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InterviewSolution
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The solubility of `BaSO_(4)` in water is `2.42xx10^(-3)g L^(-1)` at 298 K. The value of its solubility product `(K_(sp))` will be (Given molar mass of `BaSO_(4)=233 g L^(-1)`)A. `1.08xx10^(-10) "mol"^(2)L^(-2)`B. `1.08xx10^(-12) "mol"^(2)L^(-2)`C. `1.08xx10^(-14) "mol"^(2)L^(-2)`D. `1.08xx10^(-8) "mol"^(2)L^(-2)` |
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Answer» Correct Answer - A Solubility (S) `=2.42xx10^(-3) g L^(-1)` `=(2.42xx10^(-3))/(233) "mol" L^(-1)` `=0.0104xx10^(-3) "mol" L^(-1)` `=1.04xx10^(-5) "mol" L^(-1)` `{:(BaSO_(4) ,hArr,Ba^(2+),+,SO_(4)^(2-)),(,,S,,S):}` `K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=SxxS=S^(2)` `=(1.04xx10^(-5) "mol" L^(-1))^(2)` `=1.08xx10^(-10) "mol"^(2)L^(-2)` |
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