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The units of solubility product of silver chromate `(AgCrO_(4))` will beA. `mol^(2)L^(-2)`B. `mol^(3)L^(-3)`C. `mol L^(-1)`D. `mol^(-1)L` |
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Answer» Correct Answer - B The solubility product is an equilibrium constant. `Ag_(2)CrO_(4)(s)hArr2Ag^(+)(aq.)+CrO_(4)^(2-)(aq.)` `K_(sp) = C_(Ag^(+))^(2)C_(CrO_(4)^(2-))` `= (mol L^(-1))^(2)(mol L^(-1))= mol^(3)L^(-3)` |
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