Question
The solubility of CaCO3 is 7 mg/litre. Calculate the solubility product of BaCO3 from this information and from the fact that when Na2CO3 is added slowly to a solution containing equimolar concentration of Ca2+ and Ba2+, no precipitate of CaCO3 is formed until 90% of Ba2+ has been precipitated as BaCO3.
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Solution
$K_{sp}\text{ of CaC}\text{O}_{3} = \left( \frac{7 \times 10^{- 3}}{100} \right)^{2} = 4.9 \times 10^{- 9}$
Let initial [Ca2+] = [Ba2+] = x M.
From question final [Ba2+] = $x \times \frac{10}{100} = 0.1xM$
But $\left\lbrack CO_{3}^{2 -} \right\rbrack$is same for both. Hence,
$\frac{K_{sp}\left( BaCO_{3} \right)}{K_{sp}\left( CaCO_{3} \right)} = \frac{\left\lbrack Ba^{2 +} \right\rbrack\left\lbrack CO_{3}^{2 -} \right\rbrack}{\left\lbrack Ca^{2 +} \right\rbrack\left\lbrack CO_{3}^{2 -} \right\rbrack} \Rightarrow \frac{K_{sp}\left( BaCO_{3} \right)}{4.9 \times 10^{- 9}} = \frac{0.1x}{x} \therefore K_{sp}\left( BaCO_{3} \right) = 4.9 \times 10^{- 10}$
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