A $1.0 \mathrm{~L}$ of aqueous solution contains $1 \times 10^{-8} \mathrm{M} \mathrm{NaBr}, 1 \times…
A $1.0 \mathrm{~L}$ of aqueous solution contains $1 \times 10^{-8} \mathrm{M} \mathrm{NaBr}, 1 \times 10^{-8} \mathrm{M} \mathrm{NaCl}$ and $1 \times 10^{-8} \mathrm{M}$ NaI. To this solution, $1 \times 10^{-10} \mathrm{M}$ aqueous $\mathrm{AgNO}_3$ solution is added drop wise. The order of precipitation of $\operatorname{Ag} X(X=\mathrm{Cl}, \mathrm{Br}, \mathrm{I})$ is
$\begin{aligned} & \left(K_{\mathrm{sp}}(\mathrm{AgCl})=1.8=10^{-10} ; K_{\mathrm{sp}}(\mathrm{AgBr})=5 \times 10^{-13} ;\right. \\ & \left.K_{\mathrm{sp}}(\mathrm{AgI})=8.3 \times 10^{-17}\right)\end{aligned}$
$\mathrm{AgBr}, \mathrm{AgCl}, \mathrm{Agl}$
$\mathrm{AgCl}, \mathrm{AgBr}, \mathrm{Agl}$.
$\mathrm{Agl}, \mathrm{AgBr}, \mathrm{AgCl}$
$\mathrm{AgBr}, \mathrm{Agl}, \mathrm{AgCl}$
Solution
$K_{\mathrm{sp}}$ of any salt is equal to the multiplication of concentration of its ions.
For $\mathrm{AgCl} \longrightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$
$K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$1.8 \times 10^{-10}=10^{-10}\left[\mathrm{Cl}^{-}\right]$
$\left[\mathrm{Cl}^{-}\right]=\frac{1.8 \times 10^{-10}}{10^{-10}}=1.8$
Similarly, for AgBr
$\left[\mathrm{Br}^{-}\right]=\frac{5 \times 10^{-13}}{10^{-10}}=5 \times 10^{-3}$
For AgI
$\mathrm{I}^{-}=\frac{8.3 \times 10^{-17}}{10^{-10}}=8.3 \times 10^{-7}$
Lesser is the concentration of ion, more easily they get precipitated.
So, order of precipitation is
$\mathrm{AgI}>\mathrm{AgBr}>\mathrm{AgCl}$