2 \mathrm{C}$ and
$
\begin{aligned}
{\left[\mathrm{H}^{+}\right] } & =\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \\
{\left[\mathrm{H}^{+}\right] } & =\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}} \cdot \mathrm{C}} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=-\log \left(\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}} \cdot \mathrm{C}\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}_{\mathrm{b}}-1 / 2 \log \mathrm{C}
\end{aligned}
$

Question 19.
Solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $1 \times 10^{-12}$. What is the solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in $0.01 \mathrm{M} \mathrm{AgNO}_3$ solution?
Answer:
Solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$,
$
\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=1 \times 10^{-2} \\
& \mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_4^{-2} \\
& \text { (s) } \\
& \text { (2s) } \\
& \text { (s) } \\
& \mathrm{AgNO}_3 \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{NO}_3^{-} \\
& (0.01 \mathrm{M}) \quad(0.01 \mathrm{M}) \quad(0.01 \mathrm{M}) \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4{ }^{2-}\right] \\
& {\left[\mathrm{Ag}^{+}\right]=2 \mathrm{~s}+0.01} \\
& 0.01 \gg 2 \mathrm{~s} \\
& {\left[\mathrm{Ag}^{+}\right]=0.01 \mathrm{M}} \\
& {\left[\mathrm{CrO}_4^{-2}\right]=\mathrm{s}} \\
& \mathrm{K}_{\mathrm{sp}}=(0.01)^2 \text {.(s) } \\
& \mathrm{s}=\frac{1 \times 10^{-12}}{\left(10^{-2}\right)^2}=\frac{1 \times 10^{-12}}{10^{-4}}=1 \times 10^{-8} \mathrm{M} \\
&
\end{aligned}
$
Question 20.
Write the expression for the solubility product of $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$
Answer:
$
\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2(\mathrm{~s}) \rightleftharpoons 3 \mathrm{Ca}^{2+}(3 \mathrm{~s})+2 \mathrm{PO}_4^{3-}(2 \mathrm{~s})
$
Solubility of $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$ is,
$
\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4^{3-}\right]^2 \\
& \mathrm{~K}_{\mathrm{sp}}=(3 \mathrm{~s})^3 \cdot(2 \mathrm{~s})^2 \\
& \mathrm{~K}_{\mathrm{sp}}=\left(27 \mathrm{~s}^3\right) \cdot\left(4 \mathrm{~s}^2\right) \\
& \mathrm{K}_{\mathrm{sp}}=108 \mathrm{~s}^5 .
\end{aligned}
$
Question 21.
A saturated solution, prepared by dissolving $\mathrm{CaF}_2(\mathrm{~s})$ in water, has $\left[\mathrm{Ca}^{2+}\right]=3.3 \times 10^{-4} \mathrm{M}$. What is the $\mathrm{K}_{\mathrm{sp}}$ of
$\mathrm{CaF}_2$ ?

