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Chemistry · Ionic Equilibrium

JEE Advanced 2018 — Paper 2 — Question 27

Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solutions on [H+]\left[\mathrm{H}^{+}\right]are given in LIST-II.

(Note: Degree of dissociation ( α\alpha ) of weak acid and weak base is ≪1\ll 1; degree of hydrolysis of salt ≪1\ll 1; [H+]\left[\mathrm{H}^{+}\right]represents the concentration of H+\mathrm{H}^{+}ions)

LIST-ILIST-II
P. (10 mL(10 \mathrm{~mL} of 0.1MNaOH+20 mL0.1 \mathrm{M} \mathrm{NaOH}+20 \mathrm{~mL} of 0.1 M acetic acid) diluted to 60 mL1. the value of [H+]\left[\mathrm{H}^{+}\right]does not change on dilution
Q. ( 20 mL of 0.1MNaOH+20 mL0.1 \mathrm{M} \mathrm{NaOH}+20 \mathrm{~mL} of 0.1 M acetic acid) diluted to 80 mL2. the value of [H+]\left[\mathrm{H}^{+}\right]changes to half of its initial value on dilution
R. (20 mL(20 \mathrm{~mL} of 0.1MHCl+20 mL0.1 \mathrm{M} \mathrm{HCl}+20 \mathrm{~mL} of 0.1 M ammonia solution) diluted to 80 mL3. the value of [H+]\left[\mathrm{H}^{+}\right]changes to two times of its initial value on dilution
S. 10 mL saturated solution of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2} in equilibrium with excess solid Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2} is diluted to 20 mL (solid Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2} is still present after dilution).4. the value of [H+]\left[\mathrm{H}^{+}\right]changes to 12\frac{1}{\sqrt{2}} times of its initial value on dilution
5. the value of [H+]\left[\mathrm{H}^{+}\right]changes to 2\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

  1. Option A:

    P→4;Q→2;R→3;S→1\mathbf{P} \rightarrow 4 ; \mathbf{Q} \rightarrow 2 ; \mathbf{R} \rightarrow 3 ; \mathbf{S} \rightarrow 1

  2. Option B:

    P→4;Q→3;R→2;S→3\mathbf{P} \rightarrow 4 ; \mathbf{Q} \rightarrow 3 ; \mathbf{R} \rightarrow 2 ; \mathbf{S} \rightarrow 3

  3. Option C:

    P→1;Q→4;R→5;S→3\mathbf{P} \rightarrow 1 ; \mathbf{Q} \rightarrow 4 ; \mathbf{R} \rightarrow 5 ; \mathbf{S} \rightarrow 3

  4. Option D:

    P→1;Q→5;R→4;S→1\mathbf{P} \rightarrow 1 ; \mathbf{Q} \rightarrow 5 ; \mathbf{R} \rightarrow 4 ; \mathbf{S} \rightarrow 1

    Correct

Answer: D

Step-by-step solution

KwKa=[CH3COO−][OH−][CH3COOH]\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} [OH−]=(KwKa×C)1/2\left[\mathrm{OH}^{-}\right]=\left(\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}} \times \mathrm{C}\right)^{1 / 2} (P)(\mathrm{P}) is a buffer, so [H+]\left[\mathrm{H}^{+}\right]does not change on dilution, as [[salt ]=[]=[acid ]]. (Q) contains only CH3COONa\mathrm{CH}_{3} \mathrm{COONa}

So CH3COO−+H2O⇌CH3COOH+OH−\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{OH}^{-} [OH−]=Kh×C⇒[H+]\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \times \mathrm{C}} \Rightarrow\left[\mathrm{H}^{+}\right]decreases by 2\sqrt{2} times (R)(\mathrm{R}) is also salt hydrolysis So,NH4++H2O⇌NH4OH+H+\mathrm{So}, \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}+\mathrm{H}^{+} [H+]=KwKb×C\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}} \times \mathrm{C}} So, C is made 12\frac{1}{2} so, [H+]\left[\mathrm{H}^{+}\right]becomes 12\frac{1}{\sqrt{2}} (S)(\mathrm{S}) it is a solubility equilibria So dilution does not effect [H+]\left[\mathrm{H}^{+}\right]or [OH−]\left[\mathrm{OH}^{-}\right]

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
pH of solution containing implicit reaction
Dilution processes of different aqueous solutions, with water, are… | JEE Advanced 2018 PYQ with Solution · DhiX AI