Answer
Molar solubility $(AgCrO_4)$ = $6.54 \times 10^{-5}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Ag_2CrO_4(s) \lt -- \gt 2Ag^{+}(aq) + 1Cr{O_4}^{2-}(aq)$
$1.12 \times 10^{-12} = [Ag^{+}]^ 2[Cr{O_4}^{2-}]^ 1$
2. Considering a pure solution: $[Ag^{+}] = 2S$ and $[Cr{O_4}^{2-}] = 1S$
$1.12 \times 10^{-12}= ( 2S)^ 2 \times ( 1S)^ 1$
$1.12 \times 10^{-12} = 4S^ 3$
$2.8 \times 10^{-13} = S^ 3$
$ \sqrt [ 3] {2.8 \times 10^{-13}} = S$
$6.54 \times 10^{-5} = S$
- This is the molar solubility value for this salt.
