Transcript Document

Acid-Base Equilibria
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
Acid-Base Equilibria
eg. HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
Keq = ?
-]
[H
O+]
[F
3
Ka =
[HF]
Acid-Base Equilibria
 For
any weak acid
Ka =
 Why
[H3O+] [conjugate base]
[weak acid]
is H2O(l) omitted from the Ka
expression?
Acid-Base Equilibria
 the
equilibrium constant for the reaction of
a weak base with water is Kb
HS-
(aq)
+ H2O(l)
Kb =
H
S
+
OH
 2 (aq)
(aq)
Acid-Base Equilibria
 For
any weak base
Kb =
[OH-] [conjugate acid]
[weak base]
eg.
Write the expression for Kb for S2-(aq)
ANSWER:
S2-
(aq)
HS
+
OH
+ H2O(l) 
(aq)
(aq)
Kb =
[OH-] [HS-]
[S2-]
5.a) Use Ka to find [H3O+] for 0.100 mol/L HF(aq)
HF(aq) + H2O(l)

H3O+(aq) + F-(aq)

Ka = 6.6 x
10-4
Ka 
-
[ H 3 O ] [F ]
[HF]
1st try - Ignore x
6.6 x 10
-4

[x] [x]
[0.100 - 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
2nd try– Include x
6.6 x 10
-4

[x] [x]
[0.100 - 0.0081]
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
6.6 x 10
-4

[x] [x]
[0.100 - 0.0078]
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)
Ka = 1.8 x 10-5

Ka 
-
[H 3 O ] [CH 3 COO ]
[CH 3 COOH]
1st try - Ignore x
1.8 x 10
-5

[x] [x]
[0.250 - 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
2nd try– Include x
1.8 x 10
-5

[x] [x]
[0.250 - 0.0021]
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 (pp. 587 – 592)
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
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)
Step #2: Equilibrium Concentrations
Let x = [H3O+] at equilibrium
[F-] = x
[HF] = 0.100 - x
Step #3: Ka expression
Ka =
[H3O+] [F-]
[HF]
Step #4: Check (can we ignore dissociation)
If [weak acid] > 500
Ka
dissociation (- x)
may be IGNORED
(0.100)
=
151
6.6 x 10-4
Acid dissociation CANNOT be
IGNORED in this question.
Step #5: Substitute into Ka expression
6.6 x 10
-4

[x] [x]
[0.100 - x]
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
b
x 
b  4ac
2
2a
x 
 6.6 x 10
-4

(6.6 x 10
-4
)  4(1)(-6.6
2
2(1)
x 
 6.6 x 10
-4

0.000264
2
Ignore
negative
roots
x  0.0078 mol/L
x 10
-5
)
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
Practice
1.
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
3.
pH = 2.413 )
What is the percent dissociation of the
vinegar in 2.?
% diss = 0.466 %
22
Practice
4.
A solution of hydrofluoric acid has a molar
concentration of 0.0100 mol/L. What is the pH of
this solution?
( [H3O+] = 0.00226
5.
pH = 2.646 )
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)
23
pH of a weak base
same method as acids
 calculate Kb

Ka x Kb  K w

Kb 
Kw
Ka
ignore dissociation if
24
pH of a weak base
Calculate the pH of 0.0100mol/L Na2CO3(aq)
25
pH of a weak base
Calculate 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
27
C6H5COOH(aq) + H2O(l)  H3O+(aq) + C6H5COO-(aq)
[H3O+] = 10-2.68 = 0.00209 mol/L
[C6H5COO-] = 0.00209 mol/L
Find Ka
Ka =
[C6H5COOH] = 0.072 – 0.00209
= 0.06991 mol/L
(0.00209)(0.00209)
= 6.2 x 10-5
(0.06991)
28
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.
[H3O+] = 10-2.68
= 0.00209 mol/L
See p. 591 #’s 5 & 6

% diss 
[H 3 O ]
x 100%
[weak acid]

