Question
A volume of 500 ml of 0.01 M – AgNO3 solution, 250 ml of 0.02 M – NaCl solution and 250 ml of 0.02 M – NaBr solution are mixed. The final concentration of bromide ion in the solution is (Ksp of AgCl and AgBr are 10−10 and 5 × 10−13, respectively.)
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Solution
$Ag^{+} + Cl^{-} \rightleftharpoons AgCl(s)$
$\frac{500 \times 0.01}{1000}\frac{250 \times 0.02}{1000} $$$= 5 \times 10^{- 3}M = 5 \times 10^{- 3}M$$
Eqn. $5 \times 10^{- 3} - (x + y)M\left( 5 \times 10^{- 3} - x \right)M$
$Ag^{+} + Br^{-} \rightleftharpoons AgBr(s) $$${= 5 \times 10^{- 3}M\text{=5} \times \text{1}\text{0}^{- 3}M }{\text{=5} \times \text{1}\text{0}^{- 3} - (x + y)M = \left( 5 \times 10^{- 3} - y \right)M}$$
As both reactions will tend towards completion, (x + y) = 5 × 10–3
Now, $\left\lbrack Ag^{+} \right\rbrack\left( 5 \times 10^{- 3} - x \right) = 10^{- 10}(1)$
$\Rightarrow \left\lbrack Ag^{+} \right\rbrack.y = 10^{- 10}$
and $\left\lbrack Ag^{+} \right\rbrack\left( 5 \times 10^{- 3} - y \right) = 5 \times 10^{- 13}(2)$
From (1)/(2), $\frac{y}{5 \times 10^{- 3} - y} = 200 \Rightarrow y = \frac{1}{201}$
$\therefore\left\lbrack Br^{-} \right\rbrack = 5 \times 10^{- 3} - y \approx 2.5 \times 10^{- 5}\text{ M}$
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