chapter 3 acids & bases

Copyright 2011 Pearson Education, Inc.

Chapter 15

Acids and

Bases

Chapter 16

Ionic Equilibrium

CHM096


CHAPTER 3: ACID-BASE EQUILIBRIUM

3.1 Definitions of Acids and Bases

• Arrhenius Theory

• Bronsted-Lowry Concept

• Conjugate Acid-Base Relationship

• Lewis Acids and Bases

3.2 Strength of Acids and Bases

• pH and Acid strength

• Self-Ionisation of Water

• Acid and Base Ionisation Costants

• Hydrolysis of Salts

3.3 Buffer Solutions and Acid-Base Titrations

• Properties of Buffer Solutions

• The Henderson-Hasselbalch Equation

• Titration Curves

• Acid-Base Indicators

Tro: Chemistry: A Molecular Approach 2


Structures of Acids

3

• Binary acids have acid hydrogens attached to a

nonmetal atom

HCl, HF

Tro, Chemistry: A Molecular Approach, 2/e


Structure of Acids

4

• Oxy acids have acid hydrogens attached to an

oxygen atom

H2SO4, HNO3



Structure of Acids

5

• Carboxylic acids have

COOH group

HC2H3O2, H3C6H5O7

• Only the first H in the

formula is acidic

the H is on the COOH



Common Acids

6 Tro, Chemistry: A Molecular Approach, 2/e


Properties of Bases

7

• Also known as alkalis

• Taste bitter

alkaloids = plant product tHAt is

alkaline

often poisonous

• Solutions feel slippery

• Change color of vegetable dyes

red litmus turns blue

• React with acids to form ionic salts

neutralization



Common Bases



Arrhenius Theory

• Bases dissociate in water to produce OH− ions and cations

ionic substances dissociate in water

NaOH(aq) → Na+(aq) + OH−(aq)

• Acids ionize in water to produce H+ ions and anions because molecular acids are not made of ions, they cannot

dissociate

they must be pulled apart, or ionized, by the water

HCl(aq) → H+(aq) + Cl−(aq) in formula, ionizable H written in front

HC2H3O2(aq) → H+(aq) + C2H3O2−(aq)



Arrhenius Theory

HCl ionizes in water,

producing H+ and Cl– ions

NaOH dissociates in water,

producing Na+ and OH– ions



Hydronium Ion

• The H+ ions produced by the acid are so reactive they cannot

exist in water

H+ ions are protons!!

• Instead, they react with water molecules to produce complex

ions, mainly hydronium ion, H3O+

H+ + H2O H3O+



Arrhenius Acid–Base Reactions

• The H+ from the acid combines with the OH− from the base

to make a molecule of H2O

it is often helpful to think of H2O as H-OH

• The cation from the base combines with the anion from the

acid to make a salt

acid + base → salt + water

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)



Limitations of Arrhenius Theory

• Does not explain why molecular substances, such as NH3, dissolve in water to form basic solutions – even though they do not contain OH– ions

• Does not explain how some ionic compounds, such as Na2CO3 or Na2O, dissolve in water to form basic solutions – even though they do not contain OH– ions

• Does not explain why molecular substances, such as CO2, dissolve in water to form acidic solutions – even though they do not contain H+ ions

• Does not explain acid–base reactions that take place in non-aqueous solution



Brønsted-Lowry Theory

• In a Brønsted-Lowry acid–base reaction, a proton (H+)

is transferred

• The acid is an H+ donor

• The base is an H+ acceptor base structure must contain an atom with an unsHAred pair of

electrons (lone pairs)

• In a Brønsted-Lowry acid-base reaction, the acid

molecule gives an H+ to the base molecule

H–A + :B :A– + H–B+



Brønsted-Lowry Acids

• Brønsted-Lowry acids are H+ donors

any material that has H can potentially be a Brønsted-Lowry acid

• When HCl dissolves in water, the HCl is the acid because HCl transfers an H+ to H2O, forming H3O

+ ions

water acts as base, accepting H+

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

acid base



Brønsted-Lowry Bases

• Brønsted-Lowry bases are H+ acceptors any material that has atoms with lone pairs can potentially be a

Brønsted-Lowry base

• When NH3 dissolves in water, the NH3(aq) is the base

because NH3 accepts an H+ from H2O, forming OH–(aq) water acts as acid, donating H+

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

base acid



Examples of Molecules with Lone Pairs

• You need to be able to recognize when an atom in a molecule has lone

pair electrons and when it doesn’t!


