91Ó°ÊÓ

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).

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

We do not include solid and liquid state in the Ksp equations

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

Let us calculate the Ksp of silver carbonate.

Given:

Solubility of silver carbonate$\text{= 0.032 M at20°C}$

To calculate the Ksp of silver carbonate, we first need to write the dissolution equation for ${\text{Ag}}_2{\text{CO}}_3:$

$$\begin{gathered} {\text{Ag}}_2{\text{CO}}_3\left(s\right)⇔2{\text{Ag}}^{+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right) \\ \text{S}\left({\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\text{= 0.032 M} \\ \text{Ksp =}{\left[{\text{Ag}}^{\text{+}}\right]}^{\text{2}}\left[{\text{CO}}_{\text{3}}^{\text{2 -}}\right] \end{gathered}$$

We squared $\left[{\text{Ag}}^{+}\right]$because we have two moles of ${\mathrm{Ag}}^{+}$in the equation.

role="math" localid="1663306882117" $$\begin{gathered} \left[{\text{Ag}}^{\text{+}}\right]\text{= 2S} \\ =2×0.032\mathrm{M} \\ =0.064\mathrm{M} \end{gathered}$$

role="math" localid="1663307079898" $$\begin{gathered} \left[C{O}_3^{2-}\right]=S \\ \left[C{O}_3^{2-}\right]=0.032\mathrm{M} \\ Ksp=(2S{)}^2×S \\ Ksp=1.31×{10}^{-4} \end{gathered}$$

Therefore, the Ksp of silver carbonate is $1.31×{10}^{-4}$.