A solution of $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ is added dropwise to a second solution in which $\left[{\mathrm{Cl}}^{-}\right]=\left[{\mathrm{F}}^{-}\right]=\left[{\mathrm{I}}^{-}\right]=\left[{\mathrm{SO}}_4^{2-}\right]=0.001 \mathrm{M}$. What is the first precipitate that forms?

$\left[{\mathrm{Pb}}^{2+}\right]$ required in each case:

(a) ${\mathrm{PbCl}}_2\Rightarrow \frac{1.5\times {10}^{-5}}{{10}^{-6}}~10$

(b) ${\mathrm{PbF}}_2\Rightarrow \frac{3.7\times {10}^{-8}}{{10}^{-6}}~{10}^{-2}$

(c) ${\mathrm{PbI}}_2\Rightarrow \frac{8.5\times {10}^{-9}}{{10}^{-6}}~{10}^{-3}$

(d) ${\mathrm{PbSO}}_4\Rightarrow \frac{1.8\times {10}^{-8}}{{10}^{-3}}~{10}^{-5}$.

Thus, the correct option is D.