JEE Main · 2019 · Shift-IImediumIEQ-068

If Ksp of Ag2CO3 is 8 10-12, the molar solubility of Ag2CO3 in 0.1 M AgNO3 is:

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

If KspK_{sp} of Ag2CO3\mathrm{Ag_2CO_3} is 8×10−128 \times 10^{-12}, the molar solubility of Ag2CO3\mathrm{Ag_2CO_3} in 0.1 M AgNO3\mathrm{AgNO_3} is:

Options
  1. a

    8×10−138 \times 10^{-13} M

  2. b

    8×10−118 \times 10^{-11} M

  3. c

    8×10−128 \times 10^{-12} M

  4. d

    8×10−108 \times 10^{-10} M

    ✓
Correct Answerd

8×10−108 \times 10^{-10} M

Detailed Solution

Step 1 — Dissociation of Ag2CO3\mathrm{Ag_2CO_3}: Ag2CO3⇌2Ag++CO32−Ksp=[Ag+]2[CO32−]=8×10−12\mathrm{Ag_2CO_3 \rightleftharpoons 2Ag^+ + CO_3^{2-}} K_{sp} = [\mathrm{Ag^+}]^2[\mathrm{CO_3^{2-}}] = 8 \times 10^{-12}

Step 2 — Common ion effect ([Ag+]initial=0.1[\mathrm{Ag^+}]_{initial} = 0.1 M from AgNO3\mathrm{AgNO_3}): Let solubility = ss mol/L (assume s≪0.1s \ll 0.1) [Ag+]=0.1+2s≈0.1 M[CO32−]=s[\mathrm{Ag^+}] = 0.1 + 2s \approx 0.1\ \mathrm{M} [\mathrm{CO_3^{2-}}] = s

Step 3 — Solve for s: Ksp=(0.1)2×s=0.01s=8×10−12s=8×10−120.01=8×10−10 MK_{sp} = (0.1)^2 \times s = 0.01s = 8 \times 10^{-12} s = \frac{8 \times 10^{-12}}{0.01} = 8 \times 10^{-10}\ \mathrm{M}

Answer: Option (4) — 8×10−108 \times 10^{-10} M

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