Answer:
$
\begin{aligned}
& \mathrm{CaF}_2(\mathrm{~s}) \rightleftharpoons \mathrm{Ca}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{~F}_{-(\mathrm{aq})} \\
& {\left[\mathrm{F}_{-}\right]=2\left[\mathrm{Ca}^{2+}\right]=2 \times 33 \times 10^{-4} \mathrm{M}} \\
& =6.6 \times 10^{-4} \mathrm{M} \\
& =\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\
& =\left(3.3 \times 10^{-4}\right)\left(6.6 \times 10^{-4}\right)^2 \\
& =1.44 \times 10^{-10}
\end{aligned}
$
Question 22.
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{AgCI}$ is $1.8 \times 10^{-10}$. Calculate molar solubility in $1 \mathrm{M} \mathrm{AgNO}_3$
Answer:
$
\begin{aligned}
& \mathrm{AgCI}(\mathrm{s}) \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \\
& \mathrm{x}=\text { solubility of } \mathrm{AgCI}^2 1 \mathrm{M} \mathrm{AgNO} \\
& \mathrm{AgNO}_3(\mathrm{aq}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{NO}_3^{-}(\mathrm{aq}) \\
& {\left[\mathrm{Ag}^{+}\right]=x+1 \simeq 1 \mathrm{M}(\because \mathrm{x}<<\mathrm{l})} \\
& {\left[\mathrm{Cl}^{-}\right]=\mathrm{x}} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \\
& 1.8 \times 10^{-10}=(1)(\mathrm{x}) \\
& \mathrm{x}=1.8 \times 10^{-10} \mathrm{M}
\end{aligned}
$
Question 23.
A particular saturated solution of silver chromate $\mathrm{Ag}_2 \mathrm{CrO}_4$ has $\left[\mathrm{Ag}^{+}\right]=5 \times 10^{-5}$ and $\left[\mathrm{CrO}_4\right]^{2-}=4.4 \times 10 \mathrm{M}$. What is the value of for $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
Answer:
$
\begin{aligned}
& \mathrm{Ag}_2 \mathrm{CrO}_4(\mathrm{~s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CrO}_4^{2-}(\mathrm{aq}) \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right] \\
& =\left(5 \times 10^{-5}\right)^2\left(4.4 \times 10^{-4}\right) \\
& =1.1 \times 10^{-1}
\end{aligned}
$
Question 24
Write the expression for the solubility product of $\mathrm{Hg}_2 \mathrm{CI}_2$.
Answer:

Question 25 .
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $1.1 \times 10^{-12}$ What is solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in $0.1 \mathrm{M} \mathrm{K}_2 \mathrm{CrO}_4$.
Answer:
$\mathrm{x}$ is the solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in $0.1 \mathrm{M} \mathrm{K}_2 \mathrm{CrO}_4$
$
\begin{aligned}
& \mathrm{K}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{~K}^{+}+\mathrm{CrO}_4^{2-} \\
& 0.1 \mathrm{M} \quad 0.2 \mathrm{M} 0.1 \mathrm{M} \\
& {\left[\mathrm{Ag}^{+}\right]=2 \mathrm{x}} \\
& {\left[\mathrm{CrO}_4^{2-}\right]=(\mathrm{x}+0.1) \approx 0.1 \quad \therefore \mathrm{x}<<0.1} \\
& \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right] \\
& 1.1 \times 10^{-12}=(2 \mathrm{x})^2(0.1) \\
& 1.1 \times 10^{-12}=0.4 \mathrm{x}^2 \\
& \mathrm{x}^2=\frac{1.1 \times 10^{-12}}{0.4} \\
& \mathrm{x}=\sqrt{\frac{1.1 \times 10^{-12}}{0.4}} \\
& \mathrm{x}=\sqrt{2.75 \times 10^{-12}} \\
& \mathrm{x}=1.65 \times 10^{-6} \mathrm{M}
\end{aligned}
$
Question 26.
Will a precipitate be formed when $0.150 \mathrm{~L}$ of $0.1 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_3\right)_2$ and $0.100 \mathrm{~L}$ of $0.2 \mathrm{M} \mathrm{NaCl}$ are mixed?
$
\left(\mathrm{PbCI}_2\right)=1.2 \times 10^{-5} .
$

