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

(assumed all the pbcl2 is dissolve as pbcl3-)

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

2.

Molar mass of PbCl2 = 278.1 g/mol

So, 278. 1 g of PbCl2 = 1 mole

2 g of PbCl2 = (2 / 278.1) mole = 0.0072 mole

Now,

25.0 mL of 1.00 M HCl

Moles of HCl = 0.025 L x 1 M = 0.025 mole

Since,

PbCl₂(s) + Cl^-(aq) PbCl₃^-(aq),

In this case, 0.0072 mole of PbCl₂ will react with 0.0072 mole of Cl^-. But moles of Cl^- is 0.025 mole, so moles of Cl- left = (0.025 - 0.0072) mole = 0.0178 mole.

So, 2.00 g sample of PbCl2 will dissolve in 25.0 mL of 1.00 M HCl.

3.

Molar mass of lead chromate (PbCrO₄) = 323.2 g/mol

So, 323.2 g of lead chromate (PbCrO₄) = 1 mol

0.2375 g of lead chromate (PbCrO₄) = (0.2375 / 323.2) mol = 0.000735 mole

So, moles of lead (Pb) = 0.000735 mole

Molar mass of Pb = 207.2 g/mol

So, 1 mole of Pb = 207.2 g

0.000735 mole of Pb = 0.000735 x 207.2 g

= 0.1523 g

= 152.3 mg

Volume = 25 mL = 0.025 L

[Pb] = 152.3 mg / 0.025 L = 6092 mg/L