Please help with this chemistry problem

Question

A researcher measures 0.70 mol lead(II) acetate salt and 0.50 mol lead(II) bromide salt. The two solid salts are combined dry and finely powdered. Then the solid mixture is combined with water, with a final total volume of 1.00 L. The researcher finds PbBr2 has Ksp = 2.1×10–6 but Pb(CH3CO2)2 does not have a solubility equilibrium constant.

i) Start by writing the two reaction equations suggested by the information given in the problem and your knowledge of general chemistry.

1. a. compare the two reactions.
2. b. Write the charge balance equation for the solution.

ii) Calculate the concentration of the three ions in solution.

1. a. lead(II) cation, Pb 2+ .
2. b. acetate ion.
3. c. bromide ion

iii) convert one of the concentrations in #ii into a different concentration unit.

iv) Which chemical species determines the pH of the solution, and is the overall solution pH acidic or basic?

v) An electrochemical cell is set up to convert the Pb2+ in the solution into solid Pb metal. If the current of 2.00 A (or 2.00 C/s) runs for 25.0 minutes, what mass of solid lead forms?

Answer 1

I) Pb(C₂H₃O₂)₂ is soluble & dissociate completely in water. So the "reaction" is simply:

Pb(C₂H₃O₂)_{2 (s)} ---> Pb²⁺_(aq) + 2 C₂H₃O₂^-_(aq)

The PbBr2 forms an equilibrium system: PbBr2 (s) ⇔ Pb²⁺_(aq) + 2 Br^-_(aq)

When mixed, we would have the following ions present (in amounts dictated by the given info): Pb²⁺, C₂H₃O₂^-, and Br^-.

II) b) The acetate ion's concentration would be twice the concentration of the lead (II) acetate. The lead (II) acetate's concentration would be 0.70 M (0.70 mol/1.00 L). So the [C₂H₃O₂^-] = 1.40 M

To determine the lead & bromide concentrations we have to use the K_sp. Note that the Pb²⁺ has 0.70 m/L from the lead (II) acetate PLUS that produced by the lead (II) bromide.

Letting x = Pb²⁺ from the lead (II) bromide, we have: K_sp = 2.1 x 10^-6 = [Pb²⁺] [Br^-]² = (x + 0.70) (2x)²

Ignoring the x for the Pb²⁺ (because it's significantly less than the 0.70). So, 2.8x² = 2.1 x 10^-6 and x = 0.354. Thus:

a) [Pb²⁺] = 0.70 + 0.354 = 1.054 = ~ 1.1 M

c) [Br^-] = 2(0.354) = 0.708 = ~ 0.71 M

III) I will choose to convert the [C₂H₃O₂^-] from mol/L to grams/l of Pb(C₂H₃O₂)₂ :

(325.2 g lead (II) acetate)/1 mol lead (II) acetate) (0.70 mol lead (II) acetate/L) = 227.6 g/L = ~ 2.3 x 10² g/L

IV) The acetate ion is the conjugate base of a weak acid and the Pb²⁺ ion is the conjugate acid of a strong base. Furthermore, the lead (II) ion is the most concentrated of the ions. Hence, the solution should be basic with a pH > 7.

V) m = (Mit)/(nF) = [(207.2) (2.00) (60 s/1min) (25.0 min)]/[2)(96485)] = 3.22 grams of Pb