Question

A solution is a mixture of $0.05\, M \, NaCl$ and $0.05\, M \, NaI$. The concentration of iodide in the solution when $AgCl$ just starts precipitating is equal to:
$\left(K_{s p} AgCl =1 \times 10^{-10} M ^2 ; K_{s p} AgI =4 \times 10^{-16} M ^2\right)$

• A. $4 \times 10^{-6} M$
• B. $2 \times 10^{-8} M$
• C. $2 \times 10^{-7} M$
• D. $8 \times 10^{-15} M$

Answer 1

Answer: (C)

$K_{\text {sp }}( AgCl )>K_{\text {sp }}( AgI )$
Hence is $AgI$ is precipitated before $AgCl$
$\left[ Ag ^{+}\right]$to precipitated $Cl ^{-}=\frac{1 \times 10^{-10}}{0.05}=2 \times 10^{-9} M$
$\left[ I ^{-}\right]$at this stage $=\frac{K_{\text {sp }}( AgI )}{\left[ Ag ^{+}\right]}$
$=\frac{4 \times 10^{-16}}{2 \times 10^{-9}}=2 \times 10^{-7} M$