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Question
The solubility product (Ksp) of BaSO4 is 1.5 X 10-9.Calculate the solubility of barium sulphate in pure water and in 0.1 M BaCl2.
Solution
(i) Let the solubility of BaSO4 in pure water be x mol/L
In pure water `K_"sp" = [Ba^(2+)][SO_4^(2-)]`
`= x xx x`
`1.5 xx 10^(-9) = x^2`
`x = 3.87 xx 10^(-5) mol//L`
(ii) Let y be the solubility of BaSO4 in 0.1 BaCl2 solution.
Therefore,
`[Ba^(2+)] = 0.1 + y ≈ 0.1 mol//L`
`[SO_4^(2-)] = y mol//L`
Now,
`K_(sp) = [Ba^(2+)][SO_4^(2-)]`
`1.5 x 10^(-9) = 0.1 xx y`
`y = 1.5 x 10^(-8) mol//L`
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