\(1 \mathrm{dm}^{3}\) 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?
\(\begin{aligned} {\left[\mathrm{K}_{\mathrm{sp}} \mathrm{Ag}_{2} \mathrm{CrO}_{4}\right.} &\left.=4 \times 10^{-12}\right] \\ {\left[\mathrm{K}_{\mathrm{sp}} \mathrm{AgCl}\right.} &\left.=1 \times 10^{-10}\right] \end{aligned}\)

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.