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The solubility product of BaSO_4 is 1.1 xx 10^(-10) at 298 K temp. calculate its water solubility. |
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Answer» Solution :`BaSO_(4(s)) hArr Ba_((aq))^(2+) + SO_(4(aq))^(2-)` If `S=BaSO_4` Solubility of `BaSO_4` in mol `L^(-1)` `THEREFORE K_(SP)=S^2=1.1xx10^(-10)` `therefore S=sqrt(1.1xx10^(-10))` `=1.049xx10^(-5) "mol L"^(-1)` |
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