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

We are given the heterogeneous equilibrium

$$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2\,Cl^{-}_{(aq)}, \qquad K_{sp}=1.6\times10^{-5}$$

To predict whether a precipitate will form, we first calculate the reaction quotient $$Q$$ for the ions in the mixture. The definition is

$$Q=[Pb^{2+}]\,[Cl^{-}]^{2}\,,$$

evaluated with the actual (initial) ionic concentrations just after mixing but before any precipitation.

We mix two solutions:

• Volume of $$Pb(NO_3)_2$$ solution: $$V_1 = 300{\rm \,mL}=0.300{\rm \,L}$$ with $$M_1 = 0.134{\rm \,M}$$.
• Volume of NaCl solution: $$V_2 = 100{\rm \,mL}=0.100{\rm \,L}$$ with $$M_2 = 0.400{\rm \,M}$$.

The total volume after mixing is

$$V_{\text{tot}} = V_1 + V_2 = 0.300{\rm \,L}+0.100{\rm \,L}=0.400{\rm \,L}.$$

Now we work out the number of moles of each ion contributed by the two solutions.

Moles of $$Pb^{2+}$$ coming from $$Pb(NO_3)_2$$:

$$n_{Pb^{2+}} = M_1 \times V_1 = 0.134\,\text{mol\,L}^{-1}\times0.300\,\text{L}=0.0402\;\text{mol}.$$

Moles of $$Cl^{-}$$ coming from NaCl (each formula unit gives one chloride ion):

$$n_{Cl^{-}} = M_2 \times V_2 = 0.400\,\text{mol\,L}^{-1}\times0.100\,\text{L}=0.0400\;\text{mol}.$$

The initial concentrations after dilution are obtained by dividing these mole amounts by the total volume $$V_{\text{tot}}=0.400\text{ L}$$.

So we get

$$[Pb^{2+}]_{\text{initial}} = \frac{0.0402\;\text{mol}}{0.400\;\text{L}} = 0.1005\;\text{M},$$

$$[Cl^{-}]_{\text{initial}} = \frac{0.0400\;\text{mol}}{0.400\;\text{L}} = 0.1000\;\text{M}.$$

Substituting these values into the definition of $$Q$$, we have

$$Q = [Pb^{2+}]_{\text{initial}}\,[Cl^{-}]_{\text{initial}}^{2} = 0.1005\,\times\,(0.1000)^{2}.$$

Now, $$(0.1000)^{2}=0.01000$$, and multiplying,

$$Q = 0.1005 \times 0.01000 = 0.001005.$$

To compare easily with $$K_{sp}=1.6\times10^{-5}$$, we convert $$Q$$ into scientific notation:

$$Q = 1.005\times10^{-3}.$$

Clearly,

$$Q = 1.005\times10^{-3} \; \gt \; K_{sp} = 1.6\times10^{-5}.$$

Because $$Q \gt K_{sp}$$, the ionic product exceeds the solubility product, so the solution is supersaturated with respect to $$PbCl_2$$ and precipitation will occur.

Hence, the correct answer is Option C.