The absolute solubility product () of is measured to be at . What is the exact molar solubility (in — Ionic Equilibrium Chemistry Question
Question
The absolute solubility product ($K_{sp}$) of $Ag_2CrO_4$ is measured to be $1.1 \times 10^{-12}$ at $298 \text{ K}$. What is the exact molar solubility (in $\text{mol L}^{-1}$) of $Ag_2CrO_4$ in a prepared $0.1 \text{ M }$ aqueous solution of $AgNO_3$?
💡 Solution & Explanation
The primary dissociation equilibrium is $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$. The $0.1 \text{ M } AgNO_3$ provides a massive common ion concentration of $[Ag^+] = 0.1 \text{ M}$ (completely dwarfing the tiny contribution from the salt itself). Let the solubility of the salt be $s$. The equilibrium concentration of the chromate ion becomes $[CrO_4^{2-}] = s$. The solubility product expression is rigidly $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$. Substituting the known values: $1.1 \times 10^{-12} = (0.1)^2 \times s$. Therefore, $1.1 \times 10^{-12} = 0.01 \times s \Rightarrow s = \frac{1.1 \times 10^{-12}}{10^{-2}} = 1.1 \times 10^{-10} \text{ M}$.