The common ion effect, predicting precipitation: Read pg. 580

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The common ion effect, predicting
precipitationRead pg. 580 582 (common ion
effect section only). Do PE 20 21By the end
of the periodPE 14 19, PE 20 25
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Common ion lab

• PbCl2(s)? Pb2(aq) 2Cl(aq)

A B Explanation
1
2
3
4
Cl causes shift left
Cloudy / precipitate
No Cl or Pb
No reaction (oily)
Pb causes shift left
Cloudy / precipitate
Cl in water causes shift left
Cloudy / precipitate
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
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Example 14.17 (pg. 581)

• Molar solubility of PbI2? Ksp 7.9 x 109
• Concentration of NaI is 0.10, thus I 0.10
• NaI(s) ? Na(aq) I(aq)

PbI2(s)
Pb2(aq)
I (aq)
1
2
0
0.10
x
2x
0.10 2x
x
Ksp Pb2(aq) I (aq)2
Ksp x 0.10 2x2 7.9 x 109
x is small, thus we can ignore 2x in 0.10 2x
Ksp x 0.102 7.9 x 109 , x 7.9 x 107 M
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PE 20 (pg. 582)

• Molar solubility of AgI? Ksp 8.3 x 1017
• Concentration of NaI is 0.20, thus I 0.20
• NaI(s) ? Na(aq) I(aq)

AgI(s)
Ag(aq)
I (aq)
1
1
0
0.20
x
x
0.20 x
x
Ksp Ag(aq) I (aq)
Ksp x 0.20 x 8.3 x 1017
x is small, thus we can ignore it in 0.20 x
Ksp x 0.20 8.3 x 1017, x 4.2 x 1016
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PE 21 (pg. 582)

• Molar solubility of Fe(OH)3? Ksp 1.6 x 1039
• Concentration of OH is 0.050
• Fe(OH)3(s) ? Fe3(aq) 3OH(aq)

Fe(OH)3
Fe3(aq)
OH(aq)
1
3
0
0.050
x
3x
0.050 3x
x
Ksp Fe3(aq) OH (aq)3
Ksp x 0.050 3x 1.6 x 1039
x is small, thus we can ignore 3x in 0.050 3x
Ksp x 0.0503 1.6 x 1039, x 1.3 x 1035
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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, Pb2Cl2 3.4 x 105
• Ion product is large to reduce, equilibrium
  must shift left precipitate forms

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PE 22 (pg. 583)

• Step 1 write the balanced equilibrium
• CaSO4(s) ? Ca2(aq) SO42(aq)
• Step 2 Write the Ksp equation
• Ksp Ca2SO42
• Ksp 2.4 x 105
• Ca2SO42 0.00250.030 7.5 x 105
• ion product is greater than Ksp, thus a
  precipitate will form

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PE 23 (pg. 583)

• Step 1 write the balanced equilibrium
• Ag2CrO4(s) ? 2Ag(aq) CrO42(aq)
• Step 2 Write the Ksp equation
• Ksp Ag2CrO42
• Ksp 1.2 x 1012
• Ag2CrO42 4.8 x 10523.4 x 104
• 7.8 x 1013
• ion product is less than Ksp, thus no
  precipitate will form (more could be dissolved)

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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

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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

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PE 24 (pg. 583)

• Step 1 write the balanced equilibrium
• PbSO4(s) ? Pb2(aq) SO42(aq)
• Step 2 Write the Ksp equation
• Ksp Pb2SO42 6.3 x 107
• Initial concentrations
• Pb2 0.0010 mol/L x 0.1 L 1.0 x 104 mol
• 1.0 x 104 mol / 0.2 L 0.00050 M
• SO42 0.0020 mol/L x 0.1 L 2.0 x 104 mol
• 2.0 x 104 mol / 0.2 L 0.0010 M
• Pb2SO42 0.00050.001 5.0 x 107
• The ion product is smaller than Ksp,
  thus no precipitate will form

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PE 25 (pg. 583)

• Step 1 write the balanced equilibrium
• PbCl2(s) ? Pb2(aq) 2Cl(aq)
• Step 2 Write the Ksp equation
• Ksp Pb2Cl2 1.7 x 105
• Initial concentrations
• Pb2 0.10 mol/L x 0.050 L 5.0 x 103 mol
• 5.0 x 103 mol / 0.070 L 0.0714 M
• Cl 0.040 mol/L x 0.020 L 8.0 x 104 mol
• 8.0 x 104 mol / 0.070 L 0.0114 M
• Pb2Cl2 0.07140.01142 9.3 x 106
• The ion product is smaller than Ksp,
  thus no precipitate will form

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Try 14.77 and 14.78 on pg. 591
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