Question

The equilibrium constant for the reaction $AgC{l}_s\rightleftharpoons A{g}_{\left(aq\right)}^{+}+C{l}_{\left(aq\right)}^{-}$ is $x\times {10}^{-10}$. Then, the value of $x$ is :
(Given $:\Delta G_{f\left(AgCl\right)}^0=-109.7kJ;$ $\Delta G_{f\left(A{g}^{+}\right)}^0=77.1kJ;$ $\Delta G_{f\left(C{l}^{-}\right)}^0=-131.2kJ$ )

Answer 1

As we know, at equilibrium,

$\Delta G^0$ for the reaction $=77.1+(-131.2)-(-109.7)=55.6kJ$

$\Delta G^0=-2.303RT\log K$

$\log K=\frac{-\Delta G^0}{2.303RT}=\frac{-55.6\times {10}^3}{2.303\times 8.314\times 298}=-9.75$

$K=K_{sp}=1.8\times {10}^{-10}\simeq 2\times {10}^{-10}$

So, value of $x$ is $2$.