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Ionic Equilibrium question

2018 · Shift 2 · Q15
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  5. /2018 · Shift 2 · Q15

Ionic Equilibrium question

2018 · Shift 2 · Q15

JEE AdvancedChemistryIonic EquilibriumMCQ+3 / −1
Dilution processes of different aqueous solutions, with water, are given in LIST - I. The effects of dilution of the solutions on [H+]\left[ {{H^ + }} \right][H+] are given in LIST - II

(Note: Degree of dissociation (a) of weak acid and weak base is <<1;\lt \lt 1;<<1; degree of hydrolysis of salt <<1;\lt \lt 1;<<1;[H+]\left[ {{H^ + }} \right][H+] represents the concentration of H+{H^ + }H+ ions)

LIST-I LIST-II
P. (10 mL of 0.1 M NaOHNaOHNaOH + 20 mL of
0.1 M acetic acid) diluted to 60 mL		 1. the value of [H+] does not change
on dilution
Q. (20 mL of 0.1 M NaOHNaOHNaOH + 20 mL of
0.1 M acetic acid) diluted to 80 mL		 2. the value of [H+] changes to half
of its initial value on dilution
R. (20 mL of 0.1 M HCL + 20 mL of
0.1 M ammonia solution) diluted to
80 mL 		3. the value of [H+] changes to two
times of its initial value on dilution
S. 10 mL saturated solution of Ni(OH)2Ni(OH)_2Ni(OH)2​
in equilibrium with excess solid
Ni(OH)2Ni(OH)_2Ni(OH)2​ is diluted to 20 mL (solid
Ni(OH)2Ni(OH)_2Ni(OH)2​ is still present after dilution).		 4. the value of [H+] changes to 12{1 \over {\sqrt 2 }}2​1​
times of its initial value on dilution
5. the value of [H+] changes to 2\sqrt 22​
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. A
    P−4;Q−2;R−3;S−1P - 4;Q - 2;R - 3;S - 1P−4;Q−2;R−3;S−1
  2. B
    P−4;Q−3;R−2;S−3P - 4;Q - 3;R - 2;S - 3P−4;Q−3;R−2;S−3
  3. C
    P−1;Q−4;R−5;S−3P - 1;Q - 4;R - 5;S - 3P−1;Q−4;R−5;S−3
  4. D
    P−1;Q−5;R−4;S−1P - 1;Q - 5;R - 4;S - 1P−1;Q−5;R−4;S−1
View written solutionFree

Correct answer: D

  1. Process P: 10 mL of 0.1 M NaOH+20 mL of 0.1 M10\,\text{mL of }0.1\,M\,NaOH + 20\,\text{mL of }0.1\,M10mL of 0.1MNaOH+20mL of 0.1M acetic acid, then diluted to 60 mL60\,\text{mL}60mL
  • Moles of NaOHNaOHNaOH: 0.1×0.010=0.001 mol0.1\times 0.010 = 0.001\,\text{mol}0.1×0.010=0.001mol
  • Moles of acetic acid: 0.1×0.020=0.002 mol0.1\times 0.020 = 0.002\,\text{mol}0.1×0.020=0.002mol

Reaction: CH3COOH+OH−→CH3COO−+H2OCH_3COOH + OH^- \to CH_3COO^- + H_2OCH3​COOH+OH−→CH3​COO−+H2​O

After reaction:

  • Acetic acid left =0.002−0.001=0.001= 0.002 - 0.001 = 0.001=0.002−0.001=0.001 mol
  • Acetate formed =0.001= 0.001=0.001 mol

So we have a buffer with [CH3COOH]=[CH3COO−][CH_3COOH] = [CH_3COO^-][CH3​COOH]=[CH3​COO−] Hence, [H+]=Ka[H^+] = K_a[H+]=Ka​ which is independent of dilution since the ratio acid/salt remains unchanged.

