The precipitate of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(\mathrm{~K}_{\mathrm{sp}}=1.9 \times…

The precipitate of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(\mathrm{~K}_{\mathrm{sp}}=1.9 \times 10^{-12}ight)$ is obtained when equal volumes of the following are mixed.
  1. $10^{-4} \mathrm{M} \mathrm{Ag}^{+}+10^{-4} \mathrm{M} \mathrm{CrO}_{4}^{2-}$
  2. $10^{-2} \mathrm{M} \mathrm{Ag}^{+}+10^{-3} \mathrm{M} \mathrm{CrO}_{4}^{2-}$
  3. $10^{-5} \mathrm{M} \mathrm{Ag}^{+}+10^{-3} \mathrm{M} \mathrm{CrO}_{4}^{2-}$
  4. $10^{-4} \mathrm{M} \mathrm{Ag}^{+}+10^{-5} \mathrm{M} \mathrm{CrO}_{4}^{2-}$

Solution

Precipitation occurs when the ionic product exceeds $\mathrm{K}_{\mathrm{sp}}$ value. When equal volumes of two solutions are mixed, the concentration of each is reduced to half. Therefore, In first case, Ionic product,
I.P. $=\left(\frac{1}{2} \times 10^{-4}ight)^{2}\left(\frac{1}{2} \times 10^{-4}ight)=\frac{1}{8} \times 10^{-12}$
$=1.25 \times 10^{-13}$
As, I.P. $ < \mathrm{K}_{\text {sp }}$
$\therefore$ No precipitation occurs.
In second case,
I.P. $=\left(\frac{1}{2} \times 10^{-2}ight)^{2}\left(\frac{1}{2} \times 10^{-3}ight)$
$=\frac{1}{8} \times 10^{-7}=1.25 \times 10^{-8}$
As, I.P. $>\mathrm{K}_{\mathrm{sp}}$
$\therefore$ Precipitation occurs.
In third case,
I.P. $=\left(\frac{1}{2} \times 10^{-5}ight)^{2}\left(\frac{1}{2} \times 10^{-3}ight)$
$=\frac{1}{8} \times 10^{-13}=1.25 \times 10^{-14}$
As, I.P. $ < \mathrm{K}_{s p}$
$\therefore$ No precipitation occurs.
In fourth case,
I.P. $=\left(\frac{1}{2} \times 10^{-4}ight)^{2}\left(\frac{1}{2} \times 10^{-5}ight)$
$=\frac{1}{8} \times 10^{-13}=1.25 \times 10^{-14}$
As, I.P. $ < \mathrm{K}_{\mathrm{sp}} \therefore$ No precipitation occurs.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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