Question

Hello, this is for my analytical chemistry class. How do you solve the following question: Calculate the solubility of Pb(IO₃)₂ in a solution of 0.020M Mg(NO₃)₂ using activities. Assume that since Pb(IO₃)₂ is only sparingly soluble that it does not contribute to the ionic strength. Please use the extended Debye-Huckel equation. α_Pb2+ = 450pm α_IO3-​ = 450pm Pb(IO₃)₂ K_sp = 2.5*10^-13

Thank you.

Answer 1

Since Pb(IO3)2 is only sparingly soluble, we will assume that we can ignore its contribution to the ionic strength; thus

It is possible to calculate activity coefficients using the extended Debye-Hückel equation

logγA=−0.51×z_A²×μ−−√1+3.3×αA×μ−−√(6.9.13)(6.9.13)log⁡γA=−0.51×z_A²×μ1+3.3×αA×μ Equation (1)

where z_A is the ion’s charge, α_A is the effective diameter of the hydrated ion in nanometers , µ is the solution’s ionic strength and 0.51 and 3.3 are constants appropriate for an aqueous solution at 25^oC.

μ=1/2∑ⁱc_iz_i²

μ =1/2{(0.020M)(+2)²+(0.040M)(−1)²}=0.060M

As is true for any assumption, we will need to verify that it does not introduce too much error into our calculation.

Next, we use Equation (1) to calculate the activity coefficients for Pb2+ and IO3–.

logγPb2+=−0.51×(+2)²×√0.060 / 1+3.3×0.45×√0.060=−0.366

log⁡γPb²⁺= −0.366

γPb²⁺= 0.431

logγIO₃⁻=−0.51×(−1)²×√0.060 / 1+3.3×0.45×√0.060 = −0.0916

logγ⁡IO₃⁻ = −0.0916

γIO₃⁻= 0.810

Defining the equilibrium concentrations of Pb2+ and IO3– in terms of the variable x

Concentrations	Pb(IO3)2(s)	⇋	Pb2+(aq)	+	2IO3–
Initial	solid		0		0
Change	solid		+x		+2x
Equilibrium	solid		x		2x

and substituting into the thermodynamic solubility product for Pb(IO3)2 leaves us with

Ksp=aPb²⁺a2IO₃⁻=γPb²⁺[Pb²⁺]γ²IO₃⁻[IO₃⁻]²=2.5×10−13

Ksp=(0.431)(x)(0.810)²(2x)²=2.5×10⁻¹³

Ksp=1.131x³=2.5×10⁻¹³

Solving for x gives 6.0 × 10^–5, or a molar solubility of 6.0 × 10^–5 mol/L. Ignoring activity, as we did in our earlier calculation, gives the molar solubility as 4.0 × 10^-5 mol/L. Failing to account for activity in this case underestimates the molar solubility of Pb(IO3)2 by 33%.

The solution’s equilibrium composition is

[Pb²⁺]=6.0×10⁻⁵M

[IO₃⁻]=1.2×10⁻⁴M

[Mg²⁺]=0.020M

[NO₃⁻]=0.040M

Because the concentrations of Pb²⁺ and IO₃^– are much smaller than the concentrations of Mg²⁺ and NO^3–, our decision to ignore the contribution of Pb2+ and IO3– to the ionic strength is reasonable.