Chemical And Ionic Equilibrium Result Question 41
41. Dilution processes of different aqueous solutions, with water, are given in List-I. The effects of dilution of the solution on $[H ^+]$ are given in List-II.
Note: Degree of dissociation $( \alpha)$ of weak acid and weak base is $< < 1$ ; degree of hydrolysis of salt $< < 1 ;[H^+]$ represents the concentration of $\mathrm{H}^+$ ions
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) $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 1$
Show Answer
Answer:
Correct Answer: 41. ( d )
Solution:
- For $P$, i.e. $(10 \mathrm{~mL}$ of $0.1 $ $\mathrm{M}$ $ \mathrm{NaOH}+20 \mathrm{~mL}$ of $0.1$ M acetic acid) is diluted to $60$ 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.
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 $0.1$ M $\mathrm{CH}_3 \mathrm{COOH}$ ) is diluted to $80$ mL
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 $40$ mL solution. Hence, it is the salt of weak acid and strong base. So,
$ \left[\mathrm{H}^{+}\right]_{\text {initial }}=\sqrt{\dfrac{K_W K_a}{C}} $
After dilution to $80$ mL , the new ’ $C$ ’ becomes $\dfrac{C}{2}$, So,
$ \left[\mathrm{H}^{+}\right]{\text {new }}=\sqrt{\dfrac{K_w K_a}{C / 2}} \text { or }\left[\mathrm{H}^{+}\right]{\text {initial }} \times \sqrt{2} $
For $\boldsymbol{R}$, i.e. $(\mathbf{2 0} \mathbf{~ m L}$ of $0.1 $ $\mathrm{M}$ $ \mathrm{HCl}+20 \mathrm{~mL}$ of $0.1 $ $\mathrm{M}$ $ \mathrm{NH}_3)$ is diluted to $80$ mL
The correct match is $4$ , i.e. the value of $\left[\mathrm{H}^{+}\right]$ changes to $\dfrac{1}{\sqrt{2}}$ 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 $40$ mL of solution. Hence, a salt of strong acid and weak base is formed.
For this,
$ \left[\mathrm{H}^{+}\right]_{\text {initial }}=\sqrt{\dfrac{K_w \times C}{K_b}} $
Now on dilution upto $80$ mL new conc. becomes $C / 2$.
So,
$ \left[\mathrm{H}^{+}\right]_{\text {new }}=\sqrt{\dfrac{K_w \times \dfrac{C}{2}}{K_b}} $
or
$ \left[\mathrm{H}^{+}\right]{\text {new }}=\left[\mathrm{H}^{+}\right]{\text {linitial }} \times \dfrac{1}{\sqrt{2}} $
For $\boldsymbol{S}$, i.e. $10$ mL saturated solution of $\mathrm{Ni}(\mathrm{OH})_2$ in equilibrium with excess solid $\mathrm{Ni}(\mathrm{OH})_2$ is diluted to $20$ 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 }} $
BETA
