Question

The solubility product of $ A{{g}_{2}}CrO_4 $ in water a $ 298\,K $ is $ 3.2\times {{10}^{-11}} $ . What will be the concentration of $ CrO_{4}^{2-} $ ions in the saturate solution of $ A{{g}_{2}}Cr{{O}_{4}} $ ?

• A. $ 2\times {{10}^{-4}}M $
• B. $ 5.7\times {{10}^{-5}}M $
• C. $ 5.7\times {{10}^{-6}}M $
• D. $ 3.2\times {{10}^{-11}} $

Answer 1

Answer: (A)

For saturated solution of $Ag _{2} CrO _{4}$, if solubility is ' $s'\, mol \,L ^{-1}$. Then
$\underset{s}{Ag _{2} Cr O _{4}} \rightleftharpoons \underset{2s}{2 Ag ^{+}}(a q)+\underset{ s }{ Cr } O _{4}^{2-}(a q) $
$K_{s p} =(2 s)^{2}(s)=4 s^{3} $
$K_{s p} =3.2 \times 10^{-11} $ (given)
$\therefore 3.2 \times 10^{-11} =4 s^{3} $
$s^{3} =\frac{3.2 \times 10^{-11}}{4}=8 \times 10^{-12} $
$\therefore s =\sqrt{8 \times 10^{-12}}=2 \times 10^{-4} M$