Question

8. A saturated solution of Ag Croq has a silver-ion concentration of 1.3 x 10-4M. Which is the Ksp of Ag CrO 4?

Answer 1

Answer: The $K_{sp}$ of $AgCrO_{4}$ is $1.1 \times 10^{-12}$.

Explanation:

Given: $[Ag^{+}] = 1.3 \times 10^{-4} M$

The reaction equation will be written as follows.

$Ag_{2}CrO_{4} \rightleftharpoons 2Ag^{+} + CrO^{2-}_{4}$

This shows that the concentration of $CrO^{2-}_{4}$ is half the concentration of $Ag^{+}$ ion. So,

$[CrO^{2-}_{4}] = \frac{1.3 \times 10^{-4}}{2}\\= 0.65 \times 10^{-4} M$

The expression for $K_{sp}$ of this reaction is as follows.

$K_{sp} = [Ag^{+}]^{2}[CrO^{2-}_{4}]$

Substitute values into the above expression as follows.

$K_{sp} = [Ag^{+}]^{2}[CrO^{2-}_{4}]\\= (1.3 \times 10^{-4})^{2} \times 0.65 \times 10^{-4}\\= 1.1 \times 10^{-12}$

Thus, we can conclude that the $K_{sp}$ of $AgCrO_{4}$ is $1.1 \times 10^{-12}$.