Answer:
When two or more solution are mixed, the resulting concentrations are differnet from the original. Total volume $0.250 \mathrm{~L}$
$
\begin{aligned}
& \mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightleftharpoons \mathrm{Pb}^{2+}+2 \mathrm{NO}_3^{-} \\
& \begin{array}{lll}
0.1 \mathrm{M} & 0.1 \mathrm{M} & 0.2 \mathrm{M}
\end{array} \\
&
\end{aligned}
$
Number of moles
$\mathrm{Pb}^{2+}=$ molarity $\times$ Volume of the solution in lit $=0.1 \times 0.15$
$
\begin{aligned}
& {\left[\mathrm{Pb}^{2+}\right]_{\text {mix }}=\frac{0.1 \times 0.15}{0.25}=0.06 \mathrm{M}} \\
& \mathrm{NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \\
& 0.2 \mathrm{M} \quad 0.2 \mathrm{M} \quad 0.2 \mathrm{M}
\end{aligned}
$
No.of moles $\mathrm{Cl}^{-}=0.2 \times 0.1$
$
\left[\mathrm{Cl}^{-}\right]_{\text {mix }}=\frac{0.2 \times 0.1}{0.25}=0.08 \mathrm{M}
$
Precipitation of $\mathrm{PbCl}_2$ (s) occurs if $\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2>\mathrm{K}_{\mathrm{sp}}$
$
\begin{aligned}
& {\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2=(0.06)(0.08)^2} \\
& =3.84 \times 10^{-4}
\end{aligned}
$
Since ionic product $\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2>\mathrm{K}_{\mathrm{sp}} \mathrm{PbCl}_2$ is precipitated.
Question 27.
of $\mathrm{Al}(\mathrm{OH})_3$ is $1 \times 10^{-15} \mathrm{M}$. At what $\mathrm{pH}$ does $1.0 \times 10^{-13} \mathrm{MAI}^{3+}$ precipitate on the addition of buffer of $\mathrm{NH}_4 \mathrm{CI}$ and $\mathrm{NH}_4 \mathrm{OH}$ solution?
Answer:
$
\begin{aligned}
& \mathrm{Al}(\mathrm{OH})_3 \mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3 \\
& \mathrm{Al}(\mathrm{OH})_3 \text { precipitates when } \\
& {\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3>\mathrm{K}_{\mathrm{sp}}} \\
& \left(1 \times 10^{-3}\right)\left[\mathrm{OH}^{-}\right]^3>\mathrm{K}_{\mathrm{sp}} \\
& {\left[\mathrm{OH}^{-}\right]^3>1 \times 10^{-12}} \\
& {\left[\mathrm{OH}^{-}\right]>1 \times 10^{-4} \mathrm{M}} \\
& {\left[\mathrm{OH}^{-}\right]=1 \times 10^{-4} \mathrm{M}} \\
& \mathrm{pOH}=-1 \mathrm{Og}{ }_{10}\left[\mathrm{OH}^{-}\right]=-\log \left(1 \times 10^{-4}\right)=4 \\
& \mathrm{pH}=14-4=10
\end{aligned}
$
Thus, $\mathrm{Al}(\mathrm{OH})_3$ precipitates at a $\mathrm{pH}$ of 10
Evaluate Yourself
Question 1.
Classify the following as acid (or) base using Arrhenius concept
1. $\mathrm{HNO}_3$
2. $\mathrm{Ba}(\mathrm{OH})_2$
3. $\mathrm{HlPO}_4$
4. $\mathrm{CH}_3 \mathrm{COOH}$
Answer:
1. $\mathrm{HNO}_3$ :
Nitric acid, dissociates to give hydrogen ions in water. $\mathrm{HNO}_3$ is acid.
2. $\mathrm{Ba}(\mathrm{OH})_2$ :
Barium hydroxide, dissociates to give hydroxyl ions in water. $\mathrm{Ba}(\mathrm{OH})_2$ is base.
3. $\mathrm{H}_3 \mathrm{PO}_4$ :
Orthophosphoric acid, dissociates to give hydrogen ions in water. $\mathrm{H}_3 \mathrm{PO}_4$ is acid.
4. $\mathrm{CH}_3 \mathrm{COOH}$ :
Acetic acid, dissociates to give hydrogen ions in water. $\mathrm{CH}_3 \mathrm{COOH}$ is acid.

