If 450 ml of some Pb(NO3)2 solution is mixed with 250 ml of 1.10 x 10-2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2forms? (Assume Ksp = 2.00 x 10-5 M at this temperature.) Enter the concentration in M.
PbCl2 <--------------> Pb 2+ + 2Cl-
Given, 250 mL of 1.10 x 10^-2 M NaCl
So, [Cl-] = 1.10 x 10^-2 M
Ksp = [Pb2+][Cl-]^2
Ksp = [x][2x]^2
2.00 x 10^-5 = (x) ( 2x + 1.10 x 10^-2)^2
2.00 x 10^-5 = (x) (4x^2 + 0.044x + 1.21 x 10^-4)
2.00 x 10^-5 = 4x^3 + 0.044x^2 + (1.21 x 10^-4)x
4x^3 + 0.044x^2 + (1.21 x 10^-4)x - (2.00 x 10^-5) = 0
Solving the polynomial equation we get
x = [Pb2+]= 0.0136 M
So, the concentration of Pb(MO3)2 must be less than 0.0136
M
Total volume = 250 + 450 = 700 mL = 0.700 L
Moles Pb2+ = 0.0136 M x 0.700 L = 9.52 x 10^-3 mol
concentration Pb(NO3)2 = 9.52 x 10^-3 mol / 0.450 L=0.0211 M
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