0.00209
0.072
x 100%
= 2.9 %
29
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
0.00035
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%
30
More Practice:
 Weak
Acids:
pp. 591, 592 #’s 6 -8
 Weak Bases:
p. 595 #’s 11 - 16 (Kb’s on p. 592)
1.a)
b)
2.a)
b)
1.4 x 10-10
0.0014 %
2.5 x 10-9
0.0080 %
3.a)
b)
4.a)
b)
1.6 x 10-9
0.0080 %
2.7 x 10-9
0.042 %
31
Acid-Base Stoichiometry
Solution Stoichiometry (Review)
1. Write a balanced equation
m
2. Calculate moles given ( n 
OR n = CV)
M
3. Mole ratios
4. Calculate required quantity
n OR
n OR m = nM
V
C
C
V
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) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq)
nH SO =
2
4
nNaOH =
CNaOH =
33
Acid-Base Stoichiometry
pp. 600, 601 – Sample Problems
p. 602 #’s 17 - 20
34
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 Base
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
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+ OHH3O+
Cl- H2O
SB
SA
37
NIE:
OH- + H3O+ → 2 H2O
0.0200 mol/L
0.0200 L
0.00500 mol/L
0.0100 L
n = CV
4.00 x 10-4 mol OH-
5.0 x 10-5 mol H3O+
3.5 x 10-4 mol excess OH38
C
n
V total

3.5 x 10
4
mol
0.0300 L
= 0.01167 mol/L
[OH-] = 0.01167 mol/L
pOH = 1.933
pH = 12.067
39
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)
Colour #1
 H3O+(aq) + In-(aq)
Colour #2
 HIn
is the acid form of the indicator.
 Adding H3O+ causes colour 1 (LCP)
 Adding OH- removes the H3O+ & causes
colour #2
41
methyl orange
HMo(aq) + H2O(l)
red
 H3O+(aq) + Mo-(aq)
yellow
bromothymol blue
HBb(aq) + H2O(l)
yellow
 H3O+(aq) + Bb-(aq)
blue
42
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.
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.
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
Acid-Base Titration
standard 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 Titrations
H3PO4(aq) + OH-(aq) → H2PO4-(aq) + H2O(l)
H2PO4-(aq) + OH-(aq) → HPO42-(aq) + H2O(l)
HPO42-(aq) + OH-(aq)  PO43-(aq) + H2O(l)
H3PO4(aq) + 3 OH-(aq)  PO43-(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
49
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
Step #2: ICE table
[HF]
[H3O+]
[F-]
I
0.100 mol/L
0
0
C
-x
+x
+x
E
0.100 - x
x
x
0.0100mol/L CO23-(aq)
CO32-(aq) + H2O(l) HCO3-(aq) + OH(aq)
I
C
E
[PO43-]
0.0100 mol/L
-x
0.0100 - x
[HPO42-]
[OH-]
0
0
+x
+x
x
x
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.
Excess reactant
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
Acid-Base Stoichiometry
Solution stoichiometry (4 question sheet)
 Excess reagent problems (use NIE)
 Titrations
 Titration curves
 Indicators
 STSE: Acids Around Us

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
WorkSheet #10 answers:
1. pH = 13.000
2. [H3O+] = 4.12 x 10-2 mol/L
-13
[OH ] = 2.43 x 10 mol/L
3. pH = 13.125
4. pH = 7
p. 586 #’s 1 – 4

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.
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]
Check:
(0.297)
7
=
1.02
x
10
2.9 x 10-8
dissociation (- x)
may be IGNORED
2.9 x 10
-8

[x] [x]
[0.297]
X = 9.28 x 10-5
pH = 4.03
16. NIE:
OH- + H3O+ → 2 H2O
0.484 mol/L
0.07000 L
0.03388 mol OH-
0.125 mol/L
0.02500 L
0.003125 mol H3O+
0.030755 mol excess OH[OH-] = 0.3237 mol/L
pOH = 0.490
pH = 13.510
17.
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
19. Kb = 3.93 x 10-4
% diss = 6.27
18.