What is molar solubility of lead(II) chloride?

Question

In a 0.15 M solution of hydrochloric acid at 25degC??????

Answer 1

To solve the problem I used a reaction table.

.................. PbCl2(s)....... <--> ....... Pb2+(aq)........ 2Cl-(aq)

Initial............... - ................................... 0.................... .15M

Reacting......... - ................................. +x .................... +2x

Equilibrium..... - ................................... x .................. .15+2x

[Pb2+] = x  [Cl-] = .15+2x

Ksp = [Pb2+][Cl-]^2

1.7*10^-5 = (x)(.15+2x)^2

1.7*10^-5 = x(.0225 + .6x + 4x^2)

0= -1.7*10^-5 + .0225x + .6x^2 + 4x^3

This problem has imaginary and irrational roots. To solve this equation you need some sort of computer (I used a TI 86) because the mathematics needed to solve it go beyond the scope of my study. My answer was:  

x= 7.41*10^-4

x = [Pb2+] = [PbCl2] = 7.41*10^-4M