So for P: [H+] does not change[H^+]\text{ does not change}[H+] does not change Thus, P→1P \to 1P→1

  1. Process Q: 20 mL of 0.1 M NaOH+20 mL of 0.1 M20\,\text{mL of }0.1\,M\,NaOH + 20\,\text{mL of }0.1\,M20mL of 0.1MNaOH+20mL of 0.1M acetic acid, then diluted to 80 mL80\,\text{mL}80mL
  • Moles of NaOHNaOHNaOH: 0.1×0.020=0.002 mol0.1\times 0.020 = 0.002\,\text{mol}0.1×0.020=0.002mol
  • Moles of acetic acid: 0.1×0.020=0.002 mol0.1\times 0.020 = 0.002\,\text{mol}0.1×0.020=0.002mol

Complete neutralization occurs. Final solution contains only sodium acetate.

Initial volume after mixing =40 mL= 40\,\text{mL}=40mL, after dilution =80 mL= 80\,\text{mL}=80mL. So concentration of acetate becomes half.

For salt of weak acid and strong base: [OH−]=KwCKa[OH^-] = \sqrt{\frac{K_w C}{K_a}}[OH−]=Ka​Kw​C​​ Hence, [OH−]∝C[OH^-] \propto \sqrt{C}[OH−]∝C​ If concentration is halved, then [OH−]final=12[OH−]initial[OH^-]_{final} = \frac{1}{\sqrt 2}[OH^-]_{initial}[OH−]final​=2​1​[OH−]initial​ Now, [H+]=Kw[OH−][H^+] = \frac{K_w}{[OH^-]}[H+]=[OH−]Kw​​ so [H+]final=2 [H+]initial[H^+]_{final} = \sqrt 2\,[H^+]_{initial}[H+]final​=2​[H+]initial​

Thus, Q→5Q \to 5Q→5

  1. Process R: 20 mL of 0.1 M HCl+20 mL of 0.1 M20\,\text{mL of }0.1\,M\,HCl + 20\,\text{mL of }0.1\,M20mL of 0.1MHCl+20mL of 0.1M ammonia, then diluted to 80 mL80\,\text{mL}80mL
  • Moles of HClHClHCl: 0.1×0.020=0.002 mol0.1\times 0.020 = 0.002\,\text{mol}0.1×0.020=0.002mol
  • Moles of NH3NH_3NH3​: 0.1×0.020=0.002 mol0.1\times 0.020 = 0.002\,\text{mol}0.1×0.020=0.002mol

Complete reaction: NH3+HCl→NH4ClNH_3 + HCl \to NH_4ClNH3​+HCl→NH4​Cl

Only NH4ClNH_4ClNH4​Cl remains. This is a salt of weak base and strong acid.

For such a salt: [H+]=KwCKb[H^+] = \sqrt{\frac{K_w C}{K_b}}[H+]=Kb​Kw​C​​ Thus, [H+]∝C[H^+] \propto \sqrt{C}[H+]∝C​ On dilution from 404040 mL to 808080 mL, concentration becomes half. So, [H+]final=12[H+]initial[H^+]_{final} = \frac{1}{\sqrt 2}[H^+]_{initial}[H+]final​=2​1​[H+]initial​

Thus, R→4R \to 4R→4

  1. Process S: Saturated solution of Ni(OH)2Ni(OH)_2Ni(OH)2​ with excess solid present, diluted from 101010 mL to 202020 mL, solid still remains

Since excess solid is present both before and after dilution, the solution remains saturated. Thus solubility equilibrium re-establishes and ionic concentrations remain the same.

For Ni(OH)2(s)⇌Ni2++2OH−Ni(OH)_2(s) \rightleftharpoons Ni^{2+} + 2OH^-Ni(OH)2​(s)⇌Ni2++2OH− with solid present, saturated solution has fixed [OH−][OH^-][OH−] at given temperature. Hence [OH−][OH^-][OH−] does not change on dilution, so [H+]=Kw[OH−][H^+] = \frac{K_w}{[OH^-]}[H+]=[OH−]Kw​​ also does not change.

Thus, S→1S \to 1S→1

  1. Final matching

P→1,Q→5,R→4,S→1P\to 1,\quad Q\to 5,\quad R\to 4,\quad S\to 1P→1,Q→5,R→4,S→1

This matches Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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