Question

8) Calculate the following for 2.0 L solution containing [Ag⁺] = 0.100 M and [Pb²⁺] = 0.100 M. Assume no volume changes. (AgCl K_sp = 1.8 x 10^-10, PbCl₂ K_sp = 1.8 x 10^-5).

a) At what [Cl-] will each salt precipitate?

b) What percent of the Ag⁺ has precipitated before the Pb²⁺ begins to precipitate?

c) How much sodium chloride must be added (in grams) to precipitate a maximum AgCl before before any PbCl₂ begins to precipitate?

Answer 1

Answer – We are given, [Ag⁺] = 0.100 M and [Pb²⁺] = 0.100 M

AgCl K_sp = 1.8 x 10^-10, PbCl₂ K_sp = 1.8 x 10^-5

a) For AgCl –

We know Ksp expression for AgCl

Ksp = [Ag⁺] [Cl^-]

1.8 x 10^-10 = 0.100 M *[Cl^-]

[Cl^-] = 1.8 x 10^-10 / 0.100 M

        = 1.8*10^-9 M

so, we need 1.8*10^-9 M [Cl^-] to start AgCl precipitate.

For PbCl₂ –

We know Ksp expression for PbCl₂

Ksp = [Pb²⁺] [Cl^-]²

1.8 x 10^-5 = 0.100 M *[Cl^-]

[Cl^-]² = 1.8 x 10^-5 / 0.100 M

        = 1.8*10^-4 M

squar root from both side

[Cl^-] = 0.0134 M

so we need 0.0134 M [Cl^-] to start PbCl₂ precipitate.

b) For the bigan precipitation of PbCl₂ there is needed 0.0134 M Cl^-

we need to calculate for the 0.0134 M Cl^- how much Ag⁺ remaining

Ksp = [Ag⁺] [Cl^-]

1.8 x 10^-10 = [Ag⁺] * 0.0134 M

[Ag⁺] = 1.8 x 10^-10 / 0.0134

           = 1.34 *10^-8 M

so there is remaining in solution [Ag⁺] =1.34 *10^-8 M

Ag⁺ used in precipitation = 0.100 - 1.34 *10^-8 M = 0.100 M

Means almost all 100 % Ag⁺ is precipitate before the Pb²⁺ begins to precipitate.

c) We know for the start precipitation of PbCl₂ we need more than concentration of 0.0134 M Cl^-

Means at 0.0134 M Cl^- there are all Ag⁺ precipitate, since only 1.8 *10^-9 M Cl^- required to start precipitation

so, [Cl^-] =[NaCl] = 0.0134 M

Moles of NaCl = 0.0134 M x 2.0 L

                        = 0.0268 moles

Mass of NaCl = 0.0268 moles * 58.44 g/mol

                       = 1.57 g

so, 1.57 g of sodium chloride must be added (in grams) to precipitate a maximum AgCl before any PbCl₂ begins to precipitate.