Acid-Base Reactions

1
Acid-Base Reactions
Conjugates do not react!!
CHAPTER 17
Part 1
2
Stomach Acidity Acid-Base Reactions
3
ACIDS-BASE REACTIONS

• For any acid-base reaction where only one
  hydrogen ion is transferred, the equilibrium
  constant for the reaction can be calculated and
  is often called Knet.
• For the general reaction

• Knet is always Ka (reactant acid) / Ka (product
  acid)
• When Knet gtgt 1, products are favored.
• When Knet ltlt 1, reactants are favored.

See 16.5 for manipulating K
4
ACIDS-BASE REACTIONS

• There are four classifications or types of
  reactions strong acid with strong base, strong
  acid with weak base, weak acid with strong base,
  and weak acid with weak base.
• NOTE For all four reaction types the limiting
  reactant problem is carried out first. Once this
  is accomplished, one must determine which
  reactants and products remain and write an
  appropriate equilibrium equation for the
  remaining mixture.

5
STRONG ACID WITH STRONG BASE

• The net reaction is
• The product, water, is neutral.

6
STRONG ACID WITH WEAK BASE

• The net reaction is

The product is HB and the solution is acidic.
7
WEAK ACID WITH STRONG BASE

• The net reaction is

The product is A- and the solution is basic.
8
WEAK ACID WITH WEAK BASE

• The net reaction is

Notice that Knet may even be less than one.
This will occur when Ka HB gt Ka HA.
9
Acid-Base Reactions

• QUESTION You titrate 100. mL of a 0.025 M
  solution of benzoic acid with 0.100 M NaOH to the
  equivalence point (mol HBz mol NaOH). What is
  the pH of the final solution? Note HBz and NaOH
  are used up!
• HBz NaOH ---gt Na Bz- H2O

Ka 6.3 10-5
Kb 1.6 10-10
C6H5CO2H HBz
Benzoate ion Bz-
10
Acid-Base Reactions

• The product of the titration of benzoic acid, the
  benzoate ion, Bz-, is the conjugate base of a
  weak acid.
• The final solution is basic.

11
Acid-Base Reactions
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?

• Strategy find the concentration of the
  conjugate base Bz- in the solution AFTER the
  titration, then calculate pH.
• This is a two-step problem
• 1st. stoichiometry of acid-base reaction
• 2nd. equilibrium calculation

12
Acid-Base Reactions
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?

• STOICHIOMETRY PORTION
• 1. Calculate moles of NaOH required.
• (0.100 L HBz)(0.025 M) 0.0025 mol HBz
• This requires 0.0025 mol NaOH
• 2. Calculate volume of NaOH required.
• 0.0025 mol (1 L / 0.100 mol) 0.025 L
• 25 mL of NaOH required

13
Acid-Base Reactions
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?

• STOICHIOMETRY PORTION
• 25 mL of NaOH required
• 3. Moles of Bz- produced moles HBz 0.0025
  mol Bz-
• 4. Calculate concentration of Bz-.
• There are 0.0025 mol of Bz- in a TOTAL
  SOLUTION VOLUME of 125 mL
• Bz- 0.0025 mol / 0.125 L 0.020 M

14
Acid-Base Reactions

• QUESTION You titrate 100. mL of a 0.025 M
  solution of benzoic acid with 0.100 M NaOH to the
  equivalence point (mol HBz mol NaOH). What is
  the pH of the final solution? Note HBz and NaOH
  are used up!
• HBz NaOH ---gt Na Bz- H2O

Ka 6.3 10-5
Kb 1.6 10-10
C6H5CO2H HBz
Benzoate ion Bz-
15
Acid-Base Reactions
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?

• EQUILIBRIUM PORTION
• Bz- H2O HBz OH- Kb
  1.6 x 10-10
• Bz- HBz OH-
• initial
• change
• equilib

0.020 0 0 -x x x
0.020 - x x x
16
Acid-Base Reactions
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?

