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The solubility product of `BaCl_(2)` is `3.2xx10^(-9)`. What will be solubility in mol `L^(-1)`A. `4xx10^(-3)`B. `3.2xx10^(-9)`C. `1xx10^(-3)`D. `1xx10^(-9)` |
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Answer» Correct Answer - C `BaCl_(2)hArrBa^(2+)+2Cl^(-)` `K_(sp)=[Ba^(2+)][Cl^(-)]^(2)=x xx(2x)^(2)=4x^(3)` `4x^(3)=3.2xx10^(-9)` `rArr" "x=9.28xx10^(-4)=0.928xx10^(-3)~~1xx10^(-3)` |
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