Acid-Base Equilibria

1
Acid-Base Equilibria
Mar. 23

• The reaction of weak acids with water,
• OR
• the reaction of weak bases with water,
• always results in an equilibrium!!
• The equilibrium constant for the reaction of a
  weak acid with water is Ka

2
Acid-Base Equilibria
Mar. 23

• eg. HF(aq) H2O(l) º

H3O(aq) F-(aq)
Keq ?
3
Acid-Base Equilibria
Mar. 23

• For any weak acid
• Why is H2O(l) omitted from the Ka expression?

4
Acid-Base Equilibria
Mar. 23

• the equilibrium constant for the reaction of a
  weak base with water is Kb
• HS-(aq) H2O(l) º
• Kb

H2S(aq) OH-(aq)
5
Acid-Base Equilibria
Mar. 23

• For any weak base

6
Mar. 23

• eg.
• Write the expression for Kb for S2-(aq)
• ANSWER
• S2-(aq) H2O(l) º

Worksheet 5
HS-(aq) OH-(aq)
7
5.a) Use Ka to find H3O for 0.100 mol/L
HF(aq)
Mar. 24
HF(aq) H2O(l) º H3O(aq) F-(aq)
Ka 6.6 x 10-4
8
Mar. 24
1st try - Ignore x
x 2 (0.100)(6.6 x 10-4) x 2 6.6 x 10-5 x
8.1 x 10-3 mol/L
9
Mar. 24
2nd try Include x
x 2 (0.0919)(6.6 x 10-4) x 2 6.0654 x 10-5 x
7.8 x 10-3 mol/L
Different than 1st try CANNOT IGNORE DISSOCIATION
10
Mar. 24
3rd try Include new x
x 2 (0.0922)(6.6 x 10-4) x 2 6.0852 x 10-5 x
7.8 x 10-3 mol/L H3O 7.8 x 10-3 mol/L
Same as 2nd try
11
5.b) find H3O for 0.250 mol/L CH3COOH(aq
Mar. 24
CH3COOH(aq) H2O(l) º H3O(aq) CH3COO-(aq)
Ka 1.8 x 10-5
12
Mar. 24
1st try - Ignore x
x 2 (0.250)(1.8 x 10-5) x 2 4.5 x 10-6 x
2.1 x 10-3 mol/L
13
Mar. 24
2nd try Include x
x 2 (0.2479)(1.8 x 10-5) x 2 4.462 x 10-6 x
2.1 x 10-3 mol/L H3O 2.1 x 10-3 mol/L
Same as 1st try
14
Mar. 24
To ignore OR not to ignore that is the question
15
pH of a weak acid
Mar. 24

• Step 1 Write a balanced equation
• Step 2 ICE table OR assign variables
• Step 3 Write the Ka expression
• Step 4 Check (can we ignore dissociation)
• Step 5 Substitute into Ka expression to find
  H3O

16
pH of a weak acid
Mar. 24

• eg. Find pH of 0.100 mol/L HF(aq).
• Step 1 Write a balanced equation
• HF(aq) H2O(l) º H3O(aq) F-(aq)

17
Step 2 Equilibrium Concentrations
Mar. 24

Let x H3O at equilibrium F- x HF
0.100 - x
18
Step 3 Write the Ka expression
Mar. 24
19
Step 4 Check (can we ignore dissociation)
Mar. 24
dissociation (- x) may be IGNORED
151
Acid dissociation CANNOT be IGNORED in this
question.
We have to use the x
20
Step 5 Substitute into Ka expression
Quadratic Formula!!
x2 6.6 x 10-5 - 6.6 x 10-4 x
x2 6.6 x 10-4 x - 6.6 x 10-5 0
a 1 b 6.6 x 10-4 c -6.6 x 10-5
21
Ignore negative roots
22
Mar. 29
Try these

• a) Find the H3O in 0.250 mol/L HCN(aq)
• Check 4.0 x 108
• x 1.24 x 10-5
• H3O 1.24 x 10-5
• b) Calculate the pH of 0.0300 mol/L HCOOH(aq)
• Check 167
• x 2.24 x 10-3
• pH 2.651

23
Mar. 29

• HCN H2O ? H3O CN-
• Let x H3O
• x CN-
• 0.250 x HCN
• Check

4.0 x 108
Quadratic NOT needed
24
Mar. 29
x2 1.55 x 10-10
x 1.25 x 10-5 pH 4.904
25
Mar. 29

