Question

Calculate the ratio of \(\left[\mathrm{Ca}^{2+}\right]\) to \(\left[\mathrm{Fe}^{2+}\right]\) in a lake in which the water is in equilibrium with deposits of both \(\mathrm{CaCO}_{3}\) and \(\mathrm{FeCO}_{3}\). Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored.

Short answer

The ratio of [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)] in the lake is approximately 101.5, as calculated by dividing the solubility product constants for CaCO₃ and FeCO₃.

Step-by-step solution

Write the equilibrium reactions for both CaCO₃ and FeCO₃

CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq) FeCO₃(s) ⇌ Fe²⁺(aq) + CO₃²⁻(aq)

Write the equilibrium expressions for both CaCO₃ and FeCO₃

\(K_{sp}^{CaCO_3} = [\mathrm{Ca}^{2+}][\mathrm{CO_3}^{2-}]\) \(K_{sp}^{FeCO_3} = [\mathrm{Fe}^{2+}][\mathrm{CO_3}^{2-}]\)

Use the given information to find the ratio of [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)]

Since the water is in equilibrium with both CaCO₃ and FeCO₃ deposits, the concentration of CO₃²⁻ ions will be the same in both cases. Therefore, we can divide the equilibrium expressions for both salts to find the ratio of concentrations of Ca²⁺ and Fe²⁺ ions: \(\frac{[\mathrm{Ca}^{2+}]}{[\mathrm{Fe}^{2+}]} = \frac{K_{sp}^{CaCO_3}}{K_{sp}^{FeCO_3}}\)

Find the values of solubility product constants for CaCO₃ and FeCO₃

\(K_{sp}^{CaCO_3} = 3.36 \times 10^{-9}\) (from a table or given information) \(K_{sp}^{FeCO_3} = 3.31 \times 10^{-11}\) (from a table or given information)

Calculate the ratio of [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)]

\(\frac{[\mathrm{Ca}^{2+}]}{[\mathrm{Fe}^{2+}]} = \frac{3.36 \times 10^{-9}}{3.31 \times 10^{-11}} = 101.5\) The ratio of [\(\mathrm{Ca}^{2+}\)] to [\(\mathrm{Fe}^{2+}\)] in the lake is approximately 101.5.