Question 2.
Write a balanced equation for the dissociation of the following in water and identify the conjugate acid base pairs.
1. $\mathrm{NH}_4$
2. $\mathrm{H}_2 \mathrm{SO}_4$
3. $\mathrm{CH}_3 \mathrm{COOH}$.
Answer:
1. $\mathrm{NH}_4{ }^{+}$Conjugate acid-base pair

$\mathrm{NH}_4{ }^{+}$and $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_3 \mathrm{O}^{+}$are two conjugate acid - base pairs.
$\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_3 \mathrm{O}^{+}$are two conjugate acid-base pairs.
2. $\mathrm{H}_2 \mathrm{SO}_4$ :

$\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HSO}_4^{-}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_3 \mathrm{O}^{+}$are two conjugate acid-base pairs.
3. $\mathrm{CH}_3 \mathrm{COOH}$ :

$\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_3 \mathrm{O}^{+}$are two conjugate acid-base pairs.
Question 3.
Identify the Lewis acid and the Lewis base in the following reactions.

(i) 

(ii)

Answer:
(i). $\mathrm{CaO}+\mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3$
- $\mathrm{CaO}$ - Lewis base - All metals oxides are Lewis bases
- $\mathrm{CO}_2$ - Lewis acid $-\mathrm{CO}_2$ contains a polar double bond.
(ii)

- $\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_3-$ Lewis base - Electron rich species
- $\mathrm{AlCl}_3$ - Lewis acid- $\mathrm{AICI}_3$ is electron deficient molecule.
Question 4.
$\mathrm{H}_3 \mathrm{BO}_3$ accepts hydroxide ion from water as shown below
Answer:
$
\mathrm{H}_3 \mathrm{BO}_3(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})=\mathrm{B}(\mathrm{OH})_4^{-}+\mathrm{H}^{+}
$
Predict the nature of $\mathrm{H}_3 \mathrm{BO}_3$ using Lewis concept. Boric acid is also called as hydrogen borate or orthoboric acid. It is a weak mono basic Lewis acid of boron and it is written as $\mathrm{B}(\mathrm{OH})_3$. It accepts hydroxyl $\left(\mathrm{OH}^{-}\right)$ion from water. It does not dissociate to give hydronium $\left(\mathrm{H}_3 \mathrm{O}^{+}\right)$ion rather forms metaborate ion and this ions in turn give $\mathrm{H}_3 \mathrm{O}$ ion.
$\mathrm{B}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{O}\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}_3 \mathrm{O}^{+}$Hence it is considered as weak acid.
Question 5.
At a particular temperature, the $\mathrm{K}_{\mathrm{W}}$ of a neutral solution was equal to $4 \times 10^{-14}$. Calculate the concentration of $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$and $\left[\mathrm{OH}^{-}\right]$.
Answer:
Given solution is neutral
$
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]
$
Let $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\mathrm{x}$; then $\left[\mathrm{OH}^{-}\right]=\mathrm{x}$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \\
& 4 \times 10^{-14}=\mathrm{x} \cdot \mathrm{x} \\
& \mathrm{x}^2=4 \times 10^{-14} \\
& \mathrm{x}=\sqrt{4 \times 10^{-14}}=2 \times 10^{-7}
\end{aligned}
$
Question 6.

