1. Solubility
Lesson 5
Trial Ion Product

2.  When two ionic solutions are mixed and if one product has low solubility, a precipitate will form.
Pb(NO3)2(aq) NaCl(aq) → PbCl2(s) NaNO3(aq)
low solubility
The solubility chart on page 4 predicts this reaction, if the ions are > 0.10 M.
A trial ion product is required if the ion concentrations are < 0.10 M.

3. The capacity of a solution to dissolve a solid is described by the Ksp.
Pb(NO3)2
2Cl-
Pb2+ ⇌
NaCl
PbCl2(s) ⇌ Pb2+ + 2Cl-
The Ksp represents the limit of the solution to dissolve PbCl2.
Pb2+ and Cl- will dissolve until the ion concentrations are equal to the Ksp.
The solution is saturated- any more ions will form a solid.

4. 1. 200. 0 mL 0. 10 M Pb(NO3)2 is mixed with 300. 0 mL of 0
mL 0.10 M Pb(NO3)2 is mixed with mL of 0.20 M NaCl, will a precipitate occur?
PbCl2(s) ⇌ Pb Cl-
200
300
0.10 M
0.20 M
500
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

5. 2. Will a precipitate form if 20. 0 mL of 0
2. Will a precipitate form if 20.0 mL of 0.010M CaCl2 is mixed with 60.0 mL of M Na2SO4?
CaSO4(s) ⇌ Ca SO42- 20
0.010 M 60 M 80 80 M M
TIP = [Ca2+][SO42-]
TIP = [0.0025][0.0060]
= x 10-5
Ksp = x 10-5
TIP < Ksp no ppt forms

6. 3. Will a precipitate form when equal volumes of 0. 020 M AlCl3 and 0
3. Will a precipitate form when equal volumes of M AlCl3 and M AgNO3 are mixed.
The Cl- x 3
AgCl(s) ⇌ Ag Cl- 1 1
0.040 M
0.060 M 2 2
0.020 M
0.030 M
TIP = [Ag+][Cl-]
TIP = [0.020][0.030]
= x 10-4
Ksp = x
TIP > Ksp ppt forms

7. Ag2CrO4 has the greater Ag+ concentration
4. Consider the two saturated solutions AgCl and Ag2CrO4. Which has the greater Ag+ concentration?
Ag2CrO4 ⇌ 2Ag CrO42-
AgCl ⇌ Ag Cl-
s s s
s s s
Ksp = s2
Ksp = 4s3
1.8 x = s2
1.1 x = s3
s = x 10-5 M
s = 6.5 x 10-5 M
[Ag+] = x 10-5 M
[Ag+] = 2s = x 10-4 M
Ag2CrO4 has the greater Ag+ concentration