JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
A solution is \(0.1\, \mathrm{M}\) in \(\mathrm{Cl}^{-}\)and \(0.001\, \mathrm{M}\) in \(\mathrm{CrO}_{4}{ }^{2-}\). Solid \(\mathrm{AgNO}_{3}\) is gradually added to it. Assuming that the addition does not change in volume and \(\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1.7 \times 10^{-10} \,\mathrm{M}^{2}\) and \(\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right)=1.9 \,\times 10^{-12} \mathrm{M}^{3}\) Select correct statement from the following :
- A \(\mathrm{AgCl}\) will precipitate first as the amount of \(\mathrm{Ag}^{+}\)needed to precipitate is low.
- B \(\mathrm{AgCl}\) precipitates first because its \(\mathrm{K}_{\mathrm{sp}}\) is high.
- C \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) precipitates first because the amount of \(\mathrm{Ag}^{+}\)needed is low.
- D \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) precipitates first as its \(\mathrm{K}_{\mathrm{sp}}\) is low.
Answer & Solution
Correct Answer
(A) \(\mathrm{AgCl}\) will precipitate first as the amount of \(\mathrm{Ag}^{+}\)needed to precipitate is low.
Step-by-step Solution
Detailed explanation
\((i)\) \(\left[\mathrm{Ag}^{+}\right]\)required to \(\mathrm{ppt} \mathrm{AgCl}(\mathrm{s})\) \(\mathrm{Ksp}=\mathrm{IP}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=1.7 \times 10^{-10}\) \(\left[\mathrm{Ag}^{+}\right]=1.7 \times 10^{-9}\) \((ii)\)…
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