Chemistry - Ionic Equilibrium Question with Solution | TestHub

The equilibrium is$\mathrm{P}\mathrm{b}{\mathrm{I}}_2\rightleftharpoons \mathrm{P}{\mathrm{b}}^{2+}\mathrm{}+2{\mathrm{I}}^{-}$
On the basis of this equation, the concentration of$\mathrm{P}{\mathrm{b}}^{2+}$ions will be half of the concentration of${\mathrm{I}}^{-}$ions. Thus,
$\left[{\mathrm{I}}^{-}\right]=2.6\times 10^{-3}\mathrm{M}$and$\left[\mathrm{P}{\mathrm{b}}^{2+}\right]\mathrm{}=1.3\times 10^{-3}\mathrm{M}$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=\left[\mathrm{P}{\mathrm{b}}^{2+}\right][{\mathrm{I}}^{-}{]}^2=[1.3\times 10^{-3}\mathrm{M}\left]\right[2.6\times 10^{-3}\mathrm{M}{]}^2$
${\mathrm{K}}_{\mathrm{s}\mathrm{p}}=8.8\times 10^{-9}.$