The $\mathrm{K}_{\mathrm{sp}}$ for the following dissociation is $1.6 \times 10^{-5}$ $$\mathrm{PbCl}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})$$ Which of the following choices is correct for a mixture of $300 \mathrm{~mL}$ $0.134 \mathrm{M}$ $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ and $100$ mL $0.4$ M NaCl ?

(B) $\mathrm{Q}>\mathrm{K}_{\mathrm{sp}}$

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Equilibrium

The

K_{sp}

for the following dissociation is

1.6 \times 1 0^{- 5}

PbCl_{2} (s) ⇌ Pb^{2 +} (aq) + 2 Cl^{-} (aq)

Which of the following choices is correct for a mixture of

300 mL

0.134 M

Pb (NO_{3})_{2}

and

100

0.4

M NaCl ?

Answer: (B)

After mixing, the total volume is

400

mL, so the lead ion concentration is

\frac{300 \times 0.134}{400} = 0.1005

M and the chloride is

\frac{100 \times 0.4}{400} = 0.1

M. The reaction quotient is

Q = [P b^{2 +}] [C l^{-}]^{2} = 0.1005 \times (0.1)^{2} \approx 1.0 \times 1 0^{- 3}

, far above

K_{s p} = 1.6 \times 1 0^{- 5}

. Since

Q > K_{s p}

, the mixture is supersaturated and

PbC l_{2}

precipitates.

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option