Question
Using the Gibbs energy change $$\Delta {G^ \circ } = + 63.3\,kJ$$ for the following reaction,
$$A{g_2}C{O_3}\left( s \right) \rightleftharpoons $$ $$2A{g^ + }\left( {aq} \right) + CO_3^{2 - }\left( {aq} \right)$$ the $${K_{sp}}$$ of $$A{g_2}C{O_3}\left( s \right)$$ in water at $${25^ \circ }C$$ is $$\left( {R = 8.314\,J{K^{ - 1}}mo{l^{ - 1}}} \right)$$
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 :
$$8.0 \times {10^{ - 12}}$$
Solution :
$$\eqalign{
& \Delta {G^ \circ }\,\,{\text{is related to}}\,\,{K_{sp}}\,\,{\text{by the equation,}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G^ \circ } = - 2.303RT\,\,{\text{log}}\,\,{K_{sp}} \cr
& {\text{Given,}}\,\,\Delta {G^ \circ } = + 63.3\,kJ \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 63.3 \times {10^3}\,J \cr
& {\text{Thus, substitute}}\,\,\Delta {G^ \circ } = 63.3 \times {10^3}\,\,J, \cr} $$
$$R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}\,\,{\text{and}}$$ $$T = 298\,K\left[ {25 + 273\,K} \right]$$
$$\eqalign{
& {\text{from the above equation we get,}} \cr
& 63.3 \times {10^3} = - 2.303 \times 8.314 \times 298\,{\text{log}}\,\,{K_{sp}} \cr
& \therefore \,{\text{log}}\,\,{K_{sp}} = - 11.09 \cr
& \Rightarrow {K_{sp}} = {\text{antilog}}\left( { - 11.09} \right) \cr
& \,\,\,\,\,\,\,{K_{sp}} = 8.0 \times {10^{ - 12}} \cr} $$