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 K a 1
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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]
3
Acid-Base Equilibria
● the equilibrium constant for the reaction of a weak base with water is Kb
HS-(aq) + H2O(l)
H2S(aq) + OH-(aq)
4
Kb =[OH-] [H
2S]
[HS-]
Acid-Base Equilibria
● For any weak base
Kb =[OH-] [conjugate acid]
[weak base]
5
eg.
Write the expression for Kb for S2-(aq)
ANSWER:
S2-(aq) + H2O(l)
Kb =[OH-] [HS-]
[S2-]
HS-(aq) + OH-
(aq)
6
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
x = 0.0081 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
Different than 1st try:CANNOT IGNORE
DISSOCIATION
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
Same as 2nd try:
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
Same as 1st try:
14
To ignore OR not to ignore:
that is the question
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
15
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)
16
Step #2: Equilibrium Concentrations
Let x = [H3O+] at equilibrium
[F-] = x
[HF] = 0.100 - x
17
Step #3: Write the Ka expression
Ka =[H3O+] [F-]
[HF]
18
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) part 19
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●Need the Quadratic Equation!!a = 1 b = 6.6 x 10-4 c = -6.6 x 10-5
20
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
21
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:
HCN + H2O H⇋ 3O+ + CN-
Let x = [H3O+]
x = [CN-]
0.250 – x = [HCN]
Check:
23
Ka = [H3O+] [CN-] [HCN]
= 4.0 x 108 0.250
6.2 x 10-10
Quadratic NOTneeded
24
x]- [0.250
[x] [x]10 x 6.2 10-
x = 1.25 x 10-5
pH = 4.904
x2 = 1.55 x 10-10
HCOOH + H2O H⇋ 3O+ + HCOO-
0.0300 0 0
-x +x +x
0.0300 – x x x
Check:
25
Ka = [H3O+] [HCOO-] [HCOOH]
= 167 0.0300
1.8 x 10-4
Quadratic needed
26
x]- [0.0300
[x] [x]10 x 1.8 4-
A = 1 B = 1.8 x 10-4 C = -5.4 x 10-6
x = 2.24 x 10-3
pH = 2.651
x2 = 5.4 x 10-6 - 1.8 x 10-4x
x2 + 1.8 x 10-4x - 5.4 x 10-6 = 0
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, 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)
pH of a weak base same method as acids ignore dissociation if
to calculate Kb (usually given on the exam)
K x K Ka b w KK
Kbw
a
29
[weak base]
Kb
> 500
pH of a weak baseCalculate the pH of 0.0100mol/L Na2CO3(aq)
30
CO32- + H2O HCO⇋ 3
- + OH-
0.0100 0 0
-x +x +x
0.0100 – x x x
Check:
31
Kb = [OH-] [HCO3-]
[CO32-]
= 47 0.0100
2.13 x 10-4→ Quadratic needed
Kb = 1.00 x 10-14
4.7 x 10-11
= 2.13 x 10-4
32
x]- [0.0100
[x] [x]10 x 2.13 4-
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/LpOH = ?? pH = 11.13
pH of a weak baseCalculate the pH of 0.500 mol/L NaNO2(aq)
33
NO2- + H2O HNO⇋ 2 + OH-
0.500 0 0
-x +x +x
0.500 – x x x
Check:
34
Kb = [OH-] [HCO3-]
[CO32-]
= 3.6 x 1010 0.500
1.39 x 10-11 OK to ignore –x hereie. NO Quadratic
Kb = 1.00 x 10-14
7.2 x 10-4
= 1.39 x 10-11
35
x]- [0.500
[x] [x]10 x 1.39 11-
x2 = 6.95 x 10-12
[OH-] = 2.6 x 10-6 mol/LpOH = ??pH = 8.42
x = 2.6 x 10-6
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
36
C6H5COOH(aq) + H2O(l) ⇋ H3O+(aq) + C6H5COO-
(aq)
[H3O+] = 10-2.68 = 0.00209 mol/L
[C6H5COOH] = 0.072 – 0.00209
= 0.06991 mol/LFind Ka
Ka =(0.00209)(0.00209)
(0.06991)= 6.2 x 10-5
[C6H5COO-] = 0.00209 mol/L
37
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
100%xacid] [weak
]O[Hdiss % 3
= 2.9 %
38
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 = 2.15
0.00035 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%
Calculate the acid dissociation constant, Ka , and the percent dissociation for each acid:
39
Apr. 4
More Practice: ● Weak Acids:
pp. 592 # 8● Weak Bases:
p. 595 #’s 13, 15, 16 (Kb’s on p. 592)
40
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
41
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 = C x V = 0.100 mol/L x 0.0250L = 0.00250 mol
NNaOH = 2 x nH2SO4 = 0.00500 mol/L
CNaOH = n = 0.00500 mol/L = 0.137 mol/L V 0.0365 L
42
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
44
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
45
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
46
C = 0.0200 mol/LV = 0.0200 L
C = 0.00500 mol/L V = 0.0100 L
B-L Equation: OH- + H3O+ → 2 H2O
4.00 x 10-4 mol OH- 5.0 x 10-5 mol H3O+
3.5 x 10-4 mol excess OH-
47
n = CV n = CV
= 0.01167 mol/L
[OH-] = 0.01167 mol/L
pOH = 1.933
pH = 12.067
totalV
nC
48
L0.0300
mol10x3.5 4
Indicators● 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
49
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
50
methyl orange
HMo(aq) + H2O(l) ⇋ H3O+(aq) + Mo-
(aq)
red yellow
bromothymol blue
HBb(aq) + H2O(l) ⇋ H3O+(aq) + Bb-
(aq)
yellow blue
51
Related Documents