• EQUILIBRIUM PORTION
• Bz- H2O HBz OH- Kb
  1.6 x 10-10
• Bz- HBz OH-
• equilib 0.020 - x x x

Solving in the usual way, we find x OH- 1.8
x 10-6, pOH 5.75, and pH 8.25
17
Acid-Base Reactions
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH What
is the pH at the half-way point?

• HBz H2O H3O Bz- Ka
  6.3 x 10-5
• H3O HBz / Bz- Ka
• At the half-way point, HBz Bz-, so
• H3O Ka 6.3 x 10-5
• pH 4.20

18
The Common Ion Effect
18

• QUESTION What is the effect on the pH of adding
  NH4Cl to 0.25 M NH3(aq)?
• NH3(aq) H2O NH4(aq) OH-(aq)
• Here we are adding an ion COMMON to the
  equilibrium.
• Le Chatelier predicts that the equilibrium will
  shift to the ____________.

LEFT
Closer to 0
The pH will go _______. After all, NH4 is an
acid!
19
The Common Ion Effect

• Let us first calculate the pH of a 0.25 M
  NH3 solution.
• NH3 NH4 OH-
• initial 0.25 0 0
• change -x x x
• equilib 0.25 - x x x

20
The Common Ion Effect

• Assuming x is ltlt 0.25, we have
• OH- x Kb(0.25)1/2 0.0021 M
• This gives pOH 2.67 and so - - -
• pH 14.00 - 2.67 11.33 for 0.25 M NH3

21
The Common Ion Effect

• We expect that the pH will decline on adding
  NH4Cl. Lets test that at 0.10M !
• NH3 NH4 OH-
• initial
• change
• equilib

0.25 0.10 0 -x x x 0.25 - x 0.10
x x
22
The Common Ion Effect

• Because equilibrium shifts left, x is MUCH
  less than 0.0021 M, the value without NH4Cl.

OH- x (0.25 / 0.10)Kb 4.5 x 10-5 M This
gives pOH 4.35 and pH 9.65 pH drops from
11.33 to 9.65 on adding a common ion.
23
Buffer Solutions

• HCl is added to pure water.

HCl is added to a solution of a weak acid
H2PO4- and its conjugate base HPO42-.
24
Buffer Solutions

• The function of a buffer is to resist changes in
  the pH of a solution.
• Buffers are just a special case of the common ion
  effect.
• Buffer Composition
• Weak Acid Conj. Base
• HC2H3O2 C2H3O2-
• H2PO4- HPO42-
• Weak Base Conj. Acid
• NH3 NH4

25
Buffer Solutions

• Consider HOAc/OAc- to see how buffers work.
• ACID USES UP ADDED OH-.
• We know that OAc- H2O HOAc
  OH- has Kb 5.6 x 10-10
• Therefore, the reverse reaction of the WEAK
  ACID with added OH- has Kreverse 1/ Kb
  1.8 x 109
• Kreverse is VERY LARGE, so HOAc completely uses
  up the OH- !!!!

26
Buffer Solutions

• Consider HOAc/OAc- to see how buffers work.
• CONJUGATE BASE USES UP ADDED H HOAc H2O
  OAc- H3O has Ka 1.8 x 10-5.
• Therefore, the reverse reaction of the WEAK
  BASE with added Hhas Kreverse 1/ Ka 5.6
  x 104
• Kreverse is VERY LARGE, so OAc- completely uses
  up the H !

