1.

K_(sp) of Ag_2 CrO_4 is 1.1xx10^(-12). What is solubility of Ag_2 CrO_4 in ox.1M K_2 CrO_4.

Answer»

Solution :`Ag_2 CrO_4 hArr 2AG^(+)+CrO_(4)^(2-)`
x is the solubility of `Ag_2 CrO_4` in `0.1 M K_2 CrO_4`
`UNDERSET(0.1M)(K_2 CrO_4) hArr underset(0.2 M)( 2K^+)+underset(0.1 M)(CrO_(4)^(2-))`
`[Ag^+]=2x`
`[CrO_(4)^(2-)]=(x+0.1)=0.1""because xlt LT 0.1`
`K_(sp)=[Ag^(+)^(2)[CrO_(4)^(2-)]`
`1.1xx10^(-12)=(2x)^2(0.1)`
`1.1xx10^(-12)=0.4x^2`
`x^2=(1.1xx10^(-12))/(0.4)`
`x=sqrt((1.1xx10^(-12))/(0.4))`
`x=sqrt(2.75xx10^(-12))`
`x=1.65xx10^-5M`.


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