Question

The solubility product of $ Ag_{2}CrO_{4}$ is $ 32\times 10^{-12}$ . What is the concentration of $[CrO_{4}]^{2-}$ ions in that solution?

• A. $ 8\times 10^{-8}mol\,L^{-1}$
• B. $ 16\times 10^{-4}mol\,L^{-1}$
• C. $ 2\times 10^{-4}mol\,L^{-1}$
• D. $ 8\times 10^{-4}mol\,L^{-1}$

Answer 1

Answer: (C)

First write the dissociation reaction for $ Ag_{2}CrO_{4}$.
Then find relationship between concentration of ions and solubility product.
Now do the calculation to find the concentration of $[CrO_{4}]^{2-}$.
Let the solubility of $Ag _{2} CrO _{4}=x mol / L Ag _{2} CrO _{4} \rightarrow 2 Ag ^{+} CrO _{4}{ }^{2-}$ ions $2x x$
$\therefore K_{s p}=\left[ Ag ^{+}\right]^{2}\left[ CrO _{4}^{-2}\right]=(2 x)^{2}(x)$
or $K_{s p}=4 x^{3}$ or $x=\sqrt[3]{\frac{K_{s p}}{4}}$
Given, $K_{s p}=\sqrt[3]{\frac{32 \times 10^{-2}}{4}}$
$=\sqrt[3]{8 \times 10^{-12}} =2 \times 10^{-4} mol / L$