• HCOOH H2O ? H3O HCOO-
• 0.0300 0 0
• -x x x
• 0.0300 x x x
• Check

Quadratic needed
167
26
Mar. 29
x2 5.4 x 10-6 - 1.8 x 10-4x x2 1.8 x
10-4x - 5.4 x 10-6 0
A 1 B 1.8 x 10-4 C -5.4 x 10-6
x 2.24 x 10-3 pH 2.651
27
Practice
Mar. 29

• Formic acid, HCOOH, is present in the sting of
  certain ants. What is the H3O of a 0.025
  mol/L solution of formic acid? (0.00203 mol/L)
• Calculate the pH of a sample of vinegar that
  contains 0.83 mol/L acetic acid.
• ( H3O 3.87 x 10-3 pH 2.413 )
• What is the percent dissociation of the vinegar
  in 2.?
• diss 0.466

28
Practice
Mar. 29
Worksheet 6

• A solution of hydrofluoric acid has a molar
  concentration of 0.0100 mol/L. What is the pH of
  this solution?
• ( H3O 0.00226 pH 2.646 )
• The word butter comes from the Greek butyros.
  Butanoic acid, C3H7COOH, gives rancid butter its
  distinctive odour. Calculate the H3O of a 1.0
  10-2 mol/L solution of butanoic acid.
• (Ka 1.51 10-5 ) (Ans 3.89 x 10-4
  mol/L)

29
pH of a weak base
Apr. 1

• same method as acids
• ignore dissociation if
• to calculate Kb (usually given on the exam)

30
pH of a weak base
Apr. 1

• Calculate the pH of 0.0100mol/L Na2CO3(aq)

31
Apr. 1

• CO32- H2O ? HCO3- OH-
• 0.0100 0 0
• -x x x
• 0.0100 x x x
• Check

2.13 x 10-4
47
? Quadratic needed
32
Apr. 1
x2 2.13 x 10-6 - 2.13 x 10-4x x2 2.13 x
10-4x - 2.13 x 10-6 0
A 1 B 2.13 x 10-4 C -2.13 x 10-6 x
1.36 x 10-3 OH- 1.36 x 10-3 mol/L pOH
?? pH 11.13
33
pH of a weak base
Apr. 1

• Calculate the pH of 0.500 mol/L NaNO2(aq)

Worksheet 7
34
Apr. 1

• NO2- H2O ? HNO2 OH-
• 0.500 0 0
• -x x x
• 0.500 x x x
• Check

1.39 x 10-11
3.6 x 1010
OK to ignore x here ie. NO Quadratic
35
Apr. 1
x2 6.95 x 10-12
x 2.6 x 10-6
OH- 2.6 x 10-6 mol/L pOH ?? pH 8.42
36
Calculating Ka from weak acid and pH

• eg. The pH of a 0.072 mol/L solution of benzoic
  acid, C6H5COOH, is 2.68. Calculate the numerical
  value of the Ka for this acid.
• Equation
• Find H3O from pH
• Subtract from weak acid
• Substitute to find Ka

Apr. 1
See p. 591 6 8
37
C6H5COOH(aq) H2O(l) º H3O(aq)
C6H5COO-(aq)
Apr. 1
H3O 10-2.68 0.00209 mol/L
C6H5COO- 0.00209 mol/L
C6H5COOH 0.072 0.00209 0.06991 mol/L
Find Ka
6.2 x 10-5
38
Calculating Ka from weak acid and pH

• eg. The pH of a 0.072 mol/L solution of benzoic
  acid, C6H5COOH, is 2.68. Calculate the
  dissociation for this acid.

Apr. 1
H3O 10-2.68 0.00209 mol/L
See p. 591 s 5 6
2.9
39
Apr. 1
Calculate the acid dissociation constant, Ka ,
and the percent dissociation for each acid

• a) 0.250 mol/L chlorous acid, HClO2(aq) pH
  1.31
• 0.012 19.5
• b) 0.150 mol/L cyanic acid, HCNO(aq) pH 2.15
• 0.000035 4.7
• c) 0.100 mol/L arsenic acid, H3AsO4(aq) pH
  1.70
• 0.0050 20
• 0.500 mol/L iodic acid, HIO3(aq) pH 0.670
• 0.160 42.8

40
Apr. 1

• More Practice
• Weak Acids
• pp. 591, 592 s 6 -8
• Weak Bases
• p. 595 s 11 - 16 (Kbs on p. 592)

Worksheet 8
41
Acid-Base Stoichiometry
Apr. 4

• Solution Stoichiometry (Review)
• Write a balanced equation
• Calculate moles given ( OR n CV)
• Mole ratios
• Calculate required quantity
• OR OR m nM

42

• eg. 25.0 mL of 0.100 mol/L H2SO4(aq) was used to
  neutralize 36.5 mL of NaOH(aq). Calculate the
  molar concentration of the NaOH solution.

