JEE Main · 2020 · Shift-IhardIEQ-063

The Ksp for the following dissociation is 1.6 10-5: PbCl2(s) Pb2+(aq) + 2Cl-(aq) Which of the following choices is…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The KspK_{sp} for the following dissociation is 1.6×10−51.6 \times 10^{-5}: PbCl2(s)⇌Pb(aq)2++2Cl(aq)−\mathrm{PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}} Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2\mathrm{Pb(NO_3)_2} and 100 mL 0.4 M NaCl?

Options
  1. a

    Not enough data

  2. b

    Q<KspQ < K_{sp}

  3. c

    Q>KspQ > K_{sp}

    ✓
  4. d

    Q=KspQ = K_{sp} provided

Correct Answerc

Q>KspQ > K_{sp}

Detailed Solution

Step 1 — Concentrations after mixing (total volume = 400 mL): [Pb2+]=0.134×300400=40.2400=0.1005 M[Cl−]=0.4×100400=40400=0.1 M[\mathrm{Pb^{2+}}] = \frac{0.134 \times 300}{400} = \frac{40.2}{400} = 0.1005\ \mathrm{M} [\mathrm{Cl^-}] = \frac{0.4 \times 100}{400} = \frac{40}{400} = 0.1\ \mathrm{M}

Step 2 — Calculate QspQ_{sp}: Qsp=[Pb2+][Cl−]2=0.1005×(0.1)2=0.1005×0.01=1.005×10−3Q_{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2 = 0.1005 \times (0.1)^2 = 0.1005 \times 0.01 = 1.005 \times 10^{-3}

Step 3 — Compare with KspK_{sp}: Qsp=1.005×10−3≫Ksp=1.6×10−5Q>KspQ_{sp} = 1.005 \times 10^{-3} \gg K_{sp} = 1.6 \times 10^{-5} Q > K_{sp}

Precipitation will occur.

Answer: Option (3) — Q>KspQ > K_{sp}

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