Example problems solved - Chapter 8 Ionic Equilibrium 12th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On August 15, 2024

By SaraNextGen

Example problems solved
Question 1.
Identify the Lewis acid and the Lewis base in the following reactions.
$
\mathrm{Cr}^{3+}+6 \mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}
$
In the hydration of ion, each of six water molecules donates a pair of electron to $\mathrm{Cr}^{3+}$ to form the hydrated cation, hexaaquachromium (III) ion, thus, the lewis acid is $\mathrm{Cr}$ and the Lewis base $\mathrm{H}_2 \mathrm{O}$.

Question 2.
Calculate the concentration of $\mathrm{OH}^{-}$in a fruit juice which contains $2 \times 10 \mathrm{M}, \mathrm{H}_3 \mathrm{O}^{+}$Ion. Identify the nature of the solution.
Answer:
Given that $\mathrm{H}_3 \mathrm{O}^{+}=2 \times 10^{-3} \mathrm{M}$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \\
& \therefore\left[\mathrm{OH}^{-}\right]=\frac{\mathrm{K}_{\mathrm{W}}}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}=\frac{1 \times 10^{-14}}{2 \times 10^{-3}}=0.5 \times 10^{-11} \mathrm{M} \\
& 2 \times 10^{-3}>0.5 \times 10^{-11}
\end{aligned}
$
i.e., $\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \gg\left[\mathrm{OH}^{-}\right]$, hence the juice is acidic in nature

Question 3.
Calculate the $\mathrm{pH}$ of $0.001 \mathrm{M} \mathrm{HCI}$ solution
Answer:

$\mathrm{H}_3 \mathrm{O}$ from the auto ionisation of $\mathrm{H}_2 \mathrm{O}\left(10^{-7} \mathrm{M}\right)$ is negligible when compared to the $\mathrm{H}_3 \mathrm{O}$ from $10^{-3} \mathrm{M} \mathrm{HCI}$. Hence $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=0.001 \mathrm{~mol} \mathrm{dm}^{-3}$
$
\begin{aligned}
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\
& =-\log _{10}(0.001) \\
& =-\log _{10}\left(10^{-3}\right)=3
\end{aligned}
$

Question 4.
Calculate $\mathrm{pH}$ of $10^{-7} \mathrm{M} \mathrm{HCI}$
If we do not consider $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$from the ionisation of $\mathrm{H}_2 \mathrm{O}$, then $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=[\mathrm{HCl}]=10^7 \mathrm{M}$ i.e., $\mathrm{pH}=7$, which is a $\mathrm{pH}$ of a neutral solution. We know that $\mathrm{HCI}$ solution is acidic whatever may be the concentration of HCI i.e, the $\mathrm{pH}$ value should be less than 7 . In this case the concentration of the acid is very low $\left(10^{-7} \mathrm{M}\right)$. Hence, the $\mathrm{H}_3 \mathrm{O}^{+}\left(10^{-7} \mathrm{M}\right)$ formed due to the auto ionisation of water cannot be neglected. So, in this case we should consider $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$from ionisation of $\mathrm{H}_2 \mathrm{O}$
Answer:
$
\begin{aligned}
& {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-7}(\text { from } \mathrm{HCl})+10^{-7} \text { (from water) }} \\
& =10^{-7}(1+1) \\
& =2 \times 10 \\
& \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}\right] \\
& =-\log _{10}(2 \times 107)=-\left[\log 2+\log 10^{-7}\right] \\
& =-\log 2-(-7) \cdot \log _{10} \\
& =7-\log 2 \\
& =7-0.3010=6.6990 \\
& =6.70
\end{aligned}
$
Question 5.
A solution of $0.10 \mathrm{M}$ of a weak electrolyte is found to be dissociated to the extent of $1.20 \%$ at $25^{\circ} \mathrm{C}$. Find the dissociation constant of the acid.
Answer:
Given that $\alpha=1.20 \%=\frac{1.20}{100}=1.2 \times 10^{-2}$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}=\left(1.2 \times 10^{-2}\right) 2(0.1) \\
& =1.44 \times 10^{-4} \times 10^{-1}=1.44 \times 10^{-5}
\end{aligned}
$
Question 6.
Calculate the $\mathrm{pH}$ of $0.1 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}$ solution. Dissociation constant of acetic acid is $1.8 \times 10^{-5}$.
Answer:
$
\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]
$

