Question
An amount of 0.10 moles of AgCl(s) is added to one litre of water. Next, the crystals of NaBr are added until 75% of the AgCl is converted to AgBr(s), the less soluble silver halide. What is Br– at this point?
Ksp of AgCl = 2 × 10–10 and Ksp of AgBr = 4 × 10–13.
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Solution
$\underset{0.1 - 0.075}{AgCl} + \underset{x\text{ M}}{Br^{-}}(aq) \rightleftharpoons \underset{0.075}{AgBr}(s) + \underset{0.075\text{ M}}{Cl^{-}}(aq)$
$K_{eq} = \frac{\left\lbrack Cl^{-} \right\rbrack}{\left\lbrack Br^{-} \right\rbrack} = \frac{K_{sp}(AgCl)}{K_{sp}(AgBr)} \Rightarrow \frac{0.075}{\left\lbrack Br^{-} \right\rbrack} = \frac{2 \times 10^{- 10}}{4 \times 10^{- 13}} $$$\therefore\left\lbrack Br^{-} \right\rbrack = 1.5 \times 10^{- 4}\text{ M}$$
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