Question

Consider an aqueous solution that is saturated with lead iodide and lead sulfate and also contains 0.10M Li2SO4 (a strong electrolyte). Write mass and charge balance equations that describe this system. You may assume that sulfate is the conjugate base of a strong acid, and ignore any acid-base characteristics of the sulfate ion or any other species in solution.

Answer 1

The equilibriums involved in this problem are

PbI₂ (s) $\rightleftharpoons$ Pb²⁺ + 2I^-

PbSO₄ (s)  $\rightleftharpoons$ Pb²⁺ + SO₄^2-

and Li₂SO₄$\rightleftharpoons$ 2Li⁺+ SO₄^2-

For Charge balance,

The total molar concentration of positive ions = The total molar concentration of negative ions

Now, The total molar concentration of positive ions

= 2[Pb²⁺] + [Li⁺] (since each lead ion will contribute 2 and lithium ion will contribute one charge)

similarly, The total molar concentration of negative ions

= [I^-] + 2[SO₄^2-] (the question told to ignore any acid-base characteristics of the sulfate, otherwise HSO₄^- ions would also have come in charge equilibria)

so the charge balance equilibrium is

2[Pb²⁺] + [Li⁺] = [I^-] + 2[SO₄^2-]

Mass balance equilibria:

total concentration of SO₄^2- ions will be [SO₄^2-]_tot = [SO₄^2-]_PbSO4 + [SO₄^2-]_Li2SO4

and the total Pb has come from both PbI₂ and PbSO₄

For PbI₂ (s) $\rightleftharpoons$ Pb²⁺ + 2I^-

at equilibrium x 2x

so in the mass balance we have to put 2 in front of [Pb²⁺] to equalize the concentration with [I^-]

so, 2[Pb²⁺] = [I^-]

[Pb²⁺] = [I^-]/2

and from PbSO₄

[Pb²⁺] = [SO₄^2-]

So, total  [Pb²⁺] = [SO₄^2-]_PbSO4 + [I^-]/2

thus, [SO₄^2-]_PbSO4  = [Pb²⁺] - [I^-]/2

and [SO₄^2-]_Li2SO4 = 0.1 M (Li₂SO₄ dissociates completely to produce 0.1 M [SO₄^2-])

so the mass balanced equation is,

[SO₄^2-]_tot = [Pb²⁺] - [I^-]/2 + 0.1 M

i.e. [SO₄^2-]_tot + [I^-]/2= [Pb²⁺] + 0.1 M