The solubility product of Ag_2CrO_4 is 4 × 10^ -12 . What is the...

Let the solubility of $\text{Ag}_2\text{CrO}_4$ be $s$ mol/L.

The dissociation reaction is:
$\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$

The solubility product $K_{sp}$ is given by:
$K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] = (2s)^2(s) = 4s^3$

Given $K_{sp} = 4 \times 10^{-12}$, we have:
$4s^3 = 4 \times 10^{-12}$
$s^3 = 10^{-12}$
$s = 10^{-4}$ mol/L

The number of moles of $\text{Ag}_2\text{CrO}_4$ in $100$ ml (0.1$L) of saturated solution is:<br>Moles$= s \times V = 10^{-4} \times 0.1 = 10^{-5}$mol<br><br>The amount of \text{Ag}_2\text{CrO}_4$ in grams is:
Mass $= \text{Moles} \times \text{Molar mass}$
Mass $= 10^{-5} \times 331.73 = 3.3173 \times 10^{-3}$ g

Answer: $3.3173 \times 10^{-3}$ g