CH4 has no lone pair


Examples of Molecules with Lone Pairs

18

HClO

HCO3−



Amphoteric Substances

• Amphoteric substances can act as either an acid or a

base

because they have both a transferable H and an atom with lone

pair electrons

• Water acts as base, accepting H+ from HCl

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

• Water acts as acid, donating H+ to NH3

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Therefore, water is amphoteric



Brønsted-Lowry

Acid-Base Reactions

• One of the advantages of Brønsted-Lowry theory is that it allows reactions to be reversible

H–A + :B :A– + H–B+

• The original base has an extra H+ after the reaction, so it will act as an acid in the reverse process

• And the original acid has a lone pair of electrons after the reaction – so it will act as a base in the reverse process

:A– + H–B+ H–A + :B



Conjugate Pairs

• In a Brønsted-Lowry acid–base reaction, the original base

becomes an acid in the reverse reaction, and the original

acid becomes a base in the reverse process

• Each reactant and the product it becomes is called a

conjugate pair

• In an acid-base reaction, there are two aid-base conjugate

pairs.

• For the reaction: H–A + :B :A– + H–B+

:A– is the conjugate base of HA

HB+ is the conjugate acid of :B



Brønsted-Lowry

Acid–Base Reactions

H–A + :B :A– + H–B+

acid base conjugate conjugate

base acid

HCHO2 + H2O CHO2– + H3O

+


base acid

H2O + NH3 HO– + NH4+


base acid



Conjugate Pairs

In the reaction H2O + NH3 HO– + NH4+

H2O and HO– constitute an

acid/conjugate base pair

NH3 and NH4+ constitute a

base/conjugate acid pair



Practice – Write the formula for the conjugate acid of

the following

H2O

NH3

CO32−

H2PO41−

H2O H3O+

NH3 NH4+

CO32− HCO3

−

H2PO41− H3PO4



Practice – Write the formula for the conjugate base

of the following

H2O

NH3

CO32−

H2O HO−

NH3 NH2−

CO32− because CO3

2− does not have an H, it

cannot be an acid

H2PO41− HPO4

2−



Example 15.1a: Identify the Brønsted-Lowry acids and bases, and

their conjugates, in the reaction

H2SO4 + H2O HSO4– + H3O

+


base acid

H2SO4 + H2O HSO4– + H3O

+

When the H2SO4 becomes HSO4, it loses an H+ so H2SO4 must

be the acid and HSO4 its conjugate base

When the H2O becomes H3O+, it accepts an H+ so H2O must be

the base and H3O+ its conjugate acid



Example 15.1b: Identify the Brønsted-Lowry acids and bases and

their conjugates in the reaction

HCO3– + H2O H2CO3 + HO–

base acid conjugate conjugate

acid base

HCO3– + H2O H2CO3 + HO–

When the HCO3

becomes H2CO3, it accepts an H+ so HCO3

must be the base and H2CO3 its conjugate acid

When the H2O becomes OH, it donats an H+ so H2O must be

the acid and OH its conjugate base



Practice – Identify the Brønsted-Lowry acid, base, conjugate

acid, and conjugate base in the following reaction

HSO4−

(aq) + HCO3−

(aq) SO42−

(aq) + H2CO3(aq)

Base Conjugate

Acid

Acid Conjugate

Base



Practice — Write the equations for the following reacting with

water and acting as a monoprotic acid & label the conjugate

acid and base

HBr + H2O Br− + H3O+

Acid Base Conj. Conj.

base acid

HSO4− + H2O SO4

2− + H3O+

Acid Base Conj. Conj.

base acid

HBr

HSO4−



Comparing Arrhenius Theory and Brønsted-Lowry Theory

30

• Brønsted–Lowry theory

HCl(aq) + H2O(l) Cl−(aq) + H3O+(aq)

HF(aq) + H2O(l) F−(aq) + H3O+(aq)

NaOH(aq) Na+(aq) + OH−(aq)

NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)

• Arrhenius theory

HCl(aq) H+(aq) + Cl−(aq)

HF(aq) H+(aq) + F−(aq)

NaOH(aq) Na+(aq) + OH−(aq)

NH4OH(aq) NH4+(aq) + OH−(aq)



Lewis Acid–Base Theory

• Lewis Acid–Base theory focuses on transferring an electron pair lone pair bond

Does NOT require H atoms

• The electron donor is called the Lewis Base electron rich, therefore nucleophile

• The electron acceptor is called the Lewis Acid electron deficient, therefore electrophile

• Anions are better Lewis Bases than neutral atoms or molecules

N: < N:−

• Generally, the more electronegative an atom, the less willing it is to be a Lewis Base