27
Buffer Solutions
Problem What is the pH of a buffer that has
HOAc 0.700 M and OAc- 0.600 M? HOAc
H2O OAc- H3O Ka 1.8 x 10-5

• HOAc OAc- H3O
• initial
• change
• equilib

0.700 0.600 0 -x x
x 0.700 - x 0.600 x x
28
Buffer Solutions
Problem What is the pH of a buffer that has
HOAc 0.700 M and OAc- 0.600 M? HOAc
H2O OAc- H3O Ka 1.8 x 10-5
HOAc OAc- H3O equilib 0.700 -
x 0.600 x x Assuming that x ltlt 0.700
and 0.600, we have

• H3O 2.1 x 10-5 and pH 4.68

29
Buffer Solutions

• Notice that the expression for calculating the
  H concentration of the buffer is

This leads to a general equation for finding
the H or OH- concentration of a buffer.
Notice that the H or OH- concentrations
depend on K and the ratio of acid and base
concentrations.
30
Henderson-Hasselbalch Equation
Take the negative log of both sides of this
equation
OR

• This is called the Henderson-Hasselbalch
  equation.

31
Henderson-Hasselbalch Equation

• This shows that the pH is determined largely by
  the pKa of the acid and then adjusted by the
  ratio of acid and conjugate base.

32
Adding an Acid to a Buffer

• Problem What is the new pH when 1.00 mL of
  1.00 M HCl is added to
• a) 1.00 L of pure water (before HCl, pH 7.00)
• b) 1.00 L of buffer that has HOAc 0.700 M
  and OAc- 0.600 M (pH 4.68)
• Solution to Part (a)
• Calculate HCl after adding 1.00 mL of HCl to
  1.00 L of water
• M1 V1 M2 V2
• M2 1.00 x 10-3 M
• pH 3.00

33
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is
added to a) 1.00 L of pure water (after HCl,
pH 3.00) b) 1.00 L of buffer that has HOAc
0.700 M and OAc- 0.600 M (pH 4.68)

• Solution to Part (b)
• Step 1 do the stoichiometry
• H3O (from HCl) OAc- (from buffer) ---gt
  HOAc (from buffer)
• The reaction occurs completely because K is very
  large.

34
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is
added to a) 1.00 L of pure water (after HCl, pH
3.00) b) 1.00 L of buffer that has HOAc
0.700 M and OAc- 0.600 M (pH 4.68)

• Solution to Part (b) Step 1Stoichiometry
• H3O OAc- HOAc
• Before rxn 0.00100 0.600 0.700
• Change -0.00100 -0.00100 0.00100
• After rxn 0 0.599 0.701

35
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is
added to a) 1.00 L of pure water (after HCl, pH
3.00) b) 1.00 L of buffer that has HOAc
0.700 M and OAc- 0.600 M (pH 4.68)

• Solution to Part (b) Step 2Equilibrium
• HOAc H2O H3O OAc-
• HOAc OAc- H3O
• Before rxn 0.701 0.599 0
• Change -x x x
• After rxn 0.710-x 0.599x x

36
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is
added to a) 1.00 L of pure water (after HCl, pH
3.00) b) 1.00 L of buffer that has HOAc
0.700 M and OAc- 0.600 M (pH 4.68)
Solution to Part (b) Step 2Equilibrium HOAc
H2O H3O OAc-

• HOAc OAc- H3O
• After rxn 0.710-x 0.599x x
• Because H3O 2.1 x 10-5 M BEFORE adding
  HCl, we again neglect x relative to 0.701 and
  0.599.

37
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is
added to a) 1.00 L of pure water (after HCl, pH
3.00) b) 1.00 L of buffer that has HOAc
0.700 M and OAc- 0.600 M (pH 4.68)
Solution to Part (b) Step 2Equilibrium HOAc
H2O OAc- H3O

• H3O 2.1 x 10-5 M ------gt pH 4.68
• The pH has not changed significantly upon
  adding HCl to the buffer!

38
Preparing a Buffer

• You want to buffer a solution at pH 4.30.
• This means H3O 10-pH 5.0 x 10-5 M
• It is best to choose an acid such that H3O
  is about equal to Ka (or pH pKa).
• You get the exact H3O by adjusting the ratio
  of acid to conjugate base.

