Reaction:

##PbCl_2 rightleftharpoons Pb^(2+) + 2Cl^-##; ##K_(sp) = 1.6 xx 10^(-5)##

Solubility of ##PbCl_2## ##=## ##7.11 xx 10^(-4)## ##mol*L^(-1)##.

Now here ##K_(sp) = [Pb^(2+)][Cl^-]^2##. In pure water we would write ##K_(sp) = (S)(2S)^2 = 4S^3## where ##S## is the solubility of lead chloride.

However here the concentration of chloride anion has been (artificially) increased by the presence of hydrochloric acid which gives stoichiometric quantities of ##Cl^-##. So ##[Cl^-]## ##=## ##(0.15*mol*L^(-1)## + the twice the solubility of lead chloride##)##.

Our revised ##K_(sp)## expression is ##K_(sp) = {S}{0.15+2S}^2##. Now we can (reasonably) make the approximation ##{0.15+2S} ~= 0.15##; we have to justify this approximation later.

So now ##K_(sp) ~= [S][0.15]^2## and ##S = K_(sp)/(0.15)^2## ##=## ##(1.6xx10^(-5))/(0.15)^2## ##=## ##7.11 xx 10^(-4)## ##mol*L^(-1)##.

This value is indeed small compared to 0.15. We could make a 2nd approximation and plug this ##S## value back in the first equation but I am not going to bother. Would the solubility of lead chloride be greater in pure water? Why or why not?