Apr. 4
H2SO4(aq) NaOH(aq)
? H2O(l) Na2SO4(aq)
2
2
nH2SO4
nNaOH
CNaOH
43
Acid-Base Stoichiometry
Apr. 4

• pp. 600, 601 Sample Problems
• p. 602 s 17 - 20

Worksheet 9
44

• Acids Bases 9 answers
• 0.210 mol/L
• a) 22.5 mL
• b) 24.7 mL
• c) 4.8 mL
• 31.5 mL
• a) 0.0992 mol/L
• b) 0.269 mol/L
• c) 0.552 mol/L

45

• Dilution
• Given 3 of the four variables
• Only one solution
• CiVi CfVf
• Stoichiometry
• Given 3 of the four variables
• Two different solutions
• 4 step method

Apr. 6
46
Excess Acid or Base
Apr. 6

• To calculate the pH of a solution produced by
  mixing an acid with a base
• write the B-L equation (NIE)
• calculate the moles of H3O and OH-
• subtract to determine the moles of excess H3O or
  OH-
• divide by total volume to get concentration
• calculate pH

47
Apr. 6

• eg. 20.0 mL of 0.0100 M Ca(OH)2(aq) is mixed with
  10.0 mL of 0.00500 M HCl(aq).
• Determine the pH of the resulting solution.
• ANSWER
• Species present
• Ca2 OH- H3O Cl- H2O

SB
SA
48
NIE OH- H3O ? 2 H2O
Apr. 6
C 0.00500 mol/L V 0.0100 L
C 0.0200 mol/L V 0.0200 L
4.00 x 10-4 mol OH- 5.0 x 10-5 mol
H3O 3.5 x 10-4 mol excess OH-
49

• 0.01167 mol/L
• OH- 0.01167 mol/L
• pOH 1.933
• pH 12.067

Apr. 6
Worksheet 10
50
Acids Bases 10 answers
Apr. 6
1. pH 13.000 2. H3O 4.12 x 10-2
mol/L OH- 2.43 x 10-13 mol/L 3. pH
13.125 4. pH 7 p. 586 s 1 4
51
Indicators
Apr. 7

• An indicator is a weak acid that changes color
  with changes in pH
• HIn is the general formula for an indicator
• To choose an indicator for a titration, the pH of
  the endpoint must be within the pH range over
  which the indicator changes color

52
Apr. 7

• HIn(aq) H2O(l) º H3O(aq) In-(aq)
• Colour 1 Colour 2
• HIn is the acid form of the indicator.
• Adding H3O causes colour 1 (LCP)
• Adding OH- removes the H3O causes colour 2

LCP
53
Apr. 8

• methyl orange
• HMo(aq) H2O(l) º H3O(aq) Mo-(aq)
• red yellow
• bromothymol blue
• HBb(aq) H2O(l) º H3O(aq) Bb-(aq)
• yellow blue

Your Turn!! p. 23
54
Indicators p. 23

• 1.a) HMv H2O(l) º H3O(aq) Mv-(aq)
• b) HBb H2O(l) º H3O(aq) Bb-(aq)
• 2.  Indicator pH colour
• thymol blue 3.0  yellow
• methyl red 7.9  yellow
• phenolpthalein 7.1 colourless
• indigocarmine 13.5  yellow
• 3.a) pH range 2.8 4.5
• b) pH range 8.0 8.2

55
Acid-Base Titration (p. 603 ? )

• A titration is a lab technique used to determine
  an unknown solution concentration.
• A standard solution is added to a known volume
  of solution until the endpoint of the titration
  is reached.
• The endpoint occurs when there is a sharp change
  in colour
• The equivalence point occurs when the moles of
  H3O equals the moles of OH-

56
Acid-Base Titration

• The colour change is caused by the indicator
  added to the titration flask.
• An indicator is a chemical that changes color
  over a given pH range (See indicator table)
• A buret is used to add the standard solution
• standard solution - solution of known
  concentration

57
Acid-Base Titration

• primary standard - a standard solution which can
  be made by direct weighing of a stable chemical.
• Data from titrations allows us to calculate an
  unknown solution concentration.

