Write the ion-product expressions for

$\left(\text{a}\right)\mathbf{\text{lead(II)iodide;}}\left(\text{b}\right)\mathbf{\text{strontium sulfate;}}\left(\text{c}\right)\mathbf{\text{cadmium sulfide.}}$

The Qsp-ion-product expression is obtained by multiplying the concentrations of ions generated by dissolution of a chemical. When a solution is saturated, the Qsp value is referred to as the Ksp value (solubility-product constant).

${\mathbf{MX}}_{\mathbf{2}}\mathbf{>}\mathbf{>}{\mathbf{M}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{X}}^{\mathbf{-}}$

We do not include the solid and liquid states in the equations

$\mathbf{\text{Ksp =}}\left[{\text{M}}^{\text{2 +}}\right]{\left[{\text{X}}^{\text{-}}\right]}^{\mathbf{\text{2}}}$

Let us obtain the ion-product expression forlead (II) iodide.

The formula for lead (II) iodide is ${\mathrm{PbI}}_2$.

${\mathrm{PbI}}_2\left(\mathrm{s}\right)\iff {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)$

We squared $\left[{\text{I}}^{\text{-}}\right]$because we have 2 mols of in the equation.

Therefore, the ion-product expression for leadlocalid="1663304374574" $\left(\mathrm{II}\right)$iodide is $\text{Ksp =}\left[{\text{Pb}}^{\text{2 +}}\right]{\left[{\text{I}}^{\text{-}}\right]}^{\text{2}}$.

Let us obtain the ion-product expression for strontium sulphate.

The formula for silver cyanide is $SrS{O}_4$.

localid="1663305249992" ${\mathrm{SrSO}}_4\left(\mathrm{s}\right)\iff {\mathrm{Sr}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$

$\text{Ksp =}\left[{\text{Sr}}^{\text{2 +}}\right]\left[{\text{SO}}_{\text{4}}^{\text{2 -}}\right]$

Therefore, the ion-product expression for strontium sulphate islocalid="1663304658017" $\left[{\text{Sr}}^{\text{2 +}}\right]\left[{\text{SO}}_{\text{4}}^{\text{2 -}}\right]$.

Let us obtain the ion-product expression for cadmium sulfide.

The formula for cadmium sulfide is $CdS$.

localid="1663305263200" $$\begin{gathered} \mathrm{CdS}\left(\mathrm{s}\right)\iff {\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right) \\ \mathrm{Ksp}=\left[{\mathrm{Cd}}^{2+}\right]\left[{\mathrm{S}}^{2-}\right] \end{gathered}$$

Therefore, the ion-product expression for cadmium sulfide is $\left[C{d}^{2+}\right]\left[S^{2-}\right]$.