39
Preparing a Buffer Solution

• Buffer prepared from
• HCO3- weak acid
• CO32- conjugate base
• HCO3- H2O H3O CO32-

40
Preparing a Buffer

• You want to buffer a solution at pH 4.30 or
  H3O 5.0 x 10-5 M
• POSSIBLE ACIDS Ka
• HSO4- / SO42- 1.2 x 10-2
• HOAc / OAc- 1.8 x 10-5
• HCN / CN- 4.0 x 10-10
• Best choice is acetic acid / acetate.

41
Preparing a Buffer

• You want to buffer a solution at pH 4.30 or
• H3O 5.0 x 10-5 M

Solve for HOAc/OAc- ratio 2.78/ 1
Therefore, if you use 0.100 mol of NaOAc and
0.278 mol of HOAc, you will have pH 4.30.
42
Preparing a Buffer

• A final point
• CONCENTRATION of the acid and conjugate base are
  not important.
• It is the RATIO OF THE NUMBER OF MOLES of each.
• This simplifying approximation will be correct
  for all buffers with 3ltpHlt11, since the H
  will be small compared to the acid and conjugate
  base.

43
REVIEW PROBLEMS

• Calculate the pH of a 0.10 M HNO2 solution before
  and after making the solution 0.25 M in NaNO2.
• Calculate the pH of 0.500 L of a buffer solution
  composed of 0.10 M sodium acetate and 0.15 M
  acetic acid, before and after adding 1.0 grams of
  sodium hydroxide solid (no volume change).

44
REVIEW PROBLEMS

• Calculate the pH of a solution that is 0.18 M in
  Na2HPO4 and 0.12 M in NaH2PO4. 6.2 x 10-8
• Suggest an appropriate buffer system for pH 5.0.

SOLUTIONS
45
Titrations
45
pH
Titrant volume, mL
46
Acid-Base Titrations

• Adding NaOH from the buret to acetic acid in
  the flask, a weak acid.
• In the beginning the pH increases very slowly.

47
Acid-Base Titrations

• Additional NaOH is added.
• pH rises as equivalence point is approached.

48
Acid-Base Titrations

• Additional NaOH is added.
• pH increases and then levels off as NaOH is
  added beyond the equivalence point.

49
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
Equivalence point
pH of solution of benzoic acid, a weak acid
50
Acid-Base Titrations
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?

• EQUILIBRIUM PORTION
• Bz- H2O HBz OH- Kb
  1.6 x 10-10

Bz- HBz OH-
equilib 0.020 - x x x
Solving in the usual way, we find x OH-
1.8 x 10-6, pOH 5.75, and pH 8.25
51
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
Half-way point
52
Acid-Base Reactions
QUESTION You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH What
is the pH at the half-way point?

• HBz H2O H3O Bz- Ka
  6.3 x 10-5
• H3O HBz / Bz- Ka
• At the half-way point, HBz Bz-, so
• H3O Ka 6.3 x 10-5
• pH 4.20

53
Sample Problem
Titration Curve 20.00 mL 0.30 M HC2H3O2 is
titrated with 0.30 M NaOH. Calculate the pH
at
0 mL NaOH added

