Question

Calculate [ Hg₂²⁺ ] in saturated solutions of Hg₂Br2 in:

a. 0.100 M KNO₃

b. 0.001 M KBr

Note: K_sp for Hg₂Br₂ = 1.2 ⨯ 10^-18​

Answer 1

They didn't provide me with a ksp, so I looked it up and found it to be 1.30x10^-21.
Hg2Br2 <-> Hg2(2+) + 2Br-
Ksp = [Hg2(2+)][Br-]^2
KBr provides 0.001 M Br- because it is a strong acid.
1.30x10^-21 = [Hg2(2+)](0.001)^2
[Hg2(2+)] = 1.3E-15
The only thing I can see is that you probably didn't account for the activity coefficients (which they should have told you in the problem to account for), and you used the wrong ksp.

ksp Hg2Br2 = 5.6E-23, according to my book.

Accounting for activities:

Ksp = [Hg2++]γ(Hg2++) * [Br-]^2(γ(Br-))^2

To calculate the activities:

log(γ) = [(0.5 * (Zi)^2 * sqrt(I)) / (1 + sqrt(I))]

Where z is the charge and I is the ionic strength, which can be calculated by: 0.5(sum(c * z^2)).

So:

Ionic strength of Hg2Br2 is:
I = 0.5 (0.001*1^2 + 0.001*-1^2) = 0.001

γ(Hg2++) = 10^[-[0.5 * (2)^2 * sqrt(0.001)] / [1 + sqrt(0.001)]] = 0.868
γ(Br-) = 10^[-[0.5 * (1)^2 * sqrt(0.001)] / [1 + sqrt(0.001)]] = 0.965

5.6E-23 = [Hg2++](0.868) * (0.001)^2 * (0.965^2)

[Hg2++] = 6.9E-17 = ~7E-17