Question

A)Calculate The activity coefficient of Pb2+ at an ionic strength of 0.40. What does this value mean? B) determine the solubility of PbCO3 in the solution of ionic strength of 0.40. (Note the activity coefficient of the carbonate ion is the same as that for the Pb2+, why is that?) compare this solubility to the solubility of PbCO3 in pure water. C) write the mass balance and charge balance equations for a solution that is 0.20 M in NaF and saturated with CaF2.

Answer 1

diameter pb 0.45 nm or 450 pm (from internet)

log Pb = -0.51 * z² * / (1 + 3.3 * * )

z is the charge of the ion (+2), alpha is the diameter of the ion and is the ionic strenght

so

log Pb = -0.51 * (2²) * (0.4)^0.5 / (1 + 3.3 * 0.45 * (0.4^0.5) = -0.665

Pb = 10 ^-0.665 = 0.21

The activiy coefficient gives an idea of how much different is the real behavior from the ideal behavior, so for this case the behavior of the actual solution is 20% compared to the behavior of the ideal solution.

B) Ksp = 7.4 x 10^-14, from internet

PbCO3 ======= Pb + CO₃

Ksp = (*Pb) * (*CO₃)

7.4 x10^-14 = 0.21² * s²

s² = 1.68 x 10^-12

s = 1.295 x 10^-6 M, this is the solubility

Carbonate ion has the same activity coefficient at the 0.4 ionic strenght because it has the same features of Pb ( ion charge and ion size)