Question

c. Consider lead iodide, Pblz. Ignoring ion pairing, but taking to activity into account, what is the concentration of Pb2+ in a sodium nitrate solution with an ionic strengths of 0.01 M? (3 pts)

Answer 1

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 = a_Pb2+·a_I-²

The activity a_i of an ion i is: a_i = γ_i·[i], where γ_i is the activity coefficient and [i] the ion concentration. Then (2) becomes

(3)         Ksp = γ_Pb2+· [Pb²⁺]·γ_I-² [I^-]²

Now, we have to calculate both γ_i and look for the value of Ksp.

The expression that we use to determine the activity coefficient is

(4)        

log Yi = -0.51.z; VT 1+7

Where z_i is the electrical charge of the ion i, and I is the ionic strength of the solution (0.01 M). Therefore:

(5a)      

log YPb2+ -0.51.2b2+ V7 1+V7 -0.51.(+2)2 . 0.01 1+0.01 -0.1855

(5b)      l

-0.51.(-1)2 . 0.01 logy:- -0.51.zi-VT 1+7 -0.04636 1+0.01

Calculating γ_i:

(6a)       γ_Pb2+ = 10^-0.1855 = 0.652

(6b)      γ_Pb2+ = 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)     

3 1.4.10-8 C= 0.00188 M = [Pb2+] (from equation 8a) V 2.108

Hence, this is the concentration of Pb²⁺ in the solution:

[Pb²⁺] = 0.00188 M