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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 :
$\mathrm{AgCl}$ will precipitate first as the amount of $\mathrm{Ag}^{+}$needed to precipitate is low.
$\mathrm{AgCl}$ precipitates first because its $\mathrm{K}_{\mathrm{sp}}$ is high.
$\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ precipitates first because the amount of $\mathrm{Ag}^{+}$needed is low.
$\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ precipitates first as its $\mathrm{K}_{\mathrm{sp}}$ is low.
Solution
$(i)$ $\left[\mathrm{Ag}^{+}\right]$required to $\mathrm{ppt} \mathrm{AgCl}(\mathrm{s})$
$\mathrm{Ksp}=\mathrm{IP}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=1.7 \times 10^{-10}$ $\left[\mathrm{Ag}^{+}\right]=1.7 \times 10^{-9}$
$(ii)$ $\left[\mathrm{Ag}^{+}\right]$required to $\mathrm{ppt} \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s})$
$\mathrm{Ksp}=\mathrm{IP}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{-2}\right]=1.9 \times 10^{-12}$ $\left[\mathrm{Ag}^{+}\right]=4.3 \times 10^{-5}$
$\left[\mathrm{Ag}^{+}\right]$required to $\mathrm{ppt} \mathrm{AgCl}$ is low so $\mathrm{AgCl}$ will $\mathrm{ppt}$ $\mathrm{1}^{\mathrm{st}}$.
