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The solubility product constant of Ag(2)...

The solubility product constant of `Ag_(2)CrO_(4)` and `AgBr` are `1.1 xx 10^(-12)` and `5.0 xx 10^(-13)` respectively. Calculate the ratio of the molarities of their saturated solutions.

Text Solution

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Let x mol/L and y mol/L be the solubilities of `Ag_(2)CrO_(4)` and AgBr respectively.
`Ag_(2)CrO_(4) to 2Ag^(+) + CrO_(4)^(2-), K_(sp)=1.1xx10^(-12)`
2x x
`K_(sp)=[Ag^(+2)][CrO_(4)^(2-)]=(2x)^(2).x=4x^(3) implies x^(3)=K_(sp)/4 implies x=6.503xx10^(-5)M`
`AgBr to Ag^(+) + Br^(-)`,`K_(sp)=5.0xx10^(-12)`
y y
`K_(sp)=[Ag^(+)][Br^(-)]=y.y=y ^(2) implies y=7.07xx10^(-7)m`
The ratio of solubilities `=x/y = (6.5xx10^(-5))/(7.07xx10^(-7)) approx 92`
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