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?

1 Answer
Nov 9, 2015

Answer:

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#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)#.

Explanation:

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?