Question

what is the solubility of mgco₃ in a solution that contains 0.070 m mg²⁺ ions? (ksp of mgco₃ is 3.5 × 10⁻⁸)

Explanation:

Step1: Write the dissolution equation

$$\ce{MgCO_{3}(s) <=> Mg^{2 + }(aq) + CO^{2-}_{3}(aq)}$$
The solubility product constant expression is \(K_{sp}=[\ce{Mg^{2 + }}][\ce{CO^{2-}_{3}}]\)

Step2: Let the solubility of \(\ce{MgCO_{3}}\) be \(s\).

The concentration of \(\ce{Mg^{2 + }}\) from \(\ce{MgCO_{3}}\) dissolution is \(s\), but there is an initial \([\ce{Mg^{2 + }}]=0.070\space M\) (common - ion effect). So \([\ce{Mg^{2 + }}]=(0.070 + s)\space M\) and \([\ce{CO^{2-}_{3}}]=s\space M\). Since \(K_{sp}\) is very small (\(3.5\times 10^{-8}\)), \(s\) is negligible compared to \(0.070\) (i.e., \(0.070 + s\approx0.070\))

Step3: Substitute into \(K_{sp}\) expression

\(K_{sp}=(0.070)\times s\)
We know \(K_{sp}=3.5\times 10^{-8}\)
So \(s=\frac{K_{sp}}{0.070}\)

Step4: Calculate \(s\)

\(s=\frac{3.5\times 10^{-8}}{0.070}=5.0\times 10^{-7}\space M\)

Answer:

The solubility of \(\ce{MgCO_{3}}\) is \(5.0\times 10^{-7}\space M\)