First slide

Solubility product(KSP)

Question

The solubility product $K_{sp}$ of the sparingly soluble salt ${Ag}_{2} {CrO}_{4}$ is $4 \times 10^{- 12}$ . The molar solubility of the salt is

Moderate

A

$1.0 \times 10^{- 4} mol L^{- 1}$
B

$2 \times 10^{- 6} mol L^{- 1}$
C

$1.0 \times ∣ 10^{- 5} mol L^{- 1}$
D

$2 \times 10^{- 12} mol L^{- 1}$

Solution

$\begin{matrix} {Ag}_{2} {CrO}_{4} ⇌ & 2 {Ag}^{+} + {CrO}_{4}^{2 -} \\ S & 2 S & S \end{matrix}$
$K_{sp} = (2 S)^{2} S = 4 S^{3} or S = {(\frac{K_{sp}}{4})}^{1 / 3}$
$= {(\frac{4 \times 10^{- 12}}{4})}^{1 / 3} = 10^{- 4} mol L^{- 1}$

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