What is the solubility of `PbSO_(4)` in `0.01M Na_(2)SO_(4)` solution if `K_(sp)` for `PbSO_(4) = 1.25 xx 10^(-9)`?

Options

A. `1.25 xx 10^(-7) mol L^(-1)`
B. `1.25 xx 10^(-9)mol L^(-1)`
C. `1.25 xx 10^(-10) mol L^(-1)`
D. `0.10 mol L^(-1)`

Answer

A

Explanation

`{:(PbSO_(4)hArr,Pb^(2+)+,SO_(4)^(2-),,),(,S,S,,):}`
`{:(,Na_(2)SO_(4)rarr,2Na^(o+)+,SO_(4)^(2-)),("Initial",0.01,0,0),("Final",0,2xx0.01,0.01):}`
(Since `Na_(2)SO_(4)` is salt of `S_(A)//S_(B)` it is completely ionised).
`:. SO_(4)^(2-) =(x + 0.01) = 0.01` (Since `x` is very small)
`:. K_(sp) pf PbSO_(4) = [Pb^(2+)] [SO_(4)^(2-)]`
`1.25 xx 10^(-9) =x xx 0.01`
`:. x = (1.25 xx 10^(-9))/(0.01) = 1.25 xx 10^(-7) mol L^(-1)`