Question

Solid $AgN{O}_3$ is slowly added to a solution that is $0.001$ M each in $NaCl,NaBr$ and $NaI$.

(A) The concentration of $A{g}^{+}$ required to initiate precipitation of each of $AgI,AgBr$ and $AgCl$ are $1.5\times {10}^{-13}$, $3.3\times {10}^{-10}$ and $1.6\times {10}^{-7}$ respectively.

(B) The percentage of $I^{-}$ precipitated before $AgBr$ precipitates is $99.95$%.

(C) The percentage of $I^{-}$ precipitated before $C{l}^{-}$ precipitates is $99.999917$%.

Given that;

$\left(K_{sp}\right(AgI)=1.5\times {10}^{-16},$

$K_{sp}\left(AgBr\right)=3.3\times {10}^{-13},$

$K_{sp}\left(AgCl\right)=1.8\times {10}^{-10})$

Which of the above-given option(s) is/are correct?

• A. Only A
• B. A and B
• C. B and C
• D. A, B and C

Answer 1

Answer: (C)

B and C

$\left[C{l}^{-}\right]=\left[B{r}^{-}\right]=\left[I^{-}\right]$

To initiate precipitation of $AgI$,

$\left[A{g}^{+}\right]=\frac{K_{sp,AgI}}{\left[I^{-}\right]}=\frac{1.5\times {10}^{-16}}{0.001}=1.5\times {10}^{-13}M$
To initiate precipitation of $AgBr$,

$\left[A{g}^{+}\right]=\frac{K_{sp,AgBr}}{\left[B{r}^{-}\right]}=\frac{3.3\times {10}^{-13}}{0.001}=3.3\times {10}^{-10}M$
To initiate precipitation of $AgCl$,

$\left[A{g}^{+}\right]=\frac{K_{sp,AgCl}}{\left[C{l}^{-}\right]}=\frac{1.8\times {10}^{-10}}{0.001}=1.8\times {10}^{-7}M$

Hence, statement (A) is incorrect.
When $AgBr$ starts precipitating, $\left[A{g}^{+}\right]=3.3\times {10}^{-10}M$

Also, $\left[I^{-}\right]=\frac{K_{sp,AgI}}{\left[A{g}^{+}\right]}=\frac{1.5\times {10}^{-16}}{3.3\times {10}^{-10}M}=4.5454\times {10}^{-7}M$

The percentage of iodide ion precipitated $=100\times \frac{0.001-4.5454\times {10}^{-7}}{0.001}=99.95$ %

Hence, statement (B) is correct.
When $AgCl$ starts precipitating, $\left[A{g}^{+}\right]=1.8\times {10}^{-7}M$

Also, $\left[I^{-}\right]=\frac{K_{sp,AgI}}{\left[A{g}^{+}\right]}=\frac{1.5\times {10}^{-16}}{1.8\times {10}^{-7}M}=8.333\times {10}^{-10}M$

The percentage of iodide ion precipitated $=100\times \frac{0.001-8.333\times {10}^{-10}}{0.001}=99.999917$ %

Hence, statement (C) is correct.
Hence, option C is correct.