Question

A solution containing a mixture of $0.05\, M\, NaCl$ and $0.05 \,M\, Nal$ is taken. $\left(K_{ sp }\right.$ of $AgCl =10^{-10}$ and $K_{ sp }$ of $\left.AgI =4 \times 10^{-16}\right)$. When $AgNO _3$ is added to such a solution:

• A. The concentration of $Ag ^{\oplus}$ required to precipitate $Cl ^{\ominus}$ is $2 \times 10^{-9} mol L ^{-1}$
• B. The concentration of $Ag ^{\oplus}$ required to precipitate $I ^{\ominus}$ is $8 \times 10^{-15} mol \,L ^{-1}$
• C. $AgCl$ and $AgI$ will be precipitate together
• D. First $AgI$ will be precipitated

Answer 1

Answer: (A), (B), (D)

$\underset{0.05\, M}{NaCl} + \underset{0.05\,M}{NaI}$
When $AgNO _3$ is added, find the minimum concentration of $Ag ^{\oplus}$ required to start precipitation
$\left[ Ag ^{\oplus}\right]_{ AgCl }=\frac{K_{\text {sp }( NaCl )}}{\left[ Cl ^{\ominus}\right]}=\frac{10^{-10}}{0.05}=2 \times 10^{-9} M $
$\left[ Ag ^{\oplus}\right]_{ AgI }=\frac{K_{\text {sp }( Agl )}}{\left[ I ^{\ominus}\right]}=\frac{4 \times 10^{-16}}{0.05}=8 \times 10^{-15} M $
Since, there is a very high difference between the minimum concentration of $Ag ^{\oplus}$, precipitation is selective and AgI will precipitate first