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The solubility product of `Ag_(2)CrO_(4)` is `32 xx 10^(-12)`. What is the concentration of `CrO_(4)^(-)` ions in that solutionA. `2 xx 10^(-4) M`B. `16 xx 10^(-4) M`C. `8 xx 10^(-4) M`D. `8 xx 10^(-8) M` |
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Answer» Correct Answer - A `underset(S)(Ag_(2)CrO_(4)) rarr underset(2S)(2Ag^(+)) + underset(S)(CrO_(4)^(--))` `K_(sp) = (2S)^(2)S = 4S^(3)` `S = ((K_(sp))/(4))^((1)/(3)) = ((32 xx 10^(-12))/(4))^((1)/(3)) = 2 xx 10^(-4) M`. |
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