Question

One litre of an aqueous solution containing $0.1$ mole of $C{l}^{-}$; $0.1$ mole of $B{r}^{-}$. $[K_{sp}ofAgCl={10}^{-10}and{K}_{sp}ofAgBr={10}^{-14}]$ and we start adding $A{g}^{+}$ in the solution so that volume does remains same. The incorrect statement is:

• A. Precipitation of $AgCl$ starts first
• B. Precipitation of $AgBr$ starts when $\left[A{g}^{+}\right]$ becomes above ${10}^{-13}$ molar
• C. Precipitation of $AgCl$ starts when $\left[A{g}^{+}\right]$ becomes above ${10}^{-9}$ molar
• D. Precipitation of $AgBr$ starts first

Answer 1

The correct option is A Precipitation of $AgCl$ starts first

$K_{sp}$ of $AgCl={10}^{-10}$
$K_{sp}$ of $AgBr={10}^{-14}$
$AgBr\rightleftharpoons \underset{x}{A{g}^{+}}+\underset{0.1}{B{r}^{-}}$
Let, $x$ is the concentration of $\left[A{g}^{+}\right]$
$K_{sp}$ of $\left[AgBr\right]=\left[A{g}^{+}\right]\left[B{r}^{-}\right]=\left(x\right)\left(0.1\right)\Rightarrow {10}^{-14}=x\times 0.1\Rightarrow x={10}^{-13}M$
$\therefore$ Concentration of $A{g}^{+}$ required for the precipitation of $AgBr$ is ${10}^{-13}$ molar or more.
Again,
$AgCl\rightleftharpoons \underset{y}{A{g}^{+}}+\underset{0.1}{C{l}^{-}}$
$K_{sp}$ of $\left[AgCl\right]=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=\left(y\right)\left(0.1\right)\Rightarrow {10}^{-10}=y\times 0.1\Rightarrow x={10}^{-9}M$
So, when concentration of $A{g}^{+}$ ion is ${10}^{-9}M$, then only $AgCl$ starts precipitating. Thus, precipitation of $AgBr$ starts first.