The common ion effect,
predicting precipitation:
Read pg. 580 – 582
(common ion effect section only).
Do PE 20 – 21
By the end of the period:
PE 14 – 19, PE 20 – 25
Common ion lab
PbCl2(s) Pb2+(aq) + 2Cl–(aq)
A+B
Explanation
Cl– causes shift left
1 Cloudy / precipitate
No Cl or Pb
2 No reaction (oily)
Pb causes shift left
3 Cloudy / precipitate
4 Cloudy / precipitate Cl in water causes shift left
Common ion: “The ion in a mixture of ionic
substances that is common to the formulas of
at least two.”
Common ion effect: “The solubility of one salt is
reduced by the presence of another having a
common ion”
Example 14.17 (pg. 581)
Molar solubility of PbI2? Ksp = 7.9 x 10–9
Concentration of NaI is 0.10, thus [I–] = 0.10
NaI(s)  Na+(aq) + I–(aq)
Pb2+(aq)
I –(aq)
1
2
R
0
0.10
I
x
2x
C
x
0.10 + 2x
E
Ksp = [Pb2+(aq)] [I –(aq)]2
Ksp = [x] [0.10 + 2x]2 = 7.9 x 10–9
x is small, thus we can ignore 2x in 0.10 + 2x
Ksp = [x] [0.10]2 = 7.9 x 10–9 , x = 7.9 x 10–7 M
PbI2(s)
PE 20 (pg. 582)
Molar solubility of AgI? Ksp = 8.3 x 10–17
Concentration of NaI is 0.20, thus [I–] = 0.20
NaI(s)  Na+(aq) + I–(aq)
Ag+(aq)
I –(aq)
1
1
R
0
0.20
I
x
x
C
x
0.20 + x
E
Ksp = [Ag+(aq)] [I –(aq)]
Ksp = [x] [0.20 + x] = 8.3 x 10–17
x is small, thus we can ignore it in 0.20 + x
Ksp = [x] [0.20] = 8.3 x 10–17, x = 4.2 x 10–16
AgI(s)
PE 21 (pg. 582)
Molar solubility of Fe(OH)3? Ksp = 1.6 x 10–39
Concentration of OH– is 0.050
Fe(OH)3(s)  Fe3+(aq) + 3OH–(aq)
Fe3+(aq)
OH–(aq)
1
3
R
0
0.050
I
x
3x
C
x
0.050 + 3x
E
Ksp = [Fe3+(aq)] [OH –(aq)]3
Ksp = [x] [0.050 + 3x] = 1.6 x 10–39
x is small, thus we can ignore 3x in 0.050 + 3x
Ksp = [x] [0.050]3 = 1.6 x 10–39, x = 1.3 x 10–35
Fe(OH)3
Predicting when precipitation occurs
Read pg. 582. Do PE 22, 23
Similar to Kc vs. mass action expression to
predict if equilibrium exists (and which way it
will shift)
E.g. in example 14.18
PbCl2(s)  Pb2+(aq) + 2Cl–(aq)
(NaCl and Pb(NO3) are soluble according to
the solubility rules; they will not precipitate)
Ksp = 1.7 x 10-5, [Pb2+][Cl–]2 = 3.4 x 10–5
Ion product is large … to reduce, equilibrium
must shift left … precipitate forms
PE 22 (pg. 583)
Step 1: write the balanced equilibrium:
CaSO4(s)  Ca2+(aq) + SO42–(aq)
Step 2: Write the Ksp equation:
Ksp = [Ca2+][SO42–]
Ksp = 2.4 x 10–5
[Ca2+][SO42–] = [0.0025][0.030] = 7.5 x 10–5
ion product is greater than Ksp, thus a
precipitate will form
PE 23 (pg. 583)
Step 1: write the balanced equilibrium:
Ag2CrO4(s)  2Ag+(aq) + CrO42–(aq)
Step 2: Write the Ksp equation:
Ksp = [Ag+]2[CrO42–]
Ksp = 1.2 x 10–12
[Ag+]2[CrO42–] = [4.8 x 10–5]2[3.4 x 10–4]
= 7.8 x 10–13
ion product is less than Ksp, thus no precipitate
will form (more could be dissolved)
Predicting when precipitation occurs
• So far we have been dealing with one of two
situations:
• 1) dissolving a solid in a liquid (with or
without initial concentrations of ions) and
performing Ksp calculations
• 2) given the concentrations of ions predicting
if a solid (I.e. precipitate will form)
• A third situation exists that is slightly different
• Mixing two liquids
• In this case, we need to account for both the
ions and the water that is added …
Predicting when precipitation occurs
Q- E.g. will the addition of a NaCl solution to a
saturated PbCl2 solution result in a
precipitate forming?
A- It depends on the concentration of the NaCl
solution
If the NaCl solution is very dilute, the extra
water could cause more PbCl2(s) to dissolve,
than the extra Cl– causes PbCl2(s) to form
Read the example on pg. 583, do PE 24, 25
PE 24 (pg. 583)
Step 1: write the balanced equilibrium:
PbSO4(s)  Pb2+(aq) + SO42–(aq)
Step 2: Write the Ksp equation:
Ksp = [Pb2+][SO42–] = 6.3 x 10–7
Initial concentrations:
[Pb2+] = 0.0010 mol/L x 0.1 L = 1.0 x 10–4 mol
= 1.0 x 10–4 mol / 0.2 L = 0.00050 M
[SO42–] = 0.0020 mol/L x 0.1 L = 2.0 x 10–4 mol
= 2.0 x 10–4 mol / 0.2 L = 0.0010 M
[Pb2+][SO42–] = [0.0005][0.001] = 5.0 x 10–7
The ion product is smaller than Ksp,
thus no precipitate will form
PE 25 (pg. 583)
Step 1: write the balanced equilibrium:
PbCl2(s)  Pb2+(aq) + 2Cl–(aq)
Step 2: Write the Ksp equation:
Ksp = [Pb2+][Cl–]2 = 1.7 x 10–5
Initial concentrations:
[Pb2+] = 0.10 mol/L x 0.050 L = 5.0 x 10–3 mol
= 5.0 x 10–3 mol / 0.070 L = 0.0714 M
[Cl–]
= 0.040 mol/L x 0.020 L = 8.0 x 10–4 mol
= 8.0 x 10–4 mol / 0.070 L = 0.0114 M
[Pb2+][Cl–]2 = [0.0714][0.0114]2 = 9.3 x 10–6
The ion product is smaller than Ksp,
thus no precipitate will form
Try 14.77 and 14.78 on pg. 591
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