O: < S:



Lewis Acid–Base Reactions

• The base donates a pair of electrons to the acid

• Generally results in a dative bond forming

H3N: + BF3 H3N─BF3

No H transfer

• The product that forms is called an adduct

• Arrhenius and Brønsted-Lowry acid–base reactions are

also Lewis acid–base reactions



Examples of Lewis Acid–Base Reactions



Examples of Lewis Acid–Base Reactions

Ag+(aq) + 2 :NH3(aq) Ag(NH3)2

+(aq)

Lewis

Acid

Lewis

Base Adduct



Practice –

Identify the Lewis Acid and Lewis Base in Each Reaction

35

Lewis

Base

Lewis

Base

Lewis

Acid

Lewis

Acid



Strong or Weak

• A strong acid is a strong electrolyte

practically all the acid molecules ionize, →

• A strong base is a strong electrolyte

practically all the base molecules form OH– ions, either through dissociation or reaction with water, →

• A weak acid is a weak electrolyte

only a small percentage of the molecules ionize,

• A weak base is a weak electrolyte

only a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water,



Strong Acids

37

HCl H+ + Cl−

HCl + H2O H3O+ + Cl−

0.10 M HCl = 0.10 M H3O+

• The stronger the acid, the more

willing it is to donate H+

we use water as the standard base

to donate H+ to

• Strong acids donate practically all

their H+’s

100% ionized in water

strong electrolyte

• [H3O+] = [strong acid]

[X] means the molarity of X



HF H+ + F−

HF + H2O H3O+ + F−

0.10 M HF ≠ 0.10 M H3O+

Weak Acids

38

• Weak acids donate a small

fraction of their H+’s

most of the weak acid molecules do

not donate H+ to water

less than 1% ionization in water

• [H3O+] << [weak acid]



Strong & Weak Acids



Autoionization of Water

• Water is actually an extremely weak electrolyte

therefore there must be a few ions present

• About 2 out of every 1 billion water molecules form ions

through a process called autoionization or self-ionisation

H2O H+ + OH–

H2O + H2O H3O+ + OH–

• All aqueous solutions contain both H3O+ and OH–

the concentration of H3O+ and OH– are equal in water

[H3O+] = [OH–] = 10−7M @ 25 °C



Ion Product Constant of Water

• In any aqueous solution, the product of the [H3O+] and

[OH–] concentrations is always a constant number

• The number is called the Ion Product Constant of

Water and has the symbol Kw

• [H3O+] x [OH–] = Kw = 1.00 x 10−14 @ 25 °C

if you measure one of the concentrations, you can calculate the

other

• As [H3O+] increases, the [OH–] must decrease so the

product stays constant



Acidic and Basic Solutions

• All aqueous solutions contain both H3O+ and OH– ions

• Neutral solutions have equal [H3O+] and [OH–]

[H3O+] = [OH–] = 1.00 x 10−7

• Acidic solutions have a larger [H3O+] than [OH–]

[H3O+] > 1.00 x 10−7; [OH–] < 1.00 x 10−7

• Basic solutions have a larger [OH–] than [H3O+]

[H3O+] < 1.00 x 10−7; [OH–] > 1.00 x 10−7



Practice – Complete the table

[H+] vs. [OH−]

OH− H+ H+ H+ H+ H+

OH− OH−

OH− OH−

[OH−]

[H+] 100 10−1 10−3 10−5 10−7 10−9 10−11 10−13 10−14



Practice – Complete the table [H+] vs. [OH−]

OH− H+ H+ H+ H+ H+

OH− OH− OH−

OH−

[OH−]10−14 10−13 10−11 10−9 10−7 10−5 10−3 10−1 100

[H+] 100 10−1 10−3 10−5 10−7 10−9 10−11 10−13 10−14

Even though it may look like it, neither H+ nor OH− will ever be 0

The sizes of the H+ and OH− are not to scale

because the divisions are powers of 10 rather than units

Acid Base



Example 15.2b:

Calculate the [OH] at 25 °C when the [H3O+] = 1.5 x 10−9 M,

and determine if the solution is acidic, basic, or neutral

46

the units are correct; the fact tHAt the

[H3O+] < [OH] means the solution is basic

[H3O+] = 1.5 x 10−9 M

[OH]

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

[H3O+] [OH]



Practice: Determine the [H3O+] when the [OH−] = 2.5 x 10−9 M



Practice :

