JEE Main · 2021 · Shift-IhardIEQ-060

The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is: [Ksp(AgCN) = 2.2 10-16;\ Ka(HCN) = 6.2…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is:

[Ksp(AgCN)=2.2×10−16; Ka(HCN)=6.2×10−10][K_{sp}(\mathrm{AgCN}) = 2.2 \times 10^{-16};\ K_a(\mathrm{HCN}) = 6.2 \times 10^{-10}]

Options
  1. a

    1.9×10−51.9 \times 10^{-5}

    ✓
  2. b

    2.2×10−162.2 \times 10^{-16}

  3. c

    0.625×10−60.625 \times 10^{-6}

  4. d

    1.6×10−61.6 \times 10^{-6}

Correct Answera

1.9×10−51.9 \times 10^{-5}

Detailed Solution

Step 1 — Dissolution and protonation: AgCN⇌Ag++CN−;Ksp=2.2×10−16CN−+H+⇌HCN;K=1Ka=16.2×10−10\mathrm{AgCN \rightleftharpoons Ag^+ + CN^-}; \quad K_{sp} = 2.2 \times 10^{-16} \mathrm{CN^- + H^+ \rightleftharpoons HCN}; \quad K = \frac{1}{K_a} = \frac{1}{6.2 \times 10^{-10}}

Step 2 — Net reaction: AgCN+H+⇌Ag++HCNKnet=KspKa=2.2×10−166.2×10−10=3.55×10−7\mathrm{AgCN + H^+ \rightleftharpoons Ag^+ + HCN} K_{net} = \frac{K_{sp}}{K_a} = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-10}} = 3.55 \times 10^{-7}

Step 3 — At pH = 3: [H+]=10−3 M[\mathrm{H^+}] = 10^{-3}\ \mathrm{M}

Let solubility = ss: [Ag+]=s[\mathrm{Ag^+}] = s, [HCN]=sKnet=s2[H+]=s210−3=3.55×10−7s2=3.55×10−10s=3.55×10−10≈1.88×10−5≈1.9×10−5 M[\mathrm{HCN}] = s K_{net} = \frac{s^2}{[\mathrm{H^+}]} = \frac{s^2}{10^{-3}} = 3.55 \times 10^{-7} s^2 = 3.55 \times 10^{-10} s = \sqrt{3.55 \times 10^{-10}} \approx 1.88 \times 10^{-5} \approx 1.9 \times 10^{-5}\ \mathrm{M}

Answer: Option (1) — 1.9×10−51.9 \times 10^{-5}

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