1.

The solubility of Ag_(2)CrO_(4)" at "25^(@)C" is "0.0332g dm^(-3). Calculate its solubility product. (At. Masses : Ag = 108, Cr = 52, O = 16).

Answer»

Solution :The solubility of `Ag_(2)CrO_(4)=0.0332g dm^(-3)`
= `(0.0332)/(332)"mol "dm^(-3)=10^(-4)"mol "dm^(-3)`
For the solubility EQUILIBRIUM of `Ag_(2)CrO_(4)hArr2Ag^(+)+CrO_(4)^(2-)`
`K_(sp)=[Ag^(+)]^(2)[CrO_(4)^(2-)]^(2s)`
= `(2s)^(2)(s)`
= `4s^(3)`
= `4(10^(-4))^(3)`
= `4xx10^(-12)("mol "dm^(-3))^(3)`


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