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The solubility product of BaSO4 is 1.5 x 10-9 . Find out its solubility in pure water. |
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Answer» BaSO4 = Ba2+ + SO42- Ks = (S)2 = 1.5 x 10-9 S = \(\sqrt{1.5\times10^{-9}}\) = \(\sqrt{1.5\times10^{-10}}\) = 3.87 x 10-15 mol L-1 |
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