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QUESTION

# What is the solubility (in M) of PbCl2 in a 0.15 M solution of HCl? The Ksp of PbCl2 is 1.6 × 10-5?

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?