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.30 0
  0
• - x x
  x

0.30 x x
Ka

1.8 x 10-5
X 2.3 x 10-3 M H pH 2.63
54
Sample Problem
5.0 mL NaOH added

• HC2H3O2- OH- ---gt C2H3O2- HOH
• 6.0 1.5
  0
• -1.5 -1.5 1.5

4.5 0 1.5
C2H3O2-
0.060 M
HC2H3O2
0.18 M
55
Sample Problem

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.18 0
  0.060
• - x x
  x

0.18 x 0.060
Ka

1.8 x 10-5
X 5.4 x 10-5 M H pH 4.27
56
Sample Problem
20.0 mL NaOH added

• HC2H3O2 OH- ---gt C2H3O2- HOH
• 6.0 6.0
  0
• -6.0 -6.0 6.0

0 0 6.0
C2H3O2-
0.15 M
57
Sample Problem

• C2H3O2- HOH lt---gt OH-
  HC2H3O2
• 0.15 0
  0
• - x
  x x

0.15 x
x
Kb

5.6 x 10-10
X 9.2 x 10-6 M OH- pH 8.96
58
Sample Problem
30.0 mL NaOH added

• HC2H3O2 OH- ---gt C2H3O2- HOH
• 6.0 9.0 0
• -6.0 -6.0 6.0

0 3.0 6.0
OH-
0.060 M
pOH 1.22 pH 12.78
59
Sample Problem
Titration Curve 25.00 mL 0.300 M HCl is
titrated with 0.7500 M NaOH. Calculate the pH
at
0 mL NaOH added
H 0.300 M pH 0.523
60
Sample Problem
5.00 mL NaOH added

• HCl NaOH ---gt NaCl
  HOH
• 12.50 3.75
• -3.75 -3.75

8.75 0
H
0.292 M
pH 0.535
61
Sample Problem
16.67 mL NaOH added

• HCl NaOH ---gt NaCl
  HOH
• 12.50 12.50
• -12.50 -12.50

0 0
pH 7.00
62
Sample Problem
20.00 mL NaOH added

• HCl NaOH ---gt NaCl
  HOH
• 12.50 15.00
• -12.50 -12.50

0 2.50
OH-
0.0556 M
pH 12.745
63
Sample Problem
Titration Curve 20.00 mL 0.150 M NH3 is
titrated with 0.100 M HCl. Calculate the pH
at
0 mL HCl added

• NH3 HOH lt---gt NH4
  OH-
• 0.150 0 0
• - x
  x x

0.150 x
x
Kb

1.8 x 10-5
X 1.6 x 10-3 M OH- pH 11.20
64
Sample Problem
10.0 mL HCl added

• NH3 H ---gt NH4
• 3.00 1.00 0
• -1.00 -1.00 1.00

2.00 0 1.00
NH4
0.0333 M
NH3
0.0667 M
65
Sample Problem

• NH4 lt---gt NH3 H
• 0.0333 0.0667
• - x x
  x

0.0333 0.0667 x
Ka

5.6 x 10-10
X 2.8 x 10-10 M H pH 9.55
66
Sample Problem
30.0 mL HCl added
NH3 H ---gt NH4
3.00 3.00 0
-3.00 -3.00 3.00
0 0 3.00
0.0600 M
NH4
67
Sample Problem

• NH4 lt---gt NH3
  H
• 0.0600 0
  0
• - x x
  x

0.0600 x x
Ka

5.6 x 10-10
X 5.8 x 10-6 M H pH 5.24
68
Sample Problem
40.0 mL HCl added
NH3 H ---gt NH4
3.00 4.00 0
-3.00 -3.00 3.00
0 1.00 3.00
0.0167 pH 1.78
H
69
Practice Problems
1. Determine the pH of a solution made by mixing
25.0 mL of 0.20 M nitric acid with 25.0 mL of
0.10 M potassium hydroxide. 2. Determine the pH
at the equivalence point if 20.0 mL of 0.30 M HCN
is titrated with 0.20 M sodium hydroxide. 3.
Determine the pH at the equivalence point if 15.0
mL of 0.20 M nitric acid is titrated with 0.20 M
ammonia.
70
Practice Problems
4. a) Calculate the pH of a solution made by
mixing 50.0 mL of 0.15 M formic acid and 0.41 g
of sodium formate. b) Calculate the pH if 10.0
mL of 0.10 M NaOH is added. 5. What is the pH of
a solution made by mixing 25.0 mL of 0.20 M
benzoic acid and 45.0 mL of 0.10 M sodium
benzoate? 6. How many moles of sodium carbonate
must be added to 0.20 mole sodium hydrogen
carbonate in 250. mL to obtain a pH of 10.00?
71
Practice Problems
7. How many mLs of 0.30 HCl must be added to
25.0 mL of 0.500 M sodium phosphate to produce a
solution with a pH of 13.00? 8. A 20.00 mL sample
of 0.500 M HNO3 is titrated with 0.500 M KOH.
Calculate the pH of the solution a) before the
titration begins. b) when 10.00 mL of base have
been added. c) when 19.00 mL of base have been
added. d) at the equivalence point. e) when
21.00 mL of base have been added.
72
Practice Problems
9. A 20.00 mL sample of 0.5000 M formic acid is
titrated with 0.500 M sodium hydroxide. Calculate
the pH of the solution a) before the titration
begins. b) when 10.00 mL of base have been
added. c) when 19.00 mL of base have been
added. d) at the equivalence point. e) when
21.00 mL of base have been added.
73
Practice Problems
10. A 20.00 mL sample of 0.400 M ammonia is
titrated with 0.200 M HCl. Calculate the pH of
the solution a) before the titration
begins. b) when 20.00 mL of acid have been
added. c) when 39.00 mL of acid have been
added. d) at the equivalence point. e) when
41.00 mL of acid have been added.
74
Practice Problems Answers
1. 1.30 2. 11.23 3. 5.12 4. 3.66, 3.77 5.
4.15 6. .096 7. 9.1 8. .301, .77, 2.0,
7.00, 12.00 9. 2.03, 3.74, 5.03, 8.57, 12.0 10.
11.43, 9.25, 7.66, 5.07, 2.48 The End!!
75
Acid-Base Reactions

