On addition of increasing amount of $\mathrm{AgNO}_{3}$ to $0.1 \mathrm{M}$ each of $\mathrm{NaCl}$ and $\mathrm{NaBr}$ in a solution, what $\%$ of $\mathrm{Br}^{-}$ ion get precipitated when $\mathrm{Cl}^{-}$ ion starts precipitating. $\mathrm{K}_{12 \mathrm{sp}}(\mathrm{AgCl})=1.0 \times 10^{-10}$,
$\mathrm{K}_{\mathrm{sp}}(\mathrm{AgBr})=1 \times 10^{-13} \mathrm{Pp}$

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Aicharya · Accepted Answer

To precipitate the $\mathrm{AgCl}$
$\left[\mathrm{Ag}^{+}ight]$ required
$=\frac{\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})}{\left[\mathrm{Cl}^{-}ight]}=\frac{1.0 \times 10^{-10}}{0.1}=1.0 \times 10^{-9} \mathrm{M}$
$\left[\mathrm{Br}^{-}ight]$ left at this stage $=\frac{\mathrm{K}_{\mathrm{sp}}(\mathrm{AgBr})}{\left[\mathrm{Ag}^{+}ight]}$
$=\frac{1.0 \times 10^{-13}}{1.0 \times 10^{-9}}=1.0 \times 10^{-4} \mathrm{M}$
$\%$ of remaining
$\left[\mathrm{Br}^{-}ight]=\frac{1.0 \times 10^{-4}}{0.1} \times 100=0.1$
$\%$ of $\mathrm{Br}^{-}$ to be precipitated $=100-0.1=99.9$ ,

On addition of increasing amount of $\mathrm{AgNO}_{3}$ to $0.1 \mathrm{M}$ each of $\mathrm{NaCl}$ and…

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