How many grams of PbCl2 will disslove in 0.803 L of a 0.670 M Pb(NO3)2 solution ?

Question

Ksp = 1.70e−5

Answer 1

This is a common ion problem.
Let's find first the molar solubility of PbCl2 and then we can use that to find the g
Assume that the molar solubility is s mole/L

.. .. .. .. .. .. PbCl2 <=> Pb(+2) + 2Cl(-)
Initial .. .. .. .. .s .. .. .. .. 0.670
Dissolve .. .. s
Produce .. .. .. .. .. .. .. .. .. s .. .. .. 2s
At Equil .. .. .. .. .. .. .. .. 0.670+x .. 2s

Ksp= [Pb(+2)][Cl-]^2 = (0.670+s)(2s)^2

Let's assume that 0.670 >>s so that 0.670+s=0.670
Then the equation is simplified to
Ksp=0.670*4s^2 =>
s= squareroot (Ksp/2.68) =>
s= SQRT((1.7*10^-5)/2.68) =2.52*10^-3 Our assumption is fair enough

mole= M*V
also mole= mass/MW
Thus M*V= mass/MW =>
mass= M*V*MW= 2.52*(10^-3)*0.803*278.2 =0.563 g

Answer 2

Let x =mol/L PbCl2 that dissolve to estabilish the equilibrium.This will produce x mol/L Pb2+ and 2x mol/L Cl-.

But there is already some Pb2+ in the solution corresponding to 0.670 M Pb(NO3)2 that is a strong electrolyte so we have 0.670 Pb2+

Total concentration Pb2+ = 0.670+x . x is small compared to 0.670 and we can assume Pb2+=0.670

Ksp= 1.70e-5 = (Pb2+)(Cl-)^2 =0.670(2x)^2=0.670(4x^2)

x=0.00252

To get grams per 0.803 L

0.00252(0.803 L)(278.1 g/mol)= 0.563 g