Search any question & find its solution
Question:
Answered & Verified by Expert
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
$\begin{aligned} & \left(K_{\mathrm{sp}}(\mathrm{AgCl})=1.8=10^{-10} ; K_{\mathrm{sp}}(\mathrm{AgBr})=5 \times 10^{-13} ;\right. \\ & \left.K_{\mathrm{sp}}(\mathrm{AgI})=8.3 \times 10^{-17}\right)\end{aligned}$
Options:
$\begin{aligned} & \left(K_{\mathrm{sp}}(\mathrm{AgCl})=1.8=10^{-10} ; K_{\mathrm{sp}}(\mathrm{AgBr})=5 \times 10^{-13} ;\right. \\ & \left.K_{\mathrm{sp}}(\mathrm{AgI})=8.3 \times 10^{-17}\right)\end{aligned}$
Solution:
1142 Upvotes
Verified Answer
The correct answer is:
$\mathrm{Agl}, \mathrm{AgBr}, \mathrm{AgCl}$
$K_{\mathrm{sp}}$ 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
$\mathrm{AgI}>\mathrm{AgBr}>\mathrm{AgCl}$
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
$\mathrm{AgI}>\mathrm{AgBr}>\mathrm{AgCl}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.