The solubility of `Pb(OH)_(2)` in water is `6.7xx10^(-6)`M. Calculate the solubility of `Pb(OH)_(2)` in a buffer solution of `pH=8`.

Answer

A, B, C

Explanation

`Pb(OH)_(2) hArr Pb^(2+) +2overset(Theta)OH`
`K_(sp) for Pb(OH)_(2) = [Pb^(2+)] [overset(Theta)OH]^(2)`
`= 6.7 xx 10^(-6) (13.4 xx 10^(-6))^(2)`
`- 1203.05 xx 10^(-18)`
`pH` of buffer solution is `8`.
`pOH = 14 - 8 =16`
`[overset(Theta)OH] = 10^(-6)`
`Pb(OH)_(2) hArr Pb^(2+) +2 overset(Theta)OH`
`K_(sp) = [Pb^(2+)] xx [overset(Theta)OH]^(2)`
or `1.203 xx 10^(-15) = [Pb^(2+)] xx [10^(-6)]^(2)`
or `[Pb^(2+)] = 1.203 xx 10^(-3)`