In: Chemistry
The molar solubility of lead hydroxide in a 0.240 M lead nitrate solution is _____M.
Pb(OH)2 <-> Pb+2 + 2OH-
if
[Pb+2] = 0.24 (from Pb(NO3)2)
Ksp= 1.2×10–15
Ksp = [Pb+2][OH-]^2
1.2*10^-15 = (0.24)(2S)^2
S = ((1.2*10^-15)/(0.24)/4)^(1/2)
S = 3.53553*10^-8 M