Question

A solution is 0.1 M in \(\mathrm{Cl^-}\) and 0.001 M in \(\mathrm{CrO_4^{2-}}\). Solid \(\mathrm{AgNO_3}\) is gradually added to it. Assuming that the addition does not change in volume and \(K_{sp}(\mathrm{AgCl}) = 1.7\times10^{-10}\ \mathrm{M^2}\) and \(K_{sp}(\mathrm{Ag_2CrO_4}) = 1.9\times10^{-12}\ \mathrm{M^3}\).
Select correct statement from the following :

• A. AgCl precipitates first because its \(K_{sp}\) is high.
• B. \(\mathrm{Ag_2CrO_4}\) precipitates first as its \(K_{sp}\) is low.
• C. \(\mathrm{Ag_2CrO_4}\) precipitates first because the amount of \(\mathrm{Ag^+}\) needed is low.
• D. AgCl will precipitate first as the amount of \(\mathrm{Ag^+}\) needed to precipitate is low.

Hint:

Compare the \([\mathrm{Ag^+}]\) needed to start precipitation of each salt; the salt needing less \(\mathrm{Ag^+}\) precipitates first (not simply the one with lower \(K_{sp}\)).

Answer 1

The Correct Option is D

For AgCl: \([\mathrm{Ag^+}] = \frac{K_{sp}}{[\mathrm{Cl^-}]} = \frac{1.7\times10^{-10}}{0.1} = 1.7\times10^{-9}\) M.
For \(\mathrm{Ag_2CrO_4}\): \([\mathrm{Ag^+}] = \sqrt{\frac{K_{sp}}{[\mathrm{CrO_4^{2-}}]}} = \sqrt{\frac{1.9\times10^{-12}}{10^{-3}}} = \sqrt{1.9\times10^{-9}} = 4.4\times10^{-5}\) M.
AgCl needs a much smaller \([\mathrm{Ag^+}]\), so AgCl precipitates first. The reason is the lower \(\mathrm{Ag^+}\) concentration required, not the \(K_{sp}\) value alone (the two salts have different stoichiometries so \(K_{sp}\) values are not directly comparable).
Answer: (4) AgCl will precipitate first as the amount of \(\mathrm{Ag^+}\) needed to precipitate is low.