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If te solubility product of `BaSO_(4)` is `1.5 xx 10^(-9)` in water, its solubility in moles per litre, isA. `1.5 xx 10^(-9)`B. `3.9 xx 10^(-5)`C. `7.5 xx 10^(-5)`D. `1.5 xx 10^(-5)` |
Answer» Correct Answer - B `BaSO_(4) hArr Ba^(2+) + SO_(4)^(--)` Solubility constant `= S xx S` `1.5 xx 10^(-19) = S^(2), S = sqrt(1.5 xx 10^(-19)), S = 3.9 xx 10^(-5)`. |
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