JEE Challenger
More from Equilibrium
ChemHardAdv 2018 • Paper 2

Effect of Dilution on Hydrogen Ion Concentration in Various Aqueous Systems

Dilution processes for various aqueous solutions using water are described in LIST-I. The corresponding impact on [H+][\text{H}^+] concentration is provided in LIST-II.

(Note: The degree of dissociation (α)(\alpha) for weak acid and weak base is ≪1\ll 1; degree of hydrolysis of salt is ≪1\ll 1; [H+][\text{H}^+] denotes the concentration of hydrogen ions)

LIST-ILIST-IIP. (10 mL of 0.1 M NaOH +20 mL of 0.1 M acetic acid) diluted to 60 mL1. the value of [H+] does not change on dilutionQ. (20 mL of 0.1 M NaOH +20 mL of 0.1 M acetic acid) diluted to 80 mL2. the value of [H+] changes to half of its initial value on dilutionR. (20 mL of 0.1 M HCl +20 mL of 0.1 M ammonia solution) diluted to 80 mL3. the value of [H+] changes to two times of its initial value on dilutionS. 10 mL saturated solution of Ni(OH)2 in equilibrium with excess solid Ni(OH)24. the value of [H+] changes to 12 times of its initial value on dilutionis diluted to 20 mL (solid Ni(OH)2 is still present after dilution).5. the value of [H+] changes to 2 times of its initial value on dilution\begin{array}{l|l} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{P. } (10\text{ mL of } 0.1\text{ M NaOH } + 20\text{ mL of } 0.1\text{ M acetic acid) diluted to } 60\text{ mL} & \text{1. the value of } [\text{H}^+] \text{ does not change on dilution} \\ \text{Q. } (20\text{ mL of } 0.1\text{ M NaOH } + 20\text{ mL of } 0.1\text{ M acetic acid) diluted to } 80\text{ mL} & \text{2. the value of } [\text{H}^+] \text{ changes to half of its initial value on dilution} \\ \text{R. } (20\text{ mL of } 0.1\text{ M HCl } + 20\text{ mL of } 0.1\text{ M ammonia solution) diluted to } 80\text{ mL} & \text{3. the value of } [\text{H}^+] \text{ changes to two times of its initial value on dilution} \\ \text{S. } 10\text{ mL saturated solution of } \text{Ni(OH)}_2 \text{ in equilibrium with excess solid } \text{Ni(OH)}_2 & \text{4. the value of } [\text{H}^+] \text{ changes to } \frac{1}{\sqrt{2}} \text{ times of its initial value on dilution} \\ \quad \text{is diluted to } 20\text{ mL (solid } \text{Ni(OH)}_2 \text{ is still present after dilution).} & \text{5. the value of } [\text{H}^+] \text{ changes to } \sqrt{2} \text{ times of its initial value on dilution} \\ \hline \end{array}

Select the correct matching combination option below:

Options

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

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

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

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

Topics & Concepts

EquilibriumAcid-Base EquilibriumSolubility Product

Step-by-Step Solution

To determine the correct matching combination between LIST-I and LIST-II, we analyze each dilution process step-by-step:

Analysis of Process P:

  • Initial Components: Moles of NaOH=10 mL×0.1 M=1 mmol\text{Moles of NaOH} = 10 \text{ mL} \times 0.1 \text{ M} = 1 \text{ mmol} Moles of CH3COOH=20 mL×0.1 M=2 mmol\text{Moles of CH}_3\text{COOH} = 20 \text{ mL} \times 0.1 \text{ M} = 2 \text{ mmol}
  • Reaction: NaOH+CH3COOH→CH3COONa+H2O\text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}
  • Post-Reaction Composition: Moles of CH3COONa=1 mmol\text{Moles of CH}_3\text{COONa} = 1 \text{ mmol} Moles of CH3COOH remaining=2−1=1 mmol\text{Moles of CH}_3\text{COOH remaining} = 2 - 1 = 1 \text{ mmol}
  • System Type: This mixture contains a weak acid (CH3COOH)(\text{CH}_3\text{COOH}) and its conjugate salt (CH3COONa)(\text{CH}_3\text{COONa}), forming an acidic buffer solution.
  • Effect of Dilution: The [H+][\text{H}^+] concentration for an acidic buffer solution is given by: [H+]=Ka⋅[acid][salt]=Ka⋅nacidnsalt[\text{H}^+] = K_a \cdot \frac{[\text{acid}]}{[\text{salt}]} = K_a \cdot \frac{n_{\text{acid}}}{n_{\text{salt}}} Since dilution changes the total volume but keeps the ratio of moles nacidnsalt\frac{n_{\text{acid}}}{n_{\text{salt}}} constant, the concentration [H+][\text{H}^+] does not change on dilution.

