(Solved): The solubility product (KSP) for BaF2 is 1.7106. When equal volumes of 2.0102MBa(NO ...

The solubility product $\left(K_{\mathrm{SP}}\right)$ for ${\mathrm{BaF}}_2$ is $1.7\times 10^{?6}$. When equal volumes of $2.0\times 10^{?2}\mathrm{M}\mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2$ solution and $2.0\times 10^{?2}\mathrm{M}\mathrm{NaF}$ solution are mixed: $Q_{SP}>K_{SP}$ and ${\mathrm{BaF}}_2$ precipitates $Q_{SP}<K_{SP}$ and no ${\mathrm{BaF}}_2$ precipitates $Q_{SP}>K_{SP}$ and no ${\mathrm{BaF}}_2$ precipitates $Q_{SP}<K_{SP}$ and ${\mathrm{BaF}}_2$ precipitates $Q_{SP}=K_{SP}$ Part H What is the molar solubility of lead iodide, ${\mathrm{Pbl}}_2$, in a $0.10\mathrm{M}\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ solution? $K_{SP}$ for ${\mathrm{Pbl}}_2=8.0\times 10^{?9}$. $\begin{array}{l}1.3\times 10^{?3}\mathrm{M}\\ 1.4\times 10^{?4}\mathrm{M}\\ 8.0\times 10^{?7}\mathrm{M}\\ 8.9\times 10^{?8}\mathrm{M}\\ 4.0\times 10^{?8}\mathrm{M}\end{array}$