Question

An excess of $Ag_{2}CrO_{4}(s)$ is added to a $5 \times 10^{3}\, M\, K_{2}CrO_{4}$ solution. The concentration of $Ag^+$ in the solution is closest to [ Solubility product for $Ag_{2}CrO_{4} = 1.1 \times 10^{-12}]$

• A. $2.2 \times 10^{-10} M$
• B. $1.5 \times 10^{-5} M$
• C. $10 \times 10^{-6} M$
• D. $5.0 \times 10^{-3} M$

Answer 1

Answer: (B)

Given,
solubility product of $Ag_{2}CrO_{4} =11\times 10^{-12} $
Concentration of $CrO^{2-}_{4}$ ions, $[CrO_{4}]^{2-}$
$= 5 \times 10^{-3} M$
For reaction,
$Ag _{2} CrO _{4} \rightleftharpoons 2 Ag ^{+}+ CrO _{4}^{2-}$
$K_{sp} =[Ag^{2 +}]^{2} [CrO^{2-}_{4}]$
$11\times 10^{-12}=[Ag^{2 +}]^{2} [5\times10^{-13}]$
$\therefore [Ag^{+}]=1.48\times10^{-5} M \approx 1.5\times10^{-5} M$