For weak acids,
$
\begin{aligned}
{\left[\mathrm{H}^{+}\right] } & =\sqrt{\mathrm{K}_{\mathrm{a}} \times \mathrm{C}} \\
& =\sqrt{1.8 \times 10^{-5} \times 0.1} \\
& =1.34 \times 10^{-3} \mathrm{M} \quad \mathrm{pH}=-\log \left(1.34 \times 10^{-3}\right) \\
& =3-\log 1.34 \\
& =3-0.1271 \\
& =2.8729 \simeq 2.87
\end{aligned}
$
Question 7.
Find the $\mathrm{pH}$ of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. $\mathrm{K}_{\mathrm{a}}$ for acetic acid is $1.8 \times 10^{-5}$
$
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\mathrm{acid}]}
$
Given that $\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}$
$
\begin{aligned}
& \mathrm{pK}_{\mathrm{a}}=-\log \left(1.8 \times 10^{-5}\right) \\
& =5-\log 1.8 \\
& =5-0.26=4.74 \\
& \mathrm{pH}=4.74+\log \frac{0.20}{0.18} \\
& =4.74+\log \frac{10}{9} \\
& =4.74+\log 10-\log 9 \\
& =4.74+1-0.95 \\
& =5.74-0.95 \\
& =4.79
\end{aligned}
$

Question 8.
What is the $\mathrm{pH}$ of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate and making the volume equal to $500 \mathrm{ml}$. (Given: $\mathrm{K}$ for acetic acid is $8 \times 10$ )
Answer:
According to Henderson - Hessalbalch equation,
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{acid}]}$
Given that $\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}$
$[$ Salt $]=\frac{\text { Number of moles of sodium acetate }}{\text { Volume of the solution ( litre ) }}$
Number of moles of sodium acetate $=\frac{\text { mass of sodium acetate }}{\text { molar mass of sodium acetate }}=\frac{8.2}{8.2}=0.1$
$\therefore[$ Salt $]=\frac{0.1 \text { mole }}{1 / 2 \text { Litre }}=0.2 \mathrm{M}$
[acid] $=\frac{\left.\text { (mass of } \mathrm{CH}_3 \mathrm{COOH} / \text { molar mass of } \mathrm{CH}_3 \mathrm{COOH}\right)}{\text { Volume of solution in litre }}=\frac{(6 / 60)}{1 / 2}=0.2 \mathrm{M}$
$\therefore \mathrm{pH}=4.74+\log \frac{(0.2)}{(0.2)}$
$
\begin{aligned}
& \mathrm{pH}=4.74+\log 1 \\
& \mathrm{pH}=4.74+0=4.74
\end{aligned}
$

Question 9.
Calculate
- the hydrolysis constant,
- degree of hydrolysis and
- $\mathrm{pH}$ of $\mathrm{O} .1 \mathrm{M} \mathrm{CH} 3 \mathrm{COONa}$ solution ( $\mathrm{pK}$ a for $\mathrm{CH}_3 \mathrm{COOH}$ is 4.74).
Answer:
$\mathrm{CH}_3 \mathrm{COONa}$ is a salt of weak acid $\left(\mathrm{CH}_3 \mathrm{COOH}\right)$ and a strong base $(\mathrm{NaOH})$. Hence, the solutions is alkaline due to hydrolysis.
$
\mathrm{CH}_3 \mathrm{COO}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})
$
Give that $\mathrm{pK}_{\mathrm{a}}=4.74$
$
\begin{aligned}
& \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}} \\
& \text { i.e., } \mathrm{K}_{\mathrm{a}}=\text { antilog of }\left(-\mathrm{PK}_{\mathrm{a}}\right) \\
& =\text { antilog of }(-4.74) \\
& =\text { antilog of }(-5+0.26) \\
& 10^{-5} \times 1.8
\end{aligned}
$
$[\operatorname{antilog}$ of $0.26=1.82 \simeq 1.8]$
(i) $\mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{c}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 0.1}}=7.5 \times 10^{-5}$
(ii) $\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{W}}}{\mathrm{K}_{\mathrm{a}}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}=5.56 \times 10^{-10}$
(iii) $\mathrm{pH}=7+\frac{\mathrm{pK}_{\mathrm{a}}}{2}+\frac{\log \mathrm{c}}{2}=7+\frac{4.74}{2}+\frac{\log 0.1}{2}=7+2.37-0.5=8.87$
Question 10.
Establish a relationship between the solubility product and molar solubility for the following
1. $\mathrm{BaSO}_4$
2. $\mathrm{Ag}_2\left(\mathrm{CrO}_4\right)$

Answer:
$
\begin{aligned}
\mathrm{BaSO}_4(\mathrm{~s}) & \stackrel{\mathrm{H}_2 \mathrm{O}}{\rightleftharpoons} \mathrm{Ba}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \\
\mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right] \\
& =(\mathrm{s})(\mathrm{s}) \\
\mathrm{K}_{\mathrm{sp}} & =\mathrm{s}^2 \\
\mathrm{Ag}_2 \mathrm{CrO}_4 & \stackrel{\mathrm{H}_2 \mathrm{O}}{\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] \\
& =(2 \mathrm{~s})^2(\mathrm{~s}) \\
\mathrm{K}_{\mathrm{sp}} & =4 \mathrm{~s}^3
\end{aligned}
$