Solubility of calcite

We consider the reaction $Ca C O_{3} (s) = {Ca}^{2 +} (aq) + C O_{3}^{2 -} (aq)$ .

1/ Calculate the standard Gibbs free energy change of reaction for the dissolution of calcite.

Coup de pouce 1

Use the standard Gibbs free energies of formation.

Corrigé

Hess’s law states that

\begin{matrix} Δ_{𝑟} 𝐺^{\circ} = - Δ_{𝑓} 𝐺^{\circ} (Ca C O_{3} (s)) + Δ_{𝑓} 𝐺^{\circ} ({Ca}^{2 +}) + Δ_{𝑓} 𝐺^{\circ} (C O_{3}^{2 -}) \\ = 4,740 \end{matrix}

⋅104 J mol−1

2/ Deduce the solubility product of calcite at room temperature.

Coup de pouce 1

What is a solubility product?

Coup de pouce 2

How is the equilibrium constant related to the Gibbs free energy of reaction?

Corrigé

The solubility product is given by

\begin{matrix} 𝐾^{\circ} = exp (- \frac{Δ_{𝑟} 𝐺^{\circ}}{𝑅 𝑇}) \\ = 4,9 \end{matrix}

⋅10−9

Data at $𝑇 = 298 K$

The chemical species are solutes in water.

Chemical species	$Ca C O_{3} (s)$	${Ca}^{2 +} (aq)$	$C O_{3}^{2 -} (aq)$
$Δ_{𝑓} 𝐺^{\circ}$ ( $kJ {mol}^{- 1}$ )	$- 1128,8$	$- 553,5$	$- 527,9$