According to solubility rules, what is the molar solubility of MnCO3 at 25°C in pure water?

• A. MnCO3 is completely insoluble in water; its molar solubility is zero.
• B. MnCO3 is highly soluble in water, with a molar solubility greater than 1 M.
• C. MnCO3 is sparingly soluble; its molar solubility is very low due to the insolubility of most carbonates except those of alkali metals and ammonium.
• D. MnCO3 is moderately soluble in water, with a molar solubility around 0.1 M.

Answer: (C)

Verified step by step guidance

1. Identify the solubility rules relevant to MnCO3: Most carbonate salts (CO3^{2-}) are insoluble in water except those of alkali metals (Group 1) and ammonium (NH4^+). Since Mn^{2+} is a transition metal cation, MnCO3 is expected to be sparingly soluble.
2. Write the dissolution equilibrium for MnCO3 in water: \(\mathrm{MnCO_3 (s) \rightleftharpoons Mn^{2+} (aq) + CO_3^{2-} (aq)}\).
3. Express the solubility product constant (K_{sp}) for MnCO3: \(K_{sp} = [Mn^{2+}][CO_3^{2-}]\).
4. Define the molar solubility as \(s\), which is the concentration of Mn^{2+} ions dissolved at equilibrium. Since the stoichiometry is 1:1, \([Mn^{2+}] = s\) and \([CO_3^{2-}] = s\).
5. Substitute these concentrations into the \(K_{sp}\) expression to get \(K_{sp} = s \times s = s^2\). Then, solve for \(s\) by taking the square root of \(K_{sp}\): \(s = \sqrt{K_{sp}}\). This \(s\) represents the molar solubility of MnCO3.