Question

$0.10$ mol of $AgCl\left(s\right)$ is added to $1$ litre of $H_{2}O$. Next, crystals of $NaBr$ are added until $75$% of the $AgCl\left(s\right)$ is converted to $AgBr\left(s\right)$, the less soluble silver halide. What is $\left[{Br}^{-}\right]$ at this point?

$K_{sp}$ of $AgCl=1.78\times {10}^{-10}$ and $K_{sp}$ of $AgBr=5.25\times {10}^{-13}$

• A. $1.8\times {10}^{-4}M$
• B. $2.0\times {10}^{-4}M$
• C. $3.9\times {10}^{-8}M$
• D. None of these

Answer 1

The correct option is C $3.9\times {10}^{-8}M$

Initially $0.10$ moles of AgCl is present in $1$ L of solution.

At equilibrium, $0.75\times 0.10=0.075$ moles of AgBr and $0.25\times 0.10=0.025$ moles of AgCl are present.

$\underset{\text{0.025-a}}{AgCl}\rightleftharpoons \underset{\text{a+b}}{A{g}^{+}}+\underset{\text{a}}{C{l}^{-}}$......(1)

$\underset{\text{0.075-b}}{AgBr}\to \underset{\text{a+b}}{A{g}^{+}}+\underset{\text{b}}{B{r}^{-}}$......(2)

For AgCl, $K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

$1.78\times {10}^{-10}=(a+b)a$......(3)

For AgBr, $K_{sp}=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$

$5.25\times {10}^{-13}=(a+b)b$......(4)

Divide equation (3) with equation (4).

$\frac{1.78\times {10}^{-10}}{5.25\times {10}^{-13}}=\frac{(a+b)a}{(a+b)b}$

$339=\frac{a}{b}$

$a=339b$

Substitute values of a in equation 4.

$5.25\times {10}^{-13}=(339b+b)b=340b^2$

$b^2=1.54\times {10}^{-15}$

$\left[B{r}^{-}\right]=b=3.9\times {10}^{-8}$

Hence, the bromide ion concentration is $3.9\times {10}^{-8}$ M