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`.

Options

A. 8.64 M
B. `2.16xx10^(-16)` M
C. `8.64xx10^(-16)` M
D. `1.203xx10^(-3)` M

Answer

D

Explanation

`Pb(OH)_2 harr underset(s)(Pb^(2+))+underset(2s)(2OH^(-))`
Solubility of `Pb(OH)_2` in water is `6.7xx10^(-6)M` `K_(sp)=(s)(2s)^2`
`K_(sq)=[Pb^(2+)][OH^(-)]=(6.7xx10^(-6))(2xx6.7xx10^(-6))^2`
`=6xx10^(-6)xx144xx10^(-12)`
`=1.203xx10^(-15)` M
Let solubility of `Pb(OH)_2` in buffer `sol^(n)=s mol//L`
`:. [Pb^(2+)][OH^(-)]^2=K_(sp)`
`[Pb^(2+)](10^(-6))^2=1.203xx10^(-15)`
So, `[Pb^(2+)]=(1.203xx10^(-15))//10^(-12)`
`[Pb^(2+)]=1.203xx10^(-3)` M