Question

The molar solubility of \(\mathrm{MnCO}_{3}\) is \(4.2 \times 10^{-6} \mathrm{M}\). What is \(K_{\mathrm{sp}}\) for this compound?

Short answer

The solubility product constant for the compound \(\mathrm{MnCO}_{3}\) is \(1.76 \times 10^{-11}\)

Step-by-step solution

Chemical Dissociation of the Compound

Dissolve the compound in a solution. \(\mathrm{MnCO}_{3}\) dissociates or breaks down into its constituent ions according to the equation: \(\mathrm{MnCO}_{3}(s) \leftrightarrow Mn^{2+}(aq) + CO_3^{2-}(aq)\)

Set up the equation for \(K_{sp}\)

The solubility product constant is given by \(K_{sp} = [Mn^{2+}][CO_3^{2-}]\)

Substitution of the given into the equation

Since the dissociation of \(\mathrm{MnCO}_{3}\) to \(Mn^{2+}\) and \(CO_3^{2-}\) ions is in a 1:1 ratio, the concentration of \(Mn^{2+}\) and \(CO_3^{2-}\) is equal to the molar solubility. Thus, substitute the given molar solubility of 4.2e-6 M for both [\(Mn^{2+}\)] and [\(CO_3^{2-}\)]. So \(K_{sp} = (4.2 \times 10^{-6})(4.2 \times 10^{-6})\).

Calculate \(K_{sp}\)

Calculate \(K_{sp}\) by multiplying the concentrations of the ions, which gives \(K_{sp} = 1.76 \times 10^{-11}\)