• 1. Calculate the pH if 25.0 mL 0.20 M of HCl is
  added to 40.0 mL of 0.20 M NaOH.
• HCl NaOH ---gt NaCl HOH
• 5.0 8.0 0
• - 5.0 - 5.0 5.0

0 3.0 5.0
OH-
0.046 M
pOH 1.34 pH 12.66
76
Acid-Base Reactions

• 2. Calculate the pH at the equivalence point if
  25.00 mL 0.20 M of HCl is titrated with 0.20 M
  NH3.
• HCl NH3 ---gt NH4 Cl-
• 5.0 5.0 0 0
• - 5.0 - 5.0 5.0

0 0 5.0
NH4
0.10 M
77
Acid-Base Reactions

• 2. NH4 0.10 M
• NH4 lt---gt NH3 H
• O.10 0 0
• - x x x

0.10 x x
Ka

5.6 x 10-10
X 7.5 x 10-6 M H pH 5.12
78
Acid-Base Reactions

• 3. Calculate the pH at the equivalence point if
  30.00 mL 0.20 M of HC2H3O2 is titrated with 0.30
  M NaOH.
• HC2H3O2 OH- ---gt HOH C2H3O2-
• 6.0 6.0 0
• - 6.0 - 6.0
  6.0

0 0
6.0
C2H3O2-
0.12M
79
Acid-Base Reactions

• 3. C2H3O2- 0.12 M
• C2H3O2- HOH lt---gt HC2H3O2 OH-
• 0.12 0 0
• - x
  x x

0.12 x
x
Kb

5.6 x 10-10
X 8.2 x 10-6 OH- pH 8.91
80
Sample Problem
Calculate the pH of a 0.100 M HC2H3O2 solution.

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.100 0 0
• - x x
  x

0.100 x x
Ka

1.8 x 10-5
X 1.3 x 10-3 M H pH 2.89
81
Sample Problem
Calculate the pH of a solution that is 0.100 M
HC2H3O2 and 0.100 M NaC2H3O2

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.100 0
  0.100
• - x x
  x

0.100 x 0.100
Ka

1.8 x 10-5
X 1.8 x 10-5 M H pH 4.74
82
Sample Problem
1.00 L H2O has a pH 7.00 Calculate the pH if
0.010 mole HCL is added.
H
0.010
pH 2.00 Adding 0.010 mole HCl changes the pH
from 7.00 to 2.00, 5.00 pH units.
83
Sample Problem
1.00 L 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 has a
pH 4.74 Calculate the pH if 0.010 mole HCL is
added.

