Search any question & find its solution
Question:
Answered & Verified by Expert
$1 \mathrm{dm}^{3}$ solution containing $10^{-5}$ moles each of $\mathrm{Cl}^{-}$ions and $\mathrm{CrO}_{4}^{2-}$ ions is treated with $10^{-4}$ moles of silver nitrate. Which one of the following observations is made?
$\begin{aligned} {\left[\mathrm{K}_{\mathrm{sp}} \mathrm{Ag}_{2} \mathrm{CrO}_{4}\right.} &\left.=4 \times 10^{-12}\right] \\ {\left[\mathrm{K}_{\mathrm{sp}} \mathrm{AgCl}\right.} &\left.=1 \times 10^{-10}\right] \end{aligned}$
Options:
$\begin{aligned} {\left[\mathrm{K}_{\mathrm{sp}} \mathrm{Ag}_{2} \mathrm{CrO}_{4}\right.} &\left.=4 \times 10^{-12}\right] \\ {\left[\mathrm{K}_{\mathrm{sp}} \mathrm{AgCl}\right.} &\left.=1 \times 10^{-10}\right] \end{aligned}$
Solution:
1699 Upvotes
Verified Answer
The correct answer is:
Silver chloride gets precipitated first
For precipitation,
ionic product $>$ solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$
For $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$
$$
\begin{aligned}
\text { ionic product } &=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{-}\right] \\
&=\left(10^{-4}\right)^{2}\left(10^{-5}\right)=10^{-13}
\end{aligned}
$$
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}=4 \times 10^{-12}$
Here, $\mathrm{K}_{\mathrm{sp}}>\mathrm{IP}$
Thus, no precipitate is obtained.
For $\mathrm{AgCl}$,
ionic product $=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$=\left[10^{-4}\right]\left[10^{-5}\right]$
$=10^{-9}$
$\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1 \times 10^{-10}$
Here, IP > $\mathrm{K}_{\mathrm{sp}}$
So, precipitate will form.
Thus, silver chloride gets precipitated first.
ionic product $>$ solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$
For $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$
$$
\begin{aligned}
\text { ionic product } &=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{-}\right] \\
&=\left(10^{-4}\right)^{2}\left(10^{-5}\right)=10^{-13}
\end{aligned}
$$
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}=4 \times 10^{-12}$
Here, $\mathrm{K}_{\mathrm{sp}}>\mathrm{IP}$
Thus, no precipitate is obtained.
For $\mathrm{AgCl}$,
ionic product $=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$=\left[10^{-4}\right]\left[10^{-5}\right]$
$=10^{-9}$
$\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1 \times 10^{-10}$
Here, IP > $\mathrm{K}_{\mathrm{sp}}$
So, precipitate will form.
Thus, silver chloride gets precipitated first.
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.