Question
After solid SrCO3 was equilibrated with a buffer at pH 8.6, the solution was found to have [Sr2+] = 2.0 × 10−4 M, what is the Ksp of SrCO3? (Ka2 for H2CO3 = 5.0 × 10−11, log 2 = 0.3)
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Solution
$SrCO_{3}(s) \rightleftharpoons \underset{2 \times 10^{- 4}\text{ M}}{Sr^{2 +}} + \underset{2 \times 10^{- 4} - x}{CO_{3}^{2 -}}$
$\underset{\left( 2 \times 10^{- 4} - x \right)\text{ M}}{CO_{3}^{2 -}} + H_{2}O \rightleftharpoons \underset{x\text{ M}}{HCO_{3}^{-}} + \underset{4 \times 10^{- 6}\text{ M}}{OH^{-}} $$${\text{Now, }\frac{10^{- 14}}{5 \times 10^{- 11}} = \frac{x \times 4 \times 10^{- 6}}{\left( 2 \times 10^{- 4} - x \right)} \Rightarrow x = \frac{0.01}{51} }{\therefore K_{sp} = \left( 2 \times 10^{- 4} \right) \times \left( 2 \times 10^{- 4} - x \right) = \frac{4}{51} \times 10^{- 8}}$$
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