Acid-Base EquilibriaThe 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
1
Acid-Base Equilibria
eg. HF(aq) + H2O(l)
Ka =[H3O+] [F-]
[HF]
H3O+(aq) + F-
(aq)
Keq = ?
2
Acid-Base Equilibria
For any weak acid
Why is H2O(l) omitted from the Ka expression?
Ka =[H3O+] [conjugate base]
[weak acid]
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Acid-Base Equilibria
the equilibrium constant for the reaction of a weak base with water is Kb
HS-(aq) + H2O(l)
Kb =
H2S(aq) + OH-(aq)
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Acid-Base Equilibria
For any weak base
Kb =[OH-] [conjugate acid]
[weak base]
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eg.
Write the expression for Kb for S2-(aq)
ANSWER:
S2-(aq) + H2O(l)
Kb =[OH-] [HS-]
[S2-]
HS-(aq) + OH-
(aq)
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5.a) Use Ka to find [H3O+] for 0.100 mol/L HF(aq)
HF(aq) + H2O(l) H3O+(aq) + F-
(aq)
[HF]
][F ]OH[K
-3
a
Ka = 6.6 x 10-4
x]- [0.100
[x] [x]10 x 6.6 4-
x 2 = (0.100)(6.6 x 10-4)
x 2 = 6.6 x 10-5
x = 8.1 x 10-3 mol/L
1st try - Ignore x
2nd try– Include x
0.0081] - [0.100
[x] [x]10 x 6.6 4-
x 2 = (0.0919)(6.6 x 10-4)
x 2 = 6.0654 x 10-5
x = 7.8 x 10-3 mol/L
3rd try– Include new x
0.0078] - [0.100
[x] [x]10 x 6.6 4-
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
5.b) find [H3O+] for 0.250 mol/L CH3COOH(aq
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-
(aq)
COOH][CH
]COO[CH ]OH[K
3
-33
a
Ka = 1.8 x 10-5
x]- [0.250
[x] [x]10 x 1.8 5-
x 2 = (0.250)(1.8 x 10-5)
x 2 = 4.5 x 10-6
x = 2.1 x 10-3 mol/L
1st try - Ignore x
2nd try– Include x
0.0021] - [0.250
[x] [x]10 x 1.8 5-
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
pH of a weak acid
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
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pH of a weak acid
eg. Find pH of 0.100 mol/L HF(aq).Step #1: Write a balanced equation
HF(aq) + H2O(l) H3O+(aq) + F-
(aq)
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Step #2: Equilibrium Concentrations
Let x = [H3O+] at equilibrium
[F-] = x
[HF] = 0.100 - x
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Step #3: Ka expression
Ka =[H3O+] [F-]
[HF]
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Step #4: Check (can we ignore dissociation)
dissociation (- x) may be IGNORED
= 151 (0.100)
6.6 x 10-4
Acid dissociation CANNOT beIGNORED in this question.
[weak acid]
Ka
If > 500
We have to use the – x
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Step #5: Substitute into Ka expression
x]- [0.100
[x] [x]10 x 6.6 4-
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
QuadraticFormula!!
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2a
4acbbx
2
2(1)
)10x4(1)(-6.6)10x(6.610x6.6x
-52-4-4
2
0.00026410x6.6x
-4
mol/L0.0078x Ignore
negative roots
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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
Try these:
Practice1. 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)
2. 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 )
3. What is the percent dissociation of the vinegar in 2.?
% diss = 0.466 %
Practice4. 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 )
5. The word “butter” comes from the Greek butyros. Butanoic acid 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 )
(3.89 x 10-4 mol/L)
pH of a weak base same method as acids calculate Kb
ignore dissociation if
K x K Ka b w KK
Kbw
a
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pH of a weak baseCalculate the pH of 0.0100mol/L Na2CO3(aq)
25
pH of a weak baseCalculate the pH of 0.500 mol/L NaNO2(aq)
26
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 See p. 591 #6 & 8
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C6H5COOH(aq) + H2O(l) H3O+(aq) + C6H5COO-
(aq)
[H3O+] = 10-2.68 = 0.00209 mol/L
[C6H5COOH] = 0.072 – 0.00209
= 0.06991 mol/L
Find Ka
Ka =(0.00209)(0.00209)
(0.06991)= 6.2 x 10-5
[C6H5COO-] = 0.00209 mol/L
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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.
See p. 591 #’s 5 & 6[H3O+] = 10-2.68
= 0.00209 mol/L
100%xacid] [weak
]O[Hdiss % 3
= 2.9 %
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100% x 0.072
0.00209
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 =
2.150.00035 4.7%
c) 0.100 mol/L arsenic acid, H3AsO4(aq); pH = 1.70
0.0050 20%d) 0.500 mol/L iodic acid, HIO3(aq); pH = 0.670
0.160 42.8%
Calculate the acid dissociation constant, Ka , and the percent dissociation for each acid:
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More Practice: Weak Acids:
pp. 591, 592 #’s 6 -8 Weak Bases:
p. 595 #’s 11 - 16 (Kb’s on p. 592)
31
1.a) 1.4 x 10-10
b) 0.0014 %2.a) 2.5 x 10-9
b) 0.0080 %
3.a) 1.6 x 10-9
b) 0.0080 %4.a) 2.7 x 10-9
b) 0.042 %
Acid-Base Stoichiometry
Solution Stoichiometry (Review)
1. Write a balanced equation
2. Calculate moles given ( OR n = CV)
3. Mole ratios
4. Calculate required quantity
OR OR m = nM
M
mn
C
nV
V
nC
32
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.
