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)

H2S(aq) + OH-(aq)

4

Kb =[OH-] [H

2S]

[HS-]

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

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:

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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

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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: Write the 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) 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

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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:

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

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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

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[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:

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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

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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:

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Apr. 4

More Practice: ● Weak Acids:

pp. 592 # 8● Weak Bases:

p. 595 #’s 13, 15, 16 (Kb’s on p. 592)

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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

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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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