chemistry 123: physical and organic chemistry · 4/3/09 2 chemistry 123: physical and organic...

10
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 NaC 2 H 3 O 2 must be dissolved in 0.300 L of 0.25 M HC 2 H 3 O 2 to produce a solution with pH = 5.09? (Assume that the solution volume is constant at 0.300 L) HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + Equilibrium expression: [H 3 O + ] [HC 2 H 3 O 2 ] K a = [C 2 H 3 O 2 - ] = 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, [H 3 O + ] [HC 2 H 3 O 2 ] K a = [C 2 H 3 O 2 - ] = 1.8×10 -5 [H 3 O + ] = 10 -5.09 = 8.1×10 -6 [HC 2 H 3 O 2 ] = 0.25 M Solve for [C 2 H 3 O 2 - ] [H 3 O + ] [HC 2 H 3 O 2 ] = K a [C 2 H 3 O 2 - ] = 0.56 M 8.1×10 -6 0.25 = 1.8×10 -5

Upload: dodan

Post on 06-May-2018

225 views

Category:

Documents


2 download

TRANSCRIPT

Page 1: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

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

Page 2: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

2

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

Page 3: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

3

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.

Page 4: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

4

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

Page 5: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

5

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

Page 6: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

6

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

Page 7: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

7

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)

Page 8: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

8

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,

Page 9: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

4/3/09

9

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.

Page 10: Chemistry 123: Physical and Organic Chemistry · 4/3/09 2 Chemistry 123: Physical and Organic Chemistry Topic 5: Ionic Equilibrium  Winter 2009 Page 61

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