1. Calculate $\mathrm{pH}$ of $10^{-8} \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
2. Calculate the concentration of hydrogen ion in moles per litre of a solution whose $\mathrm{pH}$ is 5.4
3. Calculate the $\mathrm{pH}$ of an aqueous solution obtained by mixing $50 \mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{HCI}$ with $50 \mathrm{ml} 0.1 \mathrm{M}$ $\mathrm{NaOH}$
Answer:
1.
$
\begin{array}{cc}
\mathrm{H}_2 \mathrm{SO}_4 \stackrel{\mathrm{H}_2 \mathrm{O}}{\rightleftharpoons} 2 \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{SO}_4^{2-} \\
10^{-8} \mathrm{M} & 2 \times 10^{-8} \quad 10^{-8} \mathrm{M}
\end{array}
$
In this case the concentration of $\mathrm{H}_2 \mathrm{SO}_4$ is very low and hence $\left[\mathrm{H}_3 \mathrm{O}\right]$ from water cannot be neglected
$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=2 \times 10^{-8}\left(\text { from } \mathrm{H}_2 \mathrm{SO}_4\right)+10^{-7} \text { (from water) }} \\
& =10^{-8}(2+10) \\
& =12 \times 10^{-8}=1.2 \times 10^{-7} \\
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\
& =-\log _{10}\left(1.2 \times 10^{-7}\right) \\
& =7-\log _{10} 1.2 \\
& =7-0.0791=6.9209
\end{aligned}
$
2.
$
\begin{aligned}
& \mathrm{pH} \text { of the solution }=5.4 \\
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\text {antilog of }(-\mathrm{pH})} \\
& =\text { antilog of }(-5.4) \\
& =\text { antilog of }(-6+0.6)=\overline{6} .6 \\
& =3.981 \times 10^{-6} \\
& \text { i.e., } 3.98 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3}
\end{aligned}
$
3.
No. of moles of $\mathrm{HCl}=0.2 \times 50 \times 10^{-3}=10 \times 10^{-3}$
No. of moles of $\mathrm{NaOH}=0.1 \times 50 \times 10^{-3}=5 \times 10^{-3}$
No. of moles of $\mathrm{HCl}$ after mixing $=10 \times 10^{-3}-5 \times 10^{-3}$ $=5 \times 10^{-3}$
after mixing total volume $=100 \mathrm{~mL}$
$\therefore$ Concentration of $\mathrm{HCl}$ in moles per litre $=\frac{5 \times 10^{-3} \mathrm{~mole}}{100 \times 10^{-3} \mathrm{~L}}$

$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=5 \times 10^{-2} \mathrm{M}} \\
& \mathrm{pH}=-\log \left(5 \times 10^{-2}\right) \\
& =2-\log 5 \\
& =2-0.6990 \\
& =1.30
\end{aligned}
$
Question 7.
$\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$ Calculate the percentage of ionisation of $0.06 \mathrm{M}$ ammonium hydroxide solution.
Answer:
$
\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}}=\sqrt{\frac{1.8 \times 10^{-5}}{6 \times 10^{-2}}}=\sqrt{3 \times 10^{-4}}=1.732 \times 10^{-2}=1.732 \times 10^{-2} \times 100=1.732 \%
$

Question 8 .
1. Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.
2. Calculate the $\mathrm{pH}$ of a buffer solution consisting of $0.4 \mathrm{M} \mathrm{CH} \mathrm{COOH}_3 \mathrm{CO}$ and $0.4 \mathrm{M} \mathrm{CH}_3 \mathrm{COONa}$. What is the change in the $\mathrm{pH}$ after adding $0.01 \mathrm{~mol}$ of $\mathrm{HCI}$ to $500 \mathrm{~m} 1$ of the above buffer solution.
Assume that the addition of $\mathrm{HCI}$ causes negligible change In the volume. Given: $(\mathrm{K}=1.8 \times 105)$.
Answer:
1. Dissociation of buffer components
$
\begin{aligned}
& \mathrm{NH}_4 \mathrm{OH}(\mathrm{aq}) \rightleftharpoons \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{OH}^{-} \text {(aq) } \\
& \mathrm{NH}_4 \mathrm{CI} \rightarrow \mathrm{NH}_4^{+}+\mathrm{Cl}^{-}
\end{aligned}
$
Addition of $\mathrm{OH}^{-}$
The added $\mathrm{H}^{+}$ions are neutralized by $\mathrm{NH}_4 \mathrm{OH}$ and there is no appreciable decrease in $\mathrm{pH}$.
$
\mathrm{NH}_4 \mathrm{OH}(\mathrm{aq})+\mathrm{H}^{+} \rightleftharpoons \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(1)
$
Addition of
$
\mathrm{NH}_4^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{NH}_4 \mathrm{OH}(\mathrm{aq})
$
The added $\mathrm{OH}$ ions react with $\mathrm{NH}_4$ to produce unionized $\mathrm{NH}_4 \mathrm{OH}$. Since $\mathrm{NH}_4 \mathrm{OH}$ is a weak base, there is no appreciable increase in $\mathrm{pH}$.

