Question

a solution is prepared by mixing 50.0 ml of 0.10 m pb(no₃)₂ with 60.0 ml of 1.0 m kcl. calculate the concentrations of pb²⁺ and cl⁻ at equilibrium. ksp for pbcl₂(s) = 1.6×10⁻⁵
pb²⁺ = m
cl⁻ = m
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Explanation:

Step1: Calculate initial moles

Moles of \( \text{Pb}^{2+} \) from \( \text{Pb(NO}_3\text{)}_2 \): \( n_{\text{Pb}^{2+}} = 0.10 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \)
Moles of \( \text{Cl}^- \) from \( \text{KCl} \): \( n_{\text{Cl}^-} = 1.0 \, \text{M} \times 0.060 \, \text{L} = 0.06 \, \text{mol} \)

Step2: Reaction and limiting reactant

Reaction: \( \text{Pb}^{2+} + 2\text{Cl}^-
ightarrow \text{PbCl}_2(s) \)
Moles of \( \text{Cl}^- \) needed for \( \text{Pb}^{2+} \): \( 0.005 \, \text{mol} \times 2 = 0.01 \, \text{mol} \) (less than 0.06 mol, so \( \text{Pb}^{2+} \) is limiting).

Step3: Moles after reaction

Moles of \( \text{Pb}^{2+} \) remaining: \( 0.005 - 0.005 = 0 \, \text{mol} \) (before equilibrium, but we use \( K_{sp} \) later)
Moles of \( \text{Cl}^- \) remaining: \( 0.06 - 0.01 = 0.05 \, \text{mol} \)
Total volume: \( 0.050 + 0.060 = 0.110 \, \text{L} \)
Initial \( [\text{Cl}^-]_{\text{after reaction}} = \frac{0.05 \, \text{mol}}{0.110 \, \text{L}} \approx 0.4545 \, \text{M} \)

Step4: Use \( K_{sp} \) for equilibrium

\( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = 1.6 \times 10^{-5} \)
Let \( [\text{Pb}^{2+}] = x \), then \( [\text{Cl}^-] = 0.4545 + 2x \approx 0.4545 \) (since \( x \) is small)
\( x = \frac{K_{sp}}{[\text{Cl}^-]^2} = \frac{1.6 \times 10^{-5}}{(0.4545)^2} \approx 7.8 \times 10^{-5} \, \text{M} \)
\( [\text{Cl}^-] \approx 0.45 \, \text{M} \) (more accurately, \( 0.4545 \, \text{M} \) from remaining, but adjusted by \( K_{sp} \) gives negligible change, so \( [\text{Cl}^-] \approx 0.45 \, \text{M} \), \( [\text{Pb}^{2+}] \approx 7.8 \times 10^{-5} \, \text{M} \))

Answer:

\( [\text{Pb}^{2+}] \approx \boldsymbol{7.8 \times 10^{-5} \, \text{M}} \), \( [\text{Cl}^-] \approx \boldsymbol{0.45 \, \text{M}} \) (or more precise \( 0.45 \, \text{M} \) and \( 7.8 \times 10^{-5} \, \text{M} \))