What is the $\left[{\mathrm{OH}}^{-}\right]$ in the final solution prepared by mixing $20.0 \mathrm{mL}$ of $0.050 \mathrm{M} \mathrm{HCl}$ with $30.0 \mathrm{mL}$ of $0.10$ $\mathrm{M} \mathrm{Ba}(\mathrm{OH}{)}_2$? 

In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains ${\mathrm{Ag}}^{+}$ and ${\mathrm{Pb}}^{2+}$ at a concentration of $0.10\mathrm{M}.$ Aqueous $\mathrm{HCl}$ is added to this solution until the ${\mathrm{Cl}}^{-}$ concentration is $0.10\mathrm{M}$. What will the concentrations of ${\mathrm{Ag}}^{+}$ and ${\mathrm{Pb}}^{2+}$ be at equilibrium? $\left({\mathrm{K}}_{\mathrm{sp}}\text{ for }\mathrm{AgCl}=1.8\times {10}^{-10}\right.$,$\left.{\mathrm{K}}_{sp}^{}\text{ for }{\mathrm{PbCl}}_2=1.7\times {10}^{-5}\right)$