Question

The value of the solubility product constant for Ag2CO3 is 8.5 × 10‒12 and that of Ag2CrO4 is 1.1 × 10‒12. From this data, what is the value of Kc for the reaction, Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq) A) 9.6 × 10‒12 B) 7.7 C) 1.1 × 1023 D) 1.3 × 10‒1 E) 9.4 × 10‒24

Answer 1

Answer:

Value of $K_{c}$ for the given reaction is 7.7

Explanation:

$Ag_{2}CO_{3}(s)\rightleftharpoons 2Ag^{+}(aq.)+CO_{3}^{2-}(aq.)$

$K_{sp}(Ag_{2}CO_{3})=[Ag^{+}]^{2}[CO_{3}^{2-}]$

$Ag_{2}CrO_{4}(s)\rightleftharpoons 2Ag^{+}(aq.)+CrO_{4}^{2-}(aq.)$

$K_{sp}(Ag_{2}CrO_{4})=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

Where $K_{sp}$ represents solubility product

For the given reaction, $K_{c}=\frac{[CO_{3}^{2-}]}{[CrO_{4}^{2-}]}$ (concentration of pure solids remain constant during reaction. Hence their concentration is taken as 1 to exclude them from equilibrium constant expression)

So, $K_{c}=\frac{[Ag^{+}]^{2}[CO_{3}^{2-}]}{[Ag^{+}]^{2}[CrO_{4}^{2-}]}$

or, $K_{c}=\frac{K_{sp}(Ag_{2}CO_{3})}{K_{sp}(Ag_{2}CrO_{4})}=\frac{8.5\times 10^{-12}}{1.1\times 10^{-12}}=7.7$

Hence option (B) is correct