The solubility product of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ is $32 \times 10^{-12}$. What is the…
What is the concentration of $\mathrm{CrO}_{4}^{2-}$ ions in that solution (in $\mathrm{g} \mathrm{L}^{-1}$ )
- $2 \times 10^{-4}$
- $8 \times 10^{-4}$
- $8 \times 10^{-8}$
- $16 \times 10^{-4}$
Solution
$=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}ight)^{\frac{1}{3}}=\left(\frac{32 \times 10^{-12}}{4}ight)^{\frac{1}{3}}=2 \times 10^{-4}$ /
Asked in: JEE-TOPICTESTS-CHEMISTRY