• C2H3O2- H ---gt HC2H3O2
• 0.100 0.010 0.100
• - 0.010 - 0.010 0.010

0.090 0 0.110
C2H3O2-
0.090 M
HC2H3O2
0.110 M
84
Sample Problem

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.110 0
  0.090
• - x x
  x

0.110 x 0.090
Ka

1.8 x 10-5
X 2.2 x 10-5 M H pH 4.66
Adding 0.010 mole HCl changes the pH from 4.74 to
4.66, only 0.08 pH units.
85
Preparing a Buffer

• Preparing Buffers
• 1. Solid/Solid mix two solids. (Example 1)
• 2. Solid/Solution mix one solid and one
  solution.(Example 2)
• 3. Solution/Solution mix two solutions.
• (Example 3)
• 4. Neutralization Mix weak acid with
  strong base (Examples 4 and 5)
• or weak base with strong acid.
• (Example 6)

86
Sample Problem
1. Calculate the pH of a solution made by mixing
1.5 moles of phthalic acid and 1.2 moles of
sodium hydrogen phthalate in 500. mL of
soln. Ka 3.0 x 10-4
Conjugates do not react!!

• HA lt---gt H
  A-
• 3.0 0
  2.4
• - x x
  x

3.0 - x x
2.4 x
Ka

3.0 x 10-4
x 3.7 x 10-4 M pH 3.43
87
Sample Problem
2. How many moles of sodium acetate must be added
to 500. mL of 0.25 M acetic acid to produce a
solution with a pH of 5.50? X moles NaC2H3O2
Conjugates do not react!!

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.25 0
  x/0.500
• - 3.2x10-6 3.2x10-6
  3.2x10-6

0.25 3.2x10-6 x/0.500
Ka

1.8 x 10-5
x 0.70 mole NaC2H3O2
88
Sample Problem
3. How many mLs of 0.10 M sodium acetate must be
added to 20.0 mL of 0.20 M acetic acid to produce
a solution with a pH of 3.50? X mLs NaC2H3O2

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.20(20.0/20.0x) 0
  0.10(x/20.0x)
• - 3.2x10-4 3.2x10-4
  3.2x10-4

0.20(20.0/20.0x) 3.2x10-4
0.10(x/20.0x)
Conjugates do not react!!
89
Sample Problem
Ka
1.8 x 10-5
x 2.2 mL NaC2H3O2
90
Sample Problem
4. How many moles of potassium hydroxide must be
added to 500. mL of .250 M HCN to produce a
solution with a pH of 9.00? X moles KOH

• OH- HCN ---gt CN- HOH
• x 0.125 0
• - x - x x

0 0.125 - x x
CN-
HCN
91
Sample Problem

• HCN lt---gt H
  CN-
• (0.125 - x)/0.500
  x/0.500
• - 1.0 x 10-9 1.0 x 10-9 1.0 x
  10-9

(0.125 - x)/0.500 1.0 x 10-9 x/0.500
Ka

4.0x10-10
X 0.036 mole KOH
92
Sample Problem

• 5. How many mLs of 0.30 M sodium hydroxide must
  be added to 25.0 mL of 0.500 M acetic acid to
  produce a solution with a pH of 4.10?
• X mL NaOH
• HC2H3O2 OH- ---gt HOH C2H3O2-
• 12.5 0.30x 0
• - 0.30x - 0.30x
  0.30x

12.5 - 0.30x 0
0.30x
C2H3O2-
HC2H3O2
93
Sample Problem

• HC2H3O2 lt---gt C2H3O2-
  H
• (12.5 - 0.30x)/(25.0x) 0.30x/(25.0x)
  0
• - 7.9 x 10-5 7.9 x 10-5
  7.9 x 10-5

(12.5 - 0.30x)/(25.0x) 0.30x/(25.0x)
7.9 x 10-5
Ka

1.8 x 10-5
X 7.6 mL NaOH
94
Sample Problem

• 6. Calculate the pH of a solution made by mixing
  50.0 mL of 0.15 M NH3 and 20.0 mL of 0.10 M HCl.
• HCl NH3 ---gt NH4 Cl-
• 2.0 7.5 0 0
• - 2.0 - 2.0 2.0 NA

