Calculate `[Pb^(2+)]` in a 1M HCl solution that is saturated in PbS. `K_(a) (H_(2)S)=1.1 xx 10^(-21), K_(sp) (PbS)=8 xx 10^(-28)`

Explanation

We have
`PbS (s) Leftrightarrow Pb^(2+) +S^(2-), K_(sp)=8 xx 10^(-27)`
`and 2H^(+) +S^(2-) =H_(2)S, K.=(1)/(K_(a) (H_(2)S))=(1)/(1.1 xx 10^(-21))`
On adding we get
`PbS (s)+overset(1)(2H^(+)) Leftrightarrow underset(x)(Pb^(2+)) +underset(x)(H_(2)S)`
for which applying Eqn. 7, (chapter on Chemical Equilibrium) we get
`K=([Pb^(2+)] [H_(2)S])/([H^(+)]^2)=K_(sp) xx K.=(8 xx 10^(-28))/(1.1 xx 10^(-21))=7.3 xx 10^(-7)`
`or (x.x)/((1-2x)^2)=7.3 xx 10^(-7)`
Suppose x to be very small,
we get `x=[Pb^(2+)]=8.55 xx 10^(-4)M`