Chemical and Ionic Equilibrium 2 Question 41
41. Dilution processes of different aqueous solutions, with water, are given in List-I. The effects of dilution of the solution on $\left[\mathrm{H}^{+}\right]$are given in List-II.
Note Degree of dissociation $(\alpha)$ of weak acid and weak base is $«1$; degree of hydrolysis of salt $«1 ;\left[\mathrm{H}^{+}\right]$represents the concentration of $\mathrm{H}^{+}$ions
| List-I | List-II | |
|---|---|---|
| P. | $(10 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}+$ $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ acetic acid) diluted to $60 \mathrm{~mL}$ |
1. the value of $\left[\mathrm{H}^{+}\right]$does not change on dilution |
| Q. | $(20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}+$ $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ acetic acid) diluted to $80 \mathrm{~mL}$ |
2. the value of $\left[\mathrm{H}^{+}\right]$changes to half of its initial value on dilution |
| R. | $(20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}+20$ $\mathrm{mL}$ of $0.1 \mathrm{M}$ ammonia solution) diluted to $80 \mathrm{~mL}$ |
3. the value of $\left[\mathrm{H}^{+}\right]$changes to two times of its initial value on dilution. |
| S. | $10 \mathrm{~mL}$ saturated solution of $\mathrm{Ni}(\mathrm{OH}){2}$ in equilibrium with exces solid $\mathrm{Ni}(\mathrm{OH}){2}$ is diluted to $20 \mathrm{~mL}$ (solid $\mathrm{Ni}(\mathrm{OH})_{2}$ is still present after dilution). |
4. the value of $\left[\mathrm{H}^{+}\right]$changes to $\frac{1}{\sqrt{2}}$ times of its initial value on dilution |
| 5. the value of $\left[\mathrm{H}^{+}\right]$changes to $\sqrt{2}$ times of its initial value on dilution |
Match each process given in List-I with one or more effect(s) in List-II. The correct option is
(2018 Adv.)
(a) $\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 1$
(b) $\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 3$
(c) $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 3$
(d) $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 1$
Fill in the Blanks
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Solution:
- For $P$, i.e. $(10 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}+20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ acetic acid) is diluted to $60 \mathrm{~mL}$
The correct match is 1 , i.e. the value of $\left[\mathrm{H}^{+}\right]$does not change on dilution due to the formation of following buffer.
$$ \mathrm{NaOH}+\mathrm{CH}{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}{3} \mathrm{COO}^{-} \mathrm{Na}^{+}+\mathrm{H}_{2} \mathrm{O} $$
Initial millimol $\quad 1 \quad 2$
Final millimol $\quad 1 \quad 1$
Final volume $-30 \mathrm{~mL}(20+10)$ in which millimoles of $\mathrm{CH}{3} \mathrm{COOH}$ and $\mathrm{CH}{3} \mathrm{COO}^{-} \mathrm{Na}^{+}$are counted.
For $Q$, i.e. $(20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}+20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{CH}_{3} \mathrm{COOH}$ ) is diluted to $80 \mathbf{~ m L}$
The correct match is 5 , i.e. the value of $\left[\mathrm{H}^{+}\right]$changes to $\sqrt{2}$ times of its initial value on dilution.
As per the condition given in $Q$ the resultant solution before dilution contain 2 millimoles of $\mathrm{CH}_{3} \mathrm{COO}^{-} \mathrm{Na}^{+}$in $40 \mathrm{~mL}$ solution. Hence, it is the salt of weak acid and strong base. So,
$$ \left[\mathrm{H}^{+}\right]{\text {initial }}=\sqrt{\frac{K{W} K_{a}}{C}} $$
After dilution to $80 \mathrm{~mL}$, the new ’ $C$ ’ becomes $\frac{C}{2}$, So,
$$ \left[\mathrm{H}^{+}\right]{\text {new }}=\sqrt{\frac{K{w} K_{a}}{C / 2}} \text { or }\left[\mathrm{H}^{+}\right]_{\text {initial }} \times \sqrt{2} $$
For $R$, i.e. $\left(20 \mathrm{~mL}\right.$ of $0.1 \mathrm{M} \mathrm{HCl}+20 \mathrm{~mL}$ of $\left.0.1 \mathrm{M} \mathrm{NH}_{3}\right)$ is diluted to $80 \mathrm{~mL}$
The correct match is 4 , i.e. the value of $\left[\mathrm{H}^{+}\right]$changes to $\frac{1}{\sqrt{2}}$ times of its initial value of dilution.
As per the condition given in $R$ the resultant solution before dilution contains 2 millimoles of $\mathrm{NH}_{4} \mathrm{Cl}$ in $40 \mathrm{~mL}$ of solution. Hence, a salt of strong acid and weak base is formed.
For this,
$$ \left[\mathrm{H}^{+}\right]{\text {initial }}=\sqrt{\frac{K{w} \times C}{K_{b}}} $$
Now on dilution upto $80 \mathrm{~mL}$ new conc. becomes $C / 2$.
$$ \begin{array}{ll} \text { So, } & {\left[\mathrm{H}^{+}\right]{\text {new }}=\sqrt{\frac{K{w} \times \frac{C}{2}}{K_{b}}}} \ \text { or } & {\left[\mathrm{H}^{+}\right]{\text {new }}=\left[\mathrm{H}^{+}\right]{\text {initial }} \times \frac{1}{\sqrt{2}}} \end{array} $$
For $\boldsymbol{S}$, i.e. $10 \mathbf{~ m L}$ saturated solution of $\mathrm{Ni}(\mathrm{OH}){2}$ in equilibrium with excess solid $\mathrm{Ni}(\mathrm{OH}){2}$ is diluted to $20 \mathrm{~mL}$ and solid $\mathrm{Ni}(\mathrm{OH})_{2}$ is still present after dilution.
The correct match is 1 .
$$ \mathrm{Ni}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Ni}^{2+}+2 \mathrm{OH}^{-} $$
as per the condition given it is a sparingly soluble salt. Hence, on dilution the concentration of $\mathrm{OH}^{-}$ions remains constant in saturated solution.
So for this solution,
$$ \left[\mathrm{H}^{+}\right]{\text {new }}=\left[\mathrm{H}^{+}\right]{\text {initial }} $$
