Write the ion-product expressions forsilver carbonate;

barium fluoride; copper (II)sulphide.

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

role="math" localid="1663273315157" ${\mathbf{\text{MX}}}_{\mathbf{2}}\mathbf{>}\mathbf{>}{\mathbf{M}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{X}}^{\mathbf{-}}$

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 for barium fluoride.

The formula for barium fluoride is$Ba{F}_2$.

$Ba{F}_2\left(s\right)\iff B{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)$

$\text{Ksp =}\left[{\text{Ba}}^{\text{2 +}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}$

We squared localid="1663276813279" $\left[{\text{F}}^{-}\right]$because we have two mols of ${\text{F}}^{-}$.

Therefore, the ion-product expression for barium fluoride is $\text{Ksp =}\left[{\text{Ba}}^{\text{2 +}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}$

Let us obtain the ion-product expression for copper $\left(\mathrm{II}\right)$sulphide.

The formula for copper $\left(\mathrm{II}\right)$sulphide is $\mathrm{CuS}$.

$CuS\left(s\right)\iff C{u}^{2+}\left(aq\right)+S^{2-}\left(aq\right)$

width="116" height="18" role="math">Ksp =Cu2 +S2 -

Therefore, the ion-product expression for copper$\left(\mathrm{II}\right)$sulphide is role="math" localid="1663276004217" $\text{Ksp =}\left[{\text{Cu}}^{\text{2 +}}\right]\left[{\text{S}}^{\text{2 -}}\right]$.