Question

Using the Gibbs change, $\Delta G^o=+63.3kJ$, for the following reaction, the $K_{sp}$ of $A{g}_{2}C{O}_3\left(s\right)$ in water at ${25}^{o}C$ is :

$A{g}_{2}C{O}_3\left(g\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$ $(R=8.314J{K}^{-1}mo{l}^{-1})$

• A. $3.2\times {10}^{-26}$
• B. $8.0\times {10}^{-12}$
• C. $2.9\times {10}^{-3}$
• D. $7.9\times {10}^{-2}$

Answer 1

The correct option is B $8.0\times {10}^{-12}$

The relationship between the Gibbs free energy and the solubility product is

$\Delta G^o=-2.303RTlog{K}_{sp}$

Substitute values in the above expression.

$63.3\times {10}^3=-2.303\times 8.31\times 298log{K}_{sp}$

$-11.09=log{K}_{sp}$

$8\times {10}^{-12}=K_{sp}$

Hence, the $K_{sp}$ of $A{g}_{2}C{O}_3\left(s\right)$ in water at ${25}^{o}C$ is $8.0\times {10}^{-12}$.