2. $\mathrm{pH}$ of buffer

$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \\
& \begin{array}{ccc}
0.4-\alpha & \alpha & \alpha
\end{array} \\
&
\end{aligned}
$
Addition of $0.01 \mathrm{~mol} \mathrm{HCI}$ to $500 \mathrm{ml}$ of buffer
Added $\left[\mathrm{H}^{+}\right]$
$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COONa}(\mathrm{aq}) \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq}) \\
& \quad 0.4 \\
& {\left[\mathrm{H}^{+}\right]=\frac{\mathrm{K}_{\mathrm{a}}\left[\mathrm{CH}_3 \mathrm{COOH}\right]}{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}} \\
& {\left[\mathrm{CH}_3 \mathrm{COOH}\right]=0.4-\alpha \simeq 0.4} \\
& {\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=0.4+\alpha \simeq 0.4} \\
& \therefore\left[\mathrm{H}^{+}\right]=\frac{\mathrm{K}_{\mathrm{a}}(0.4)}{(0.4)} \\
& \therefore \mathrm{pH}=-\log \left(1.8 \times 10^{-5}\right)=4.74
\end{aligned}
$

Addition of $0.01 \mathrm{~mol} \mathrm{HCl}$ to $500 \mathrm{ml}$ of buffer
$
\text { Added } \begin{aligned}
{\left[\mathrm{H}^{+}\right]=\frac{0.01 \mathrm{~mol}}{500 \mathrm{~mL}} } & =\frac{0.01 \mathrm{~mol}}{1 / 2 \mathrm{~L}} \\
& =0.02 \mathrm{M}
\end{aligned}
$
$
\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})
$
$
\mathrm{CH}_3 \mathrm{COONa} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Na}^{+}
$
0.4
$
\mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{HCl} \rightarrow \mathrm{CH}_3 \mathrm{COOH}+\mathrm{Cl}^{-}
$
$\begin{array}{llll}(0.02) & 0.02 & 0.02 & 0.02\end{array}$
$
\begin{aligned}
& \therefore\left[\mathrm{CH}_3 \mathrm{COOH}\right]=0.4-\alpha+0.02=0.42-\alpha \simeq 0.42 \\
& {\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=0.4+\alpha-0.02=0.38+\alpha \simeq 0.38 } \\
& {\left[\mathrm{H}^{+}\right] }=\frac{\left(1.8 \times 10^{-5}\right)(0.42)}{(0.38)} \\
& {\left[\mathrm{H}^{+}\right] }=1.99 \times 10^{-5} \\
& \mathrm{pH}=-\log \left(1.99 \times 10^{-5}\right) \\
&=5-\log 1.99 \\
&=5-0.30 \\
&=4.70
\end{aligned}
$

Question 9.
1. How can you prepare a buffer solution of $\mathrm{pH} 9$. You are provided with $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ solution and ammonium chloride crystals. (Given: $\mathrm{pKb}$ for $\mathrm{NH}_4 \mathrm{OH}$ is 4.7 at $25^{\circ} \mathrm{C}$ )
2. What volume of $0.6 \mathrm{M}$ sodium formate solution is required to prepare a buffer solution of $\mathrm{pH} 4.0$ by mixing it with $100 \mathrm{ml}$ of $0.8 \mathrm{M}$ formic acid. (Given: $\mathrm{pK}_{\mathrm{a}}$ for formic acid is 3.75.)
Answer:
1.
$
\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { [salt }]}{\text { [base] }}
$
We know that $\mathrm{pH}+\mathrm{pOH}=14$
$
\begin{aligned}
& 9+\mathrm{pOH}=14 \\
& =\mathrm{pOH}=14-9=5 \\
& 5=4.7+\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]} \\
& 0.3=\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{0.1} \\
& \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{0.1}=\text { antilog of }(0.3) \\
& {\left[\mathrm{NH}_4 \mathrm{Cl}\right]=0.1 \mathrm{M} \times 1.995}
\end{aligned}
$

