Question

Find the concentration of Hg22 in 0.10 M KI saturated with Hg2I2. Include activity coefficients in your solubility-product expression. The Ksp of Hg2I2 is 4.6× 10–29.

Answer 1

the mecuryiodide will dissolve and dissociate as:
Hg₂I₂(s) -> Hg₂²*(aq) + 2 I⁻(aq)

the solubility product and activity are related as
Ksp = a_Hg₂²*∙ ( a_I⁻ }²
Given ; Ksp = 4.6×10⁻²⁹

we know that that the activity of an ion in an electrolyte solution = activity coefficient X concentration
Ksp = γ_Hg₂²* ∙ [Hg₂²*] ∙ (γ_I⁻)² ∙[I⁻]²

Now ionic strength is
I = (1/2) ∙ ∑ c_i ∙ (z_i)²
(with c_i ion concentration, z_i charge of the ion)

we can neglect the amount of dissolved mercury iodide, So the ionic strength is determined by the dissolved potassium iodide.
I = (1/2) X ( [K⁺]∙(+1)² + [I⁻]∙(-1)²) = (1/2) X( 0.1M X1 + 0.1M X1 ) = 0.1 M

As we have weak electrolyte we can use Debye Huckel equation
log₁₀(γ_i) = - A ∙ (z_i)² ∙√I / (1 + B∙a∙√I)

For water at 298K
A = 0.51 (in √(L/mol) )
B = 3.29 ( √(L/mol)∙nm⁻¹ )

The parameter a is the effective ionic diameter, and constant for chemicals
a_I⁻ = 300 pm = 0.3 nm
a_Hg₂²* = 400 pm = 0.4 nm

So the acidity of an iodide is:
γ_I⁻ = 10^{ - A X (z_I⁻)² X√I / (1 + BXa_I⁻X√I) }
= 10^{ - 0.51 X (-1)² X√0.1 / (1 + 3.29X0.3X√01) }
= 0.7535
and the activity of mercury ions is
γ_Hg₂²* = 10^{ - A ∙ (z_Hg₂²⁺)² X √I / (1 + B Xa_Hg₂²⁺X√I) }
= 10^{ - 0.51 X (+2)² X√0.1 / (1 + 3.29 X 0.4 X √01) }
= 0.3503

According to solubility product equation the concentration of mercury (I) ions is given by:
[Hg₂²*] = Ksp / ( γ_Hg₂²* ∙(γ_I⁻)² ∙[I⁻]² )

the raise of iodide ion concentration due to saturation of the solution with Hg₂I₂ may be ignored, that means
[I⁻] = 0.1 M

Hence,
[Hg₂²⁺] = 4.6×10⁻²⁹ / ( 0.3503 ∙ (0.7535)² ∙ (0.1)² ) = 2. 31×10⁻²⁶ M