Question

# If 450 ml of some Pb(NO3)2 solution is mixed with 400 ml of 1.10 x 10-2...

If 450 ml of some Pb(NO3)2 solution is mixed with 400 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 PbCl2 forms? (Assume Ksp = 2.00 x 10-5 M at this temperature.) Enter the concentration in M.

Ksp = 2 X 10^-5 For PbCl2

PbCl2 --> Pb+2 + 2Cl-

Ksp = [Pb+2] [ Cl-]2

2 X 10^-5 = [Pb+2] [ Cl-]2

Now moles of NaCl added = molarity of NaCl X volume of NaCl in Litres = 1.1 X 10^-2 X 0.4 = 0.0044 moles

Total volume of solution = 450 + 400 = 850 mL = 0.85 L

So concentration of Chloride ions due to NaCl = 0.0044 / 0.85 = 0.00517 molar

Now let" a" moles of PbCl2 dissolved so it will give a moles of Pb+2 and 2a moles of Cl-

2.00 x 10-5 = [a] [2a + 0.00517]

2.00 x 10-5 = 2a2 + 0.00517a

on solving for a

2a2 + 0.00517a - 2X10^-5 = 0

a = 0.00212

So moles of Pb+2 = Molarity of Pb+2 X total volume = 0.00212 X 0.85 = 0.0018

so concentration of Pb+2 required = Moles / volume in litres = 0.0018 / 0.45 = 0.004 Molar

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