Write the ion-product expressions for

$\mathbf{\left(}\mathbf{\text{a}}\mathbf{\right)}\mathbf{\text{iron}}\mathbf{\left(}\mathbf{\text{III}}\mathbf{\right)}\mathbf{\text{hydroxide;}}\mathbf{\left(}\mathbf{\text{b}}\mathbf{\right)}\mathbf{\text{barium phosphate;}}\mathbf{\left(}\mathbf{\text{c}}\mathbf{\right)}\mathbf{\text{tin}}\mathbf{\left(}\mathbf{\text{II}}\mathbf{\right)}\mathbf{\text{sulfide.}}$

The 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 Qsp value is referred to as the Ksp value (solubility-product constant).

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

We do not include the solid and liquid states in the Ksp 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$\text{Iron}\left(\text{III}\right)\text{hydroxide}$.

The formula for$\text{Iron}\left(\text{III}\right)\text{hydroxide}$is$Fe(OH{)}_3$.

$Fe(OH{)}_3\left(s\right)\iff F{e}^{3+}\left(aq\right)+3O{H}^{-}\left(aq\right)$

$\text{Ksp =}\left[{\text{Fe}}^{\text{3 +}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}$

We cubed $\left[{\text{OH}}^{\text{-}}\right]$because we have two mols of ${\text{OH}}^{\text{-}}$in the equation.

Therefore, the ion-product expression for$\text{Iron}\left(\text{III}\right)\text{hydroxide}$ is $\left[{\text{Fe}}^{\text{3 +}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}$.

Let us obtain the ion-product expression for Barium fluoride.

The formula for Barium- phosphate is $B{a}_3{\left(P{O}_4\right)}_2$.

$B{a}_3{\left(P{O}_4\right)}_2\left(s\right)\iff 3B{a}^{2+}\left(aq\right)+2P{O}_4^{3-}\left(aq\right)$

$\text{Ksp =}{\left[{\text{Ba}}^{\text{2 +}}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}^{\text{3 -}}\right]}^{\text{2}}$

We cubed$\left[{\text{Ba}}^{\text{2 +}}\right]$because we have three mols of ${\text{Ba}}^{\text{2 +}}$in the equation and we squared $\left[{\text{PO}}_{\text{4}}^{\text{3 -}}\right]$because we have two mols of $\left[{\text{PO}}_{\text{4}}^{\text{3 -}}\right]$in the equation.

Therefore, the ion-product expression for barium phosphate is ${\left[{\text{Ba}}^{\text{2 +}}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}^{\text{3 -}}\right]}^{\text{2}}$.

The formula for$\text{Tin}\left(\text{II}\right)\text{sulfide}$ is $SnS$.

$SnS\left(s\right)\iff S{n}^{2+}\left(aq\right)+S^{2-}\left(aq\right)$

$\text{Ksp =}\left[{\text{Sn}}^{\text{2 +}}\right]\left[{\text{S}}^{\text{2 -}}\right]$

Therefore, the ion-product expression for $\text{Tin}\left(\text{II}\right)\text{sulfide}$is $\left[{\text{Sn}}^{\text{2 +}}\right]\left[{\text{S}}^{\text{2 -}}\right]$.