SolubilityLesson 5

Trial Ion Product

 

We have learned that when two ionic solutions are mixed and if one product has low solubility, then there is a reaction where a precipitate will form.  Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)

low solubility

The solubility chart on page 4 predicts this reaction, but only if the solution is 0.10 M or greater. If the molarity is less than 0.10 M, then the reaction may or may not happen.

A trial ion product must be calculated to predict the reaction of all solutions less than 0.10 M.

The capacity of a solution to dissolve a solid is described by the Ksp.

PbCl2(s) ⇌ Pb2+ + 2Cl-

The Ksp represents the limit of the solution to dissolve PbCl2.You can add Pb2+ and Cl- until the ion concentrations are equal to the Ksp. The solution is saturated and addition PbCl2 must sit on the bottom not dissolved.

Pb2+ 2Cl-

Pb(NO3)2

NaCl

1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M

NaCl, will a precipitate occur?

1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M

NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl-

Write a dissociation equation for the compound with low solubility.

1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M

NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl-

0.10 M 0.20 M

Write a dissociation equation for the compound with low solubility.List the initial molarities of each ion.

1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M

NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl-

200 0.10 M 300 0.20 M500 500

0.040 M 0.12 M

Write a dissociation equation for the compound with low solubility.List the initial molarities of each ion.Reduce each molarity by the dilution factor: V1/V2.

1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M

NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl-

200 0.10 M 300 0.20 M500 500

0.040 M 0.12 M

TIP = [Pb2+][Cl-]2

TIP = [0.040][0.12] 2

= 5.8 x 10-4

Write a dissociation equation for the compound with low solubility.List the initial molarities of each ion.Reduce each molarity by the dilution factor: V1/V2.Write the Ksp or TIP (trial ion product) and solve.

1. 200.0 mL 0.10 M Pb(NO3)2 is mixed with 300.0 mL of 0.20 M

NaCl, will a precipitate occur? PbCl2(s) ⇌ Pb2+ + 2Cl-

200 0.10 M 300 0.20 M500 500

0.040 M 0.12 M

TIP = [Pb2+][Cl-]2

TIP = [0.040][0.12] 2

= 5.8 x 10-4

Ksp = 1.2 x 10-5 TIP > Ksp ppt forms

Write a dissociation equation for the compound with low solubility.List the initial molarities of each ion.Reduce each molarity by the dilution factor: V1/V2.Write the Ksp or TIP (trial ion product) and solve.Compare to real Ksp

2. Will a precipitate form if 20.0 mL of 0.010M CaCl2 is

mixed with 60.0 mL of 0.0080 M Na2SO4?

CaSO4(s) ⇌ Ca2+ + SO42-

20 0.010 M 60 0.0080 M80 80

0.0025 M 0.0060 M

TIP = [Ca2+][SO42-]

TIP = [0.0025][0.0060] = 1.5 x 10-5

Ksp = 7.1 x 10-5 TIP < Ksp no ppt forms

AgCl(s) ⇌ Ag+ + Cl-

1 0.040 M 1 0.040 M2 20.020 M 0.020 M

TIP = [Ag+][Cl-]TIP = [0.020][0.020]

= 4.0 x 10-4

Ksp = 1.8 x 10-10

TIP > Ksp ppt forms

The Cl- is doubled

3. Will a precipitate form when equal volumes of 0.020 M CaCl2 and 0.040 M AgNO3 are mixed.

4. Consider the two saturated solutions AgCl and Ag2CrO4. Which has the greater Ag+ concentration?

AgCl ⇌ Ag+ + Cl- Ag2CrO4 ⇌ 2Ag+ + CrO42-

s s s s 2s s

Ksp = s2 Ksp = 4s3

1.8 x 10 -10 = s2 1.1 x 10-12 =4s3

s = 1.3 x 10-5 M s = 6.5 x 10-5 M

[Ag+] = 1.3 x 10-5 M [Ag+] = 2s = 1.3 x 10-4 M

Ag2CrO4 has the greater Ag+ concentration