Question
A volume of 1.0 L of solution which was in equilibrium with solid mixture of AgCl and Ag2CrO4 was found to contain 1 × 10−4 moles of Ag+ ions, 1.0 × 10−6 moles of Cl− ions and 8.0 × 10−4 moles of CrO42− ions. The Ag+ ions are added slowly to the above mixture (keeping the volume constant) till 8.0 × 10−7 moles of AgCl got precipitated. How many moles of Ag2CrO4 were precipitated simultaneously?
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Solution
Ksp (AgCl) = $10^{- 4} \times 10^{- 6} = 10^{- 10}$
Ksp(AgCrO4) = $\left( 10^{- 4} \right)^{2} \times 8 \times 10^{- 4} = 8 \times 10^{- 12}$
After precipitation of AgCl, find the concentration of $Cl^{-}.\left\lbrack Cl^{-} \right\rbrack_{\text{final}} = 1.0 \times 10^{- 6} - 8 \times 10^{- 7} \times 2 \times 10^{- 7}\text{ M}$
Now, $\frac{\left\lbrack CrO_{4}^{2 -} \right\rbrack_{\text{final}}}{\left\lbrack Cl^{-} \right\rbrack_{\text{final}}^{2}} = \frac{K_{sp}\left( AgCrO_{4} \right)}{K_{sp}^{2}(AgCl)}$
$\frac{\left\lbrack CrO_{4}^{2 -} \right\rbrack_{\text{final}}}{\left( 2 \times 10^{- 7} \right)^{2}} = \frac{8 \times 10^{- 12}}{\left( 10^{- 10} \right)^{2}} \Rightarrow \left\lbrack CrO_{4}^{2 -} \right\rbrack_{\text{final}} = 3.2 \times 10^{- 5}\text{ M}$
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