$=0.1995 \mathrm{M}$
$=0.2 \mathrm{M}$
Amount of $\mathrm{NH}_4 \mathrm{CI}$ required to prepare 1 litre $0.2 \mathrm{M}$ solution = Strength of $\mathrm{NH}_4 \mathrm{CI} \times$ molar
mass of $\mathrm{NH}_4 \mathrm{CI}$
$=0.2 \times 535$
$=10.70 \mathrm{~g}$
$10.70 \mathrm{~g}$ ammonium chloride is dissolved in water and the solution is made up to one litre to get $0.2 \mathrm{M}$ solution. On mixing equal volume of the given $\mathrm{NH}_4 \mathrm{OH}$ solution and the prepared $\mathrm{NH}_4 \mathrm{CI}$ solution will give a buffer solution with required $\mathrm{pH}$ value $(\mathrm{pH}=9)$.

2.
$
\begin{aligned}
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& 4=3.75+\log \frac{\text { [sodium formate] }}{[\text { formic acid] }} \\
& \text { [Sodium formate] = number of moles of } \mathrm{HCOONa} \\
& =0.6 \times \mathrm{V} \times 10^{-3} \\
& {[\text { formic acid] }=\text { number of moles of } \mathrm{HCOOH}} \\
& =0.8 \times 100 \times 10^{-3} \\
& {[\text { formic acid] }=\text { number of moles of } \mathrm{HCOOH}} \\
& =0.8 \times 100 \times 10^{-3} \\
& =80 \times 10^{-3} \\
& 4=3.75+\log \frac{0.6 \mathrm{~V}}{80} \\
& 0.25=\log \frac{0.6 \mathrm{~V}}{80} \\
& \text { antilog of } 0.25=\frac{0.6 \mathrm{~V}}{80} \\
& 0.6 \mathrm{~V}=1.778 \times 80 \\
& =1.78 \times 80 \\
& =142.4 \\
& \mathrm{~V}=\frac{142.4 \mathrm{~mL}}{0.6}=237.33 \mathrm{~mL}
\end{aligned}
$

Question 10.
Calculate the
1. hydrolysis constant
2. degree of hydrolysis and
3. $\mathrm{pH}$ of $0.05 \mathrm{M}$ sodium carbonate solution $\mathrm{pK}_{\mathrm{a}}$ for $\mathrm{HCO}_3{ }^{-}$is 10.26 .
Answer:
1. Hydrolysis constant
$
h=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{c}}}
$
Given
$
\begin{aligned}
& \mathrm{K}_{\mathrm{W}}=1 \times 10^{-14} \\
& \mathrm{c}=0.05 \mathrm{M} \\
& \mathrm{PK}_{\mathrm{a}}=10.26
\end{aligned}
$

$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}} \\
& \mathrm{K}_{\mathrm{a}}=\operatorname{antilog} \text { of }\left(-\mathrm{pK}_{\mathrm{a}}\right) \\
& \mathrm{K}_{\mathrm{a}}=\text { antilog of }(-10.26) \\
& \mathrm{K}_{\mathrm{a}}=5.49 \times 10^{-11} \\
& \mathrm{~h}=\sqrt{\frac{1 \times 10^{-14}}{5.49 \times 10^{-11} \times 0.05}}=\sqrt{3.642 \times 10^{-3}} \\
& \mathrm{~h}=6.034 \times 10^{-2} \\
& 2 . \text { Degree of hydrolysis } \\
& \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\frac{1 \times 10^{-14}}{5.49 \times 10^{-11}}=1.82 \times 10^{-4}
\end{aligned}
$
3.
$
\begin{aligned}
\mathrm{pH} & =7+\frac{\mathrm{pK}_{\mathrm{a}}}{2}+\frac{\log \mathrm{C}}{2} \\
& =7+\frac{10.26}{2}+\frac{\log (0.05)}{2}=7+5.13+\left(\frac{-1.30}{2}\right)=7+5.13-0.65 \\
\mathrm{pH} & =11.48
\end{aligned}
$