P→1\mathbf{P} \rightarrow 1

Analysis of Process Q:

  • Initial Components: Moles of NaOH=20 mL×0.1 M=2 mmol\text{Moles of NaOH} = 20 \text{ mL} \times 0.1 \text{ M} = 2 \text{ mmol} Moles of CH3COOH=20 mL×0.1 M=2 mmol\text{Moles of CH}_3\text{COOH} = 20 \text{ mL} \times 0.1 \text{ M} = 2 \text{ mmol}
  • Reaction: Complete neutralization occurs, forming 2 mmol2 \text{ mmol} of CH3COONa\text{CH}_3\text{COONa} (salt of a weak acid and strong base).
  • Concentration before further dilution: V1=20 mL+20 mL=40 mL  ⟹  C1=2 mmol40 mL=0.05 MV_1 = 20 \text{ mL} + 20 \text{ mL} = 40 \text{ mL} \implies C_1 = \frac{2 \text{ mmol}}{40 \text{ mL}} = 0.05 \text{ M}
  • Concentration after dilution: V2=80 mL  ⟹  C2=2 mmol80 mL=C12V_2 = 80 \text{ mL} \implies C_2 = \frac{2 \text{ mmol}}{80 \text{ mL}} = \frac{C_1}{2}
  • Effect of Dilution: For the salt of a weak acid and strong base, the [H+][\text{H}^+] concentration is given by: [H+]=Kw⋅KaC[\text{H}^+] = \sqrt{\frac{K_w \cdot K_a}{C}} Therefore, the initial concentration of hydrogen ions is: [H+]i=Kw⋅KaC1[\text{H}^+]_i = \sqrt{\frac{K_w \cdot K_a}{C_1}} Upon dilution, the final concentration becomes: [H+]f=Kw⋅KaC2=Kw⋅KaC1/2=2⋅[H+]i[\text{H}^+]_f = \sqrt{\frac{K_w \cdot K_a}{C_2}} = \sqrt{\frac{K_w \cdot K_a}{C_1 / 2}} = \sqrt{2} \cdot [\text{H}^+]_i Thus, [H+][\text{H}^+] changes to 2\sqrt{2} times of its initial value.

Q→5\mathbf{Q} \rightarrow 5

Analysis of Process R:

  • Initial Components: Moles of HCl=20 mL×0.1 M=2 mmol\text{Moles of HCl} = 20 \text{ mL} \times 0.1 \text{ M} = 2 \text{ mmol} Moles of NH3=20 mL×0.1 M=2 mmol\text{Moles of NH}_3 = 20 \text{ mL} \times 0.1 \text{ M} = 2 \text{ mmol}
  • Reaction: Complete neutralization occurs, forming 2 mmol2 \text{ mmol} of NH4Cl\text{NH}_4\text{Cl} (salt of a strong acid and weak base).
  • Concentration before further dilution: V1=20 mL+20 mL=40 mL  ⟹  C1=2 mmol40 mL=0.05 MV_1 = 20 \text{ mL} + 20 \text{ mL} = 40 \text{ mL} \implies C_1 = \frac{2 \text{ mmol}}{40 \text{ mL}} = 0.05 \text{ M}
  • Concentration after dilution: V2=80 mL  ⟹  C2=2 mmol80 mL=C12V_2 = 80 \text{ mL} \implies C_2 = \frac{2 \text{ mmol}}{80 \text{ mL}} = \frac{C_1}{2}
  • Effect of Dilution: For the salt of a strong acid and weak base, cationic hydrolysis occurs, and [H+][\text{H}^+] is given by: [H+]=Kw⋅CKb[\text{H}^+] = \sqrt{\frac{K_w \cdot C}{K_b}} Therefore, the initial concentration of hydrogen ions is: [H+]i=Kw⋅C1Kb[\text{H}^+]_i = \sqrt{\frac{K_w \cdot C_1}{K_b}} Upon dilution, the final concentration becomes: [H+]f=Kw⋅C2Kb=Kw⋅C1/2Kb=12⋅[H+]i[\text{H}^+]_f = \sqrt{\frac{K_w \cdot C_2}{K_b}} = \sqrt{\frac{K_w \cdot C_1 / 2}{K_b}} = \frac{1}{\sqrt{2}} \cdot [\text{H}^+]_i Thus, [H+][\text{H}^+] changes to 12\frac{1}{\sqrt{2}} times of its initial value.

R→4\mathbf{R} \rightarrow 4

Analysis of Process S:

  • System: A saturated solution of Ni(OH)2\text{Ni(OH)}_2 in equilibrium with solid Ni(OH)2\text{Ni(OH)}_2: Ni(OH)2(s)⇌Ni2+(aq)+2OH−(aq)\text{Ni(OH)}_2(s) \rightleftharpoons \text{Ni}^{2+}(aq) + 2\text{OH}^-(aq)
  • Effect of Dilution: The solubility product equation is: Ksp=[Ni2+][OH−]2K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 Since excess solid Ni(OH)2\text{Ni(OH)}_2 remains present after dilution, adding water dissolves more solid to maintain the equilibrium solubility state. Hence, the equilibrium concentrations [OH−][\text{OH}^-] and [Ni2+][\text{Ni}^{2+}] remain unchanged. Since [OH−][\text{OH}^-] is constant, [H+]=Kw[OH−][\text{H}^+] = \frac{K_w}{[\text{OH}^-]} does not change on dilution.

S→1\mathbf{S} \rightarrow 1

Conclusion:

The correct match is: P→1,Q→5,R→4,S→1\mathbf{P} \rightarrow 1, \quad \mathbf{Q} \rightarrow 5, \quad \mathbf{R} \rightarrow 4, \quad \mathbf{S} \rightarrow 1

This matches Option D.

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