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4/3/09
1
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 59
Topic 5: Introduction,
Preparing a Buffer Solution of a Desired pH. What mass of NaC2H3O2 must be dissolved in 0.300 L of 0.25 M HC2H3O2 to produce a solution with pH = 5.09? (Assume that the solution volume is constant at 0.300 L)
HC2H3O2 + H2O C2H3O2- + H3O+
Equilibrium expression:
[H3O+] [HC2H3O2]
Ka= [C2H3O2
-] = 1.8×10-5
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 60
Topic 5: Introduction,
[H3O+] [HC2H3O2]
Ka= [C2H3O2
-] = 1.8×10-5
[H3O+] = 10-5.09 = 8.1×10-6
[HC2H3O2] = 0.25 M
Solve for [C2H3O2-]
[H3O+] [HC2H3O2] = Ka
[C2H3O2-] = 0.56 M
8.1×10-6
0.25 = 1.8×10-5
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 61
Topic 5: Introduction,
[C2H3O2-] = 0.56 M
1 mol NaC2H3O2
82.0 g NaC2H3O2 mass C2H3O2- = 0.300 L 1 L
0.56 mol × ×
= 14 g NaC2H3O2
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 62
Topic 5: Introduction, Six Methods of Preparing Buffer Solutions
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 63
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 64
Topic 5: Introduction, Buffer Capacity and Range
♦ Buffer capacity is the amount of acid or base that a buffer can neutralize before its pH changes appreciably. • Maximum buffer capacity exists when [HA] and
[A-] are large and approximately equal to each other.
♦ Buffer range is the pH range over which a buffer effectively neutralizes added acids and bases. • Practically, range is 2 pH units around pKa.
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 65
Topic 5: Introduction, Acid-Base Indicators Color of some substances depends on the pH.
HIn + H2O
In the acid form the color appears to be the acid color. In the base form the color appears to be the base color. Intermediate color is seen in between these two states.
The complete color change occurs over about 2 pH units.
In- + H3O+
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 66
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 67
Topic 5: Introduction, Neutralization Reactions and Titration Curves
♦ Equivalence point: • The point in the reaction at which both acid and base
have been consumed. • Neither acid nor base is present in excess.
♦ End point: • The point at which the indicator changes color.
♦ Titrant: • The known solution added to the solution of unknown
concentration. ♦ Titration Curve:
• The plot of pH vs. volume.
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 68
Topic 5: Introduction, Titration of a Strong Acid with a Strong Base
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 69
Topic 5: Introduction, Titration of a Weak Acid with a Strong Base
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 70
Topic 5: Introduction, Titration of a Weak Acid with a Strong Base
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 71
Topic 5: Introduction, Titration of a Weak Polyprotic Acid
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 72
Topic 5: Introduction, The Solubility Product Constant, Ksp
CaSO4(s)
Ksp = [Ca2+][SO42-] = 9.1×10-6 at 25°C
♦ The equilibrium constant for the equilibrium established between a solid solute and its ions in a saturated solution.
Ca2+(aq) + SO42-(aq)
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 73
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 74
Topic 5: Introduction,
4/3/09
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Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 75
Topic 5: Introduction, Criteria for Precipitation and Its Completeness
AgI(s))
Mix AgNO3(aq) and KI(aq) to obtain a solution that is 0.010 M in Ag+ and 0.015 M in I-. Saturated, supersaturated or unsaturated?
Q = [Ag+][Cl-] = (0.010)(0.015) = 1.×10-4 > Ksp
Ksp = [Ag+][Cl-] = 8.5×10-17
Ag+(aq) + I-(aq)
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 76
Topic 5: Introduction,
Q is generally called the ion product.
Q > Ksp Precipitation should occur. Q = Ksp The solution is just saturated. Q < Ksp Precipitation cannot occur.
4/3/09
10
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 77
Topic 5: Introduction,
PbI2(s) → Pb2+(aq) + 2 I-(aq) Ksp= 7.1×10-9
Determine the amount of I- in the solution:
= 3×10-5 mol I-
nI- = 3 drops 1 drop
0.05 mL
1000 mL
1 L
1 L
0.20 mol KI
1 mol KI 1 mol I-
Applying the Criteria for Precipitation of a Slightly Soluble Solute. Three drops of 0.20 M KI are added to 100.0 mL of 0.010 M Pb(NO3)2. Will a precipitate of lead iodide form? (1 drop ≈ 0.05 mL)
Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium http://people.ok.ubc.ca/orcac/chem123out.html
Winter 2009 Page 78
Topic 5: Introduction,
[I-] = 0.1000 L
3 x 10-5 mol I- = 3×10-4 M I-
Determine the concentration of I- in the solution:
Apply the Precipitation Criteria:
Q = [Pb2+][I-]2 = (0.010)(3×10-4)2 = 9×10-10
< Ksp = 7.1×10-9