Answer:
Value of [tex]K_{c}[/tex] for the given reaction is 7.7
Explanation:
[tex]Ag_{2}CO_{3}(s)\rightleftharpoons 2Ag^{+}(aq.)+CO_{3}^{2-}(aq.)[/tex]
[tex]K_{sp}(Ag_{2}CO_{3})=[Ag^{+}]^{2}[CO_{3}^{2-}][/tex]
[tex]Ag_{2}CrO_{4}(s)\rightleftharpoons 2Ag^{+}(aq.)+CrO_{4}^{2-}(aq.)[/tex]
[tex]K_{sp}(Ag_{2}CrO_{4})=[Ag^{+}]^{2}[CrO_{4}^{2-}][/tex]
Where [tex]K_{sp}[/tex] represents solubility product
For the given reaction, [tex]K_{c}=\frac{[CO_{3}^{2-}]}{[CrO_{4}^{2-}]}[/tex] (concentration of pure solids remain constant during reaction. Hence their concentration is taken as 1 to exclude them from equilibrium constant expression)
So, [tex]K_{c}=\frac{[Ag^{+}]^{2}[CO_{3}^{2-}]}{[Ag^{+}]^{2}[CrO_{4}^{2-}]}[/tex]
or, [tex]K_{c}=\frac{K_{sp}(Ag_{2}CO_{3})}{K_{sp}(Ag_{2}CrO_{4})}=\frac{8.5\times 10^{-12}}{1.1\times 10^{-12}}=7.7[/tex]
Hence option (B) is correct