58
Titration Calculation
Omit first trial OVERSHOT the endpoint

• eg. Clem Student performed a titration by adding
  0.250 mol/L HCl(aq) to 10.0 mL samples of
  Ca(OH)2(aq). Use the data below to determine the
  molar concentration of Ca(OH)2(aq).
• Trial 1 2 3 4
• Final volume 8.48 mL 15.70 mL 22.91
  mL 30.14 mL
• Initial volume 1.05 mL 8.48 mL 15.70
  mL 22.91 mL
• Volume HCl(aq)
• used

7.43 mL
7.23 mL
7.21 mL
7.22 mL
59

• Equation
• 2 HCl(aq) Ca(OH)2(aq) ? CaCl2(aq) 2
  H2O(l)
• nHCl
• nCa(OH)2
• C

C 0.250 mol/L Vave 0.00722 L
C ? mol/L V 0.0100 L
60
Acid-Base Titration

• Titration Lab pp. 606, 607

Worksheet 11
61
Multi-Step Titrations (p. 609 - 611)
Apr. 12

• Polyprotic acids donate their protons one at a
  time when reacted with a base.
• eg. Write the equations for the steps that occur
  when H3PO4(aq) is titrated with NaOH(aq)
• H3PO4(aq) OH-(aq)
• H2PO4-(aq) OH-(aq)
• HPO42-(aq) OH-(aq)

62
Multi-Step Titrations
Apr. 12

• H3PO4(aq) OH-(aq) ? H2PO4-(aq) H2O(l)
• H2PO4-(aq) OH-(aq) ? HPO42-(aq) H2O(l)
• HPO42-(aq) OH-(aq) º PO43-(aq) H2O(l)
• H3PO4(aq) 3 OH-(aq) º PO43-(aq) 3 H2O(l)

63
Multi-Step Titrations
Apr. 12

• Write the balanced net ionic equations, and the
  overall equation, for the titration of Na2S(aq)
  with HCl(aq).
• p. 611 s 21.b), 22, 23
• LAST TOPIC!! Titration Curves

64
Acids and Bases

• Properties / Operational Definitions
• Acid-Base Theories and Limitations
• Arrhenius H-X and X-OH
• Modified react with water ? hydronium
• BLT proton donor/acceptor (CA and CB)
• Writing Net Ionic Equations (BLT)
• Strong vs. Weak
• pH pOH calculations
• Equilibria (Kw, Ka, Kb)
• Titrations/Indicators/Titration Curves
• Dilutions and Excess Reagent questions

65
Step 2 ICE table

• HF H3O F-
• I
• C
• E

0.100 mol/L 0 0
-x x x
0.100 - x x x
66
0.0100mol/L CO23-(aq)

• CO32-(aq) H2O(l) º HCO3-(aq) OH-(aq)

PO43- HPO42-
OH- I C E
0.0100 mol/L 0 0
-x x x
0.0100 - x x x
67
Acid-Base Stoichiometry

• Solution stoichiometry (4 question sheet)
• Excess reagent problems (use NIE)
• Titrations
• Titration curves
• Indicators
• STSE Acids Around Us

68

• A primary standard is a pure substance that is
  stable enough to be stored indefinitely without
  decomposition, can be weighed accurately without
  special precautions when exposed to air, and will
  undergo an accurate stoichiometric reaction in a
  titration.

69
15. pH of 0.297 mol/L HOCl

• HOCl(aq) H2O(l) º H3O(aq) OCl-(aq)
• Let x H3O at equilibrium
• OCl- x
• HOCl 0.297 - x

70
Check
dissociation (- x) may be IGNORED
X 9.28 x 10-5 pH 4.03
71
16. NIE OH- H3O ? 2 H2O
0.484 mol/L 0.07000 L
0.125 mol/L 0.02500 L
0.03388 mol OH- 0.003125 mol H3O 0.030755
mol excess OH- OH- 0.3237 mol/L pOH
0.490 pH 13.510
72

• Ignore dissociation
• OH- 0.0146 mol/L
• diss 2.92
• Vave 10.975 mL
• nNaOH 0.001262 mol
• nH2SO4 0.000631 mol
• C 0.0252 mol/L
• Kb 3.93 x 10-4
• diss 6.27

73

• c) 2.50 mol/L NaCN(aq) Kb 1.61 x 10-5
• Check 1.5 x 105
• x 6.34 x 10-3
• pOH 2.20 pH 11.80
• d) 0.100 mol/L K2S(aq) Kb 0.0769
• Check 1.3
• x 0.0573
• pOH 1.24 pH 12.76