Question

Find the concentration of Hg22+ in 0.0050 M KI saturated with Hg2I2. Include actiivity coefficients in your solubility-product ecpression. The Ksp of Hg2I2 is 4.6x10^-29.

Answer 1

The dissociation of Hg₂I₂ will be

Hg₂I₂ ---> Hg₂⁺² + 2I^-

The solubility product = Ksp = 4.6x10^-29 = activity coeffecient X Hg₂⁺² [Hg₂⁺²] X activity coeffecient X I^- [I^-]²

Let us represent activity coeffecient = y
The ionic strength = 1/2 X sum of ( product of concentration and charge on ions)
So ionic strenght of mercury iodide will be calcualted from KI
I = (1/2) X ( [K⁺]∙(+1)² + [I⁻]X (-1)²) = (1/2) ∙ ( 0.1 + 0.1) = 0.2 / 2 = 0.1

According to Debye Huckel equation, the activity will be
log₁₀(activity) = - A X (charge)² X √I / (1 + BX aX √I)

The parameters A and B depend on the solvent and its temperature. For water
A = 0.51
B = 3.29
The parameter a is the effective ionic diameter. These are standard values are equal to
a_I⁻ = 300 pm = 0.3 nm
a_Hg₂⁺²= 400 pm = 0.4 nm

So the activity of an iodide is:
acitivity I⁻ = 10^{ - A X(Charge of I⁻)² ∙√I / (1 + B∙a_I⁻∙√I)
= 0.754
and the activity of mercury ions is
activity of Hg₂²* = 10^{ - A ∙ (Charge of Hg₂²⁺)² ∙√I / (1 + B∙a_Hg₂²⁺∙√I) }
= 10^{ - 0.51 ∙ (+2)² ∙√0.1 / (1 + 3.29 X 0.4X √01) }
= 0.350

[Hg₂⁺²] = Ksp / ( y Hg₂⁺² X (yI⁻)² ∙[I⁻]² )

[I⁻] = 0.1 M

Hence,
[Hg₂⁺²] = 4.6 X 10^-29 / ( 0.350 X (0.754)² ∙ (0.1)² ) = 2. 32×10^-26 M