Question

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 1

Correct Answer - A::B::C
$Pb(OH)(2) hArr Pb^(2+) +2overset(Theta)OH$
$K(sp) for Pb(OH)(2) = [Pb^(2+)] [overset(Theta)OH]^(2)$
$= 6.7 xx 10^(-6) (13.4 xx 10^(-6))^(2)$
$- 1203.05 xx 10^(-18)$
$pH$ 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 $[Pb^(2+)] = 1.203 xx 10^(-3)$