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Using the Gibbs energy change, $\Delta G^\circ = +63.3$ kJ, for the following reaction, $Ag_2CO_3(s) \rightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$ the $K_{sp}$ of $Ag_2CO_3(s)$ in water at $25^\circ C$ is: (R = 8.314 J $K^{-1}mol^{-1}$)
A
$3.2\times10^{-26}$
B
$8.0\times10^{-12}$
C
$2.9\times10^{-3}$
D
$7.9\times10^{-2}$
Detailed Solution
$\Delta G^\circ = -2.303RT\log K_{sp}$
$63.3\times1000 = -2.303\times8.314\times298\log K_{sp}$
$\log K_{sp} = -11.09$
$K_{sp} = 10^{-11.09} = 8\times10^{-12}$
$63.3\times1000 = -2.303\times8.314\times298\log K_{sp}$
$\log K_{sp} = -11.09$
$K_{sp} = 10^{-11.09} = 8\times10^{-12}$