Question

2. Show that a 2.00 g sample of PbCl2 will dissolve in 25.0 mL of 1.00 M HCl.

PbCl2(s) Cl-(aq)   ---- [PbCl3-](aq), Kf = 0.014


3. A 25.00 mL sample of a lead ion solution gave 0.2375 g of lead chromate. What is the       concentration on lead in the solution in mg/L?

Answer 1

In question no. 2, some PbCl2 will remain undissolved as Cl- in the above question is limiting reagent thus some PbCl2 will remain in solid state.

In Question no. 3, using Principle of atom conservation, we can say all the Pb²⁺ of solution appear as PbCrO₄ thus mole of Pb(II) is equal to no. of mole of PbCrO4, thus

Molarity * Volume [Pb(II)] = weight of lead chromate / molar mass of lead chromate

M * 0.025 (litre) = 0.2375 / (207+52+64)

Molarity of Solution = 0.0294 M