Question

A solution containing a mixture of $0.05MNaCl$ and $0.05MNal$ is taken. $(K_{sp}$ of $AgCl=10^{-10}$ and $K_{sp}$ of $AgI=4\times 10^{-16})$. When $AgN{O}_3$ is added to such a solution:

• A. The concentration of $A{g}^{\oplus }$ required to precipitate $C{l}^{\ominus }$ is $2\times 10^{-9}mol{L}^{-1}$
• B. The concentration of $A{g}^{\oplus }$ 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)

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