JEE Main · 2020 · Shift-IhardIEQ-062

The stoichiometry and solubility product of a salt with the solubility curve given below (solubility = 10-3 M at 298 K)…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The stoichiometry and solubility product of a salt with the solubility curve given below (solubility = 10−310^{-3} M at 298 K) is, respectively: image

Options
  1. a

    X2Y, 2×10−9 M3\mathrm{X_2Y},\ 2 \times 10^{-9}\ \mathrm{M^3}

  2. b

    XY2, 4×10−9 M3\mathrm{XY_2},\ 4 \times 10^{-9}\ \mathrm{M^3}

    ✓
  3. c

    XY2, 1×10−9 M3\mathrm{XY_2},\ 1 \times 10^{-9}\ \mathrm{M^3}

  4. d

    XY, 2×10−6 M3\mathrm{XY},\ 2 \times 10^{-6}\ \mathrm{M^3}

Correct Answerb

XY2, 4×10−9 M3\mathrm{XY_2},\ 4 \times 10^{-9}\ \mathrm{M^3}

Detailed Solution

Step 1 — Identify stoichiometry from graph: The solubility curve shape indicates XY2\mathrm{XY_2} type salt (1:2 dissociation).

Step 2 — Calculate KspK_{sp} for XY2\mathrm{XY_2}: XY2⇌X2++2Y−[X2+]=s=10−3 M,[Y−]=2s=2×10−3 MKsp=s(2s)2=4s3=4×(10−3)3=4×10−9 M3\mathrm{XY_2 \rightleftharpoons X^{2+} + 2Y^-} [\mathrm{X^{2+}}] = s = 10^{-3}\ \mathrm{M}, \quad [\mathrm{Y^-}] = 2s = 2 \times 10^{-3}\ \mathrm{M} K_{sp} = s(2s)^2 = 4s^3 = 4 \times (10^{-3})^3 = 4 \times 10^{-9}\ \mathrm{M^3}

Answer: Option (2) — XY2, 4×10−9 M3\mathrm{XY_2},\ 4 \times 10^{-9}\ \mathrm{M^3}

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