H2SO4(aq) + NaOH(aq) → H2O(l) + Na2SO4(aq)22
nH2SO4 =
nNaOH =
CNaOH =
33
Acid-Base Stoichiometry
pp. 600, 601 – Sample Problems
p. 602 #’s 17 - 20
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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
35
Excess Acid or BaseTo 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
36
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
37
0.0200 mol/L0.0200 L
0.00500 mol/L0.0100 L
NIE: OH- + H3O+ → 2 H2O
n = CV 4.00 x 10-4 mol OH- 5.0 x 10-5 mol H3O+
3.5 x 10-4 mol excess OH-
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= 0.01167 mol/L
[OH-] = 0.01167 mol/L
pOH = 1.933
pH = 12.067
totalV
nC
39
L0.0300
mol10x3.5 4
Indicators An indicator is a weak acid that
changes color with changes in pH
To choose an indicator for a titration, the pH of the endpoint must be within the pH range over which the indicator changes color
40
HIn(aq) + H2O(l) 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
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methyl orange
HMo(aq) + H2O(l) H3O+(aq) + Mo-
(aq)
red yellow
bromothymol blue
HBb(aq) + H2O(l) H3O+(aq) + Bb-
(aq)
yellow blue
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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.
43
Acid-Base Titration
The endpoint occurs when there is a sharp change in colour
The equivalence point occurs when the moles of hydronium equals the moles of hydroxide
The colour change is caused by the indicator added to the titration flask.
44
Acid-Base Titration
An indicator is a chemical that changes colour over a given pH range
(See indicator table) A buret is used to deliver the standard
solution
45
Acid-Base Titrationstandard solution - solution of known
concentration
primary standard - a standard solution which can be made by direct weighing of a stable chemical.
Titration Lab – pp. 606, 607
46
Multi-Step Titrations (p. 609 - 611)
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) 47
Multi-Step TitrationsH3PO4(aq) + OH-
(aq) → H2PO4-(aq) + H2O(l)
H2PO4-(aq) + OH-
(aq) → HPO42-
(aq) + H2O(l)
HPO42-
(aq) + OH-(aq) PO4
3-(aq) + H2O(l)
H3PO4(aq) + 3 OH-(aq) PO4
3-(aq) + 3 H2O(l)
48
Multi-Step Titrations
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
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Properties / Operational Definitions Acid-Base Theories and Limitations
Arrhenius – H-X and X-OHModified – react with water → hydroniumBLT – 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
Acids and Bases
50
Step #2: ICE table
[HF] [H3O+] [F-]
I
C
E
0.100 mol/L 0 0
-x +x +x
0.100 - x x x
51
CO32-
(aq) + H2O(l) HCO3-(aq) + OH-
(aq)
0.0100mol/L CO23-
(aq)
[PO43-] [HPO4
2-] [OH-]
I
C
E
0.0100 mol/L 0 0
-x + x + x
0.0100 - x x x52
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:
Ca(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + CaCl2(aq)
nbase = 0.0002 mol
→ needs 0.004 mol HCl
nacid = 0.00005 mol
→ needs 0.000025 mol Ca(OH)2
Limiting reactant
Excess reactant
53
Acid-Base Stoichiometry Solution stoichiometry (4 question sheet) Excess reagent problems (use NIE) Titrations Titration curves Indicators STSE: Acids Around Us
54
WorkSheet #9 answers:
1. 0.210 mol/L
2. a) 22.5 mL
b) 24.7 mL
c) 4.8 mL
3. 31.5 mL
4. a) 0.0992 mol/L
b) 0.269 mol/L
c) 0.552 mol/L
55
WorkSheet #10 answers:
1. pH = 13.0002. [H3O+] = 4.12 x 10-2 mol/L
[OH-] = 2.43 x 10-13 mol/L3. pH = 13.1254. pH = 7
p. 586 #’s 1 – 456
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.
57
15. pH of 0.297 mol/L HOCl
HOCl(aq) + H2O(l) H3O+(aq) + OCl-(aq)
Let x = [H3O+] at equilibrium
[OCl-] = x
[HOCl] = 0.297 - x
Ka =[H3O+] [OCl-]
[HOCl]
58
Check:
dissociation (- x) may be IGNORED
= 1.02 x 107 (0.297)
2.9 x 10-8
[0.297]
[x] [x]10 x 2.9 8-
X = 9.28 x 10-5 pH = 4.03
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0.484 mol/L0.07000 L
0.125 mol/L0.02500 L
16. NIE: OH- + H3O+ → 2 H2O
0.03388 mol OH- 0.003125 mol H3O+
0.030755 mol excess OH-
[OH-] = 0.3237 mol/L pOH = 0.490
pH = 13.51060
17. Ignore dissociation
[OH-] = 0.0146 mol/L
% diss = 2.92 %
18. Vave = 10.975 mL
nNaOH = 0.001262 mol
nH2SO4 = 0.000631 mol
C = 0.0252 mol/L
19. Kb = 3.93 x 10-4
% diss = 6.27
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