0 5.5 2.0 NA
NH4
0.029 M
NH3
0.079 M
95
Sample Problem

• NH4 lt---gt NH3 H
• 0.029 0.079 0
• - x x
  x

0.029 0.079 x
Ka

5.6 x 10-10
X 2.1 x 10-10 M H pH 9.68
For more practice do all General problems 84-101
96
Sample Problems

• 1. Calculate the pH of a 0.10 M HNO2 solution
  before and after making the solution 0.25 M in
  NaNO2.

97
Sample Problem
Calculate the pH of a 0.10 M HNO2 solution.

• HNO2 lt---gt H NO2-
• 0.10 0
  0
• - x x
  x

0.10 - x x x
Ka

4.5 x 10-4
X 6.5 x 10-3 M H pH 2.19
98
Sample Problem
Calculate the pH of a 0.10 M HNO2 and 0.25 M
NaNO2 solution.

• HNO2 lt---gt H NO2-
• 0.10 0 0.25
• - x x
  x

0.10 - x x 0.25 x
Ka

4.5 x 10-4
X 1.8 x 10-4 M H pH 3.74
99
Sample Problems

• 2. Calculate the pH of 0.500 L of a buffer
  solution composed of 0.10 M sodium acetate and
  0.15 M acetic acid, before and after adding 1.0
  grams of sodium hydroxide solid (no
  volume change).

100
Sample Problem
Calculate the pH of 0.500 L of a buffer solution
composed of 0.10 M sodium acetate and 0.15 M
acetic acid.

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.15 0
  0.10
• - x x
  x

0.15 x 0.10
Ka

1.8 x 10-5
X 2.7 x 10-5 M H pH 4.57
101
Sample Problem
Calculate the pH after adding 1.0 grams of sodium
hydroxide solid.
HC2H3O (.500L)(.15M) .075 mole NaC2H3O (.500L)(.
10M) .050 mole NaOH (1.0g)(40.0g/mole) .025
mole
HC2H3O2 OH- ---gt HOH
C2H3O2- 0.075 0.025
0.050 - 0.025 - 0.025
0.025
0.050 0 0
.075
C2H3O2-
0.15 M
HC2H3O2
0.10
102
Sample Problem
Calculate the pH after adding 1.0 grams of sodium
hydroxide solid.

• HC2H3O2 lt---gt H
  C2H3O2-
• 0.10
  0.15
• - x x
  x

0.15 x 0.10
Ka

1.8 x 10-5
X 1.2 x 10-5 M H pH 4.92
103
Sample Problems

• 3. Calculate the pH of a solution that is 0.18 M
  in Na2HPO4 and 0.12 M in NaH2PO4.

104
Sample Problem

• H2PO4- lt---gt H
  HPO42-
• 0.12 0
  0.18
• - x x
  x

0.12 x 0.18
Ka

6.2 x 10-8
X 4.1 x 10-8 M H pH 7.38
105
Sample Problems

• 4. Suggest an appropriate buffer system for pH
  5.0.

Name of Acid Ka pKa Oxalic Acid 3.8 x
10-2 1.42 Hydrogen Sulfate Ion 1.2 x
10-2 1.92 Phosphoric Acid 7.1 x
10-3 2.15 Formic Acid 1.8 x 10-4 3.74 Hydrogen
Oxalate Ion 5.0 x 10-4 4.30 Acetic Acid 1.8 x
10-5 4.74 Dihydrogen Phosphate Ion 6.3 x
10-8 7.20 Boric Acid 6.0 x 10-10 9.22 Ammonium
Ion 9.6 x 10-10 9.25 Hydrogen Carbonate
Ion 4.7 x 10-11 10.33 Hydrogen Phosphate
Ion 4.4 x 10-13 12.36