Problem statement
In this exercise we have to apply the expression of the ionic strength of a salt (NaNO3) in solution. This is because the presence of a salt decreases the attraction between Pb2+ and I- and the concentration of Pb2+ then it is altered.
The starting point is that we have PbI2 that is added to a solution of NaNO3, with an ionic strength I = 0.01 M. We want to know the concentration of Pb2+ in this solution taking into account the activities of the ions.
Development
The solubility equation for PbI2 in solution is:
(1) PbI2 → Pb2+ + 2 I-
The expression for the equilibrium constant (Ksp) with activities (if we wouldn’t use activities, they could be changed by concentrations) is:
(2) Ksp = aPb2+·aI-2
The activity ai of an ion i is: ai = γi·[i], where γi is the activity coefficient and [i] the ion concentration. Then (2) becomes
(3) Ksp = γPb2+· [Pb2+]·γI-2 [I-]2
Now, we have to calculate both γi and look for the value of Ksp.
The expression that we use to determine the activity coefficient is
(4)
Where zi is the electrical charge of the ion i, and I is the ionic strength of the solution (0.01 M). Therefore:
(5a)
(5b) l
Calculating γi:
(6a) γPb2+ = 10-0.1855 = 0.652
(6b) γPb2+ = 10-0.04636 = 0.899
The Ksp value for PbI2 is (you can find it easy on the Internet):
(7) Ksp = 1.4 x 10-8
And now, from (1), we see that the concentrations of Pb2+ and I- are (for each atom of Pb2+ there are two atoms of I-):
(8a) [Pb2+] = c
(8b) [I-] = 2c
If we substitute the values and expressions of (6a), (6b), (7), (8a) and (8b) in (3), we have:
(9) 1.4 x 10-8 = 0.652·c·0.8992·(2c)2
1.4 x 10-8 = 2.108·c3
And, finally,
(10)
Hence, this is the concentration of Pb2+ in the solution:
[Pb2+] = 0.00188 M
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