1.0 L of solution which was in equilibrium with solid mixture of AgCl and ${Ag}_{2} {CrO}_{4}$ was found to contain 1 x10^-4 moles of Ag⁺ ions, 1.0 x 10^-6 moles of Cl^- ions and 8.0 x 10^-4moles of ${CrO}_{4}^{2 -}$ ions. At constant volume, Ag⁺ ions are added slowly to the above mixture till 8.0 x 10^-7 moles of AgCl got precipitated. How many moles of ${Ag}_{2} {CrO}_{4}$ were also precipitated? Give your answer after multiplying with 10⁶

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Detailed Solution

$\begin{array}{l} K_{s p} (AgCl) = 10^{- 4} \times 10^{- 6} = 10^{- 10} \\ K_{s p} ({Ag}_{2} {CrO}_{4}) = {(10^{- 4})}^{2} \times 8 \times 10^{- 4} = 8 \times 10^{- 12} \end{array}$

No. of moles of Cl^- remaining in solution

$\begin{array}{l} = 10 \times 10^{- 7} - 8 \times 10^{- 7} \\ = 2 \times 10^{- 7} \\ \Rightarrow [{Ag}^{+}] = \frac{10^{- 10}}{2 \times 10^{- 7}} = 5 \times 10^{- 4} \\ ∴ [{CrO}_{4}^{2 -}] = \frac{8 \times 10^{- 12}}{25 \times 10^{- 8}} = 3.2 \times 10^{- 5} \end{array}$