1. Write the ionic reaction and expression for $Q$ and $K_{sp}$

For $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$ we have $K_{sp} = [Pb^{2+}][Cl^-]^2$

For the given mixture, we first calculate the ion concentrations after mixing, then compute the ionic product $Q = [Pb^{2+}][Cl^-]^2$

If $Q>K_{sp}$, precipitation occurs.

2. Calculate moles before mixing

• From $300\,\text{mL}$ of $0.134\,\text{M}\; Pb(NO_3)_2$: $n(Pb^{2+}) = 0.300 \times 0.134 = 0.0402\;\text{mol}$

• From $100\,\text{mL}$ of $0.4\,\text{M}\; NaCl$: $n(Cl^-) = 0.100 \times 0.4 = 0.0400\;\text{mol}$

3. Find total volume after mixing

$V_{\text{total}} = 300\,\text{mL} + 100\,\text{mL} = 400\,\text{mL} = 0.400\,\text{L}$

4. Calculate initial concentrations in the mixture

$[Pb^{2+}] = \frac{0.0402}{0.400} = 0.1005\,\text{M}$

$[Cl^-] = \frac{0.0400}{0.400} = 0.100\,\text{M}$

5. Calculate the reaction quotient

$Q = [Pb^{2+}][Cl^-]^2 = (0.1005)(0.100)^2$

$Q = 0.1005 \times 0.01 = 1.005 \times 10^{-3}$

6. Compare with $K_{sp}$

Given: $K_{sp} = 1.6 \times 10^{-5}$

Now, $Q = 1.005 \times 10^{-3}$ which is much larger than $K_{sp}$.

Therefore, $Q > K_{sp}$

So the solution is supersaturated and $PbCl_2$ will precipitate.

7. Evaluate options

• A: $Q > K_{sp}$ → Correct
• B: Not enough data provided → Incorrect
• C: $Q < K_{sp}$ → Incorrect
• D: $Q = K_{sp}$ → Incorrect

8. Comparison with stored correct answer

Stored correct answer: A

My derived answer is also A, so they agree.