Determine the [H3O+] when the [OH−] = 2.5 x 10−9 M

48

the units are correct; the fact that

[H3O+] > [OH] means the solution is acidic

[OH] = 2.5 x 10−9 M

[H3O+]

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

[OH] [H3O+]



Measuring Acidity: pH

49

• The acidity or basicity of a solution is

often expressed as pH

• pH = −log[H3O+]

exponent on 10 with a positive sign

pHwater = −log[10−7] = 7

need to know the [H3O+] concentration

to find pH

• pH < 7 is acidic; pH > 7 is basic, pH = 7 is

neutral

• [H3O+] = 10−pH



[H3O+] and [OH−] in a

Strong Acid or Strong Base Solution

• There are two sources of H3O+ in an aqueous solution of a

strong acid – the acid and the water

• There are two sources of OH− in an aqueous solution of a

strong acid – the base and the water

• For a strong acid or base, the contribution of the water to the

total [H3O+] or [OH−] is negligible

the [H3O+]acid shifts the Kw equilibrium so far that [H3O

+]water is too

small to be significant

except in very dilute solutions, generally < 1 x 10−4 M



Finding pH of a Strong Acid or Strong Base

Solution

• For a monoprotic strong acid [H3O+] = [HA]

for polyprotic acids, the other ionizations can generally be ignored

0.10 M HCl HAs [H3O+] = 0.10 M and pH = 1.00

• For a strong ionic base, [OH−] = (number OH−)x[Base] for molecular bases with multiple lone pairs available, only one lone

pair accepts an H, the other reactions can generally be ignored

0.10 M Ca(OH)2 HAs [OH−] = 0.20 M and pH = 13.30



Example 15.3b: Calculate the pH at 25 °C when the [OH] = 1.3 x 10−2

M, and determine if the solution is acidic, basic, or neutral

52

pH is unitless; the fact that the pH > 7 means the

solution is basic

[OH] = 1.3 x 10−2 M

pH

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

[H3O+] [OH] pH



Practice – Determine the pH @ 25 ºC of a solution

that has [OH−] = 2.5 x 10−9 M



Practice – Determine the pH @ 25 ºC of a solution that has

[OH−] = 2.5 x 10−9 M

54

pH is unitless; the fact that the pH < 7 means

the solution is acidic

[OH] = 2.5 x 10−9 M

pH

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

[H3O+] [OH] pH



Practice – Determine the [OH−] of a solution with a

pH of 5.40

55

because the pH < 7, [OH−] should be less

tHAn 1 x 10−7; and it is

pH = 5.40

[OH−], M

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

[H3O+] pH [OH]



pOH

56

• Another way of expressing the acidity/basicity of a solution is

pOH

• pOH = −log[OH],

• [OH] = 10−pOH

pOHwater = −log[10−7] = 7

need to know the [OH] concentration to find pOH

• pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is neutral

• pH + pOH = 14.0



Relationship between pH and pOH

• pH + pOH = 14.00

at 25 °C

you can use pOH to find pH of a solution

57

[H3O ][OH ] K

w 1.0 1014

log [H3O ][OH ] log 1.0 1014

log [H3O ] log [OH ] 14.00

pH pOH 14.00



Example: Calculate the pH at 25 °C when the [OH] = 1.3 x

10−2 M, and determine if the solution is acidic, basic or neutral

58

pH is unitless; since t pH > 7, the solution is

basic

[OH] = 1.3 x 10−2 M

pH

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

pOH [OH] pH



Practice – Determine the pOH at 25 ºC of a solution that has

[H3O+] = 2.5 x 10−9 M

59

pH is unitless; the fact tHAt the pH < 7 means

the solution is acidic

[H3O+] = 2.5 x 10−9 M

pOH

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

pH [H3O+] pOH



Example 15.11: Calculate the pH at 25 °C of a 0.0015 M

Sr(OH)2 solution and determine if the solution is acidic, basic,

or neutral

pH is unitless; the fact tHAt the pH > 7 means

the solution is basic

[Sr(OH)2] = 1.5 x 10−3 M

pH

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

[H3O+] [OH] pH [Sr(OH)2]

[OH]=2[Sr(OH)2]

[OH]

= 2(0.0015)

= 0.0030 M



Practice – Calculate the pH of the following strong

acid or base solutions

0.0020 M HCl

0.0015 M Ca(OH)2

[H3O+] = [HCl] = 2.0 x 10−3 M

pH = −log(2.0 x 10−3) = 2.70

[OH−] = 2 x [Ca(OH)2] = 3.0 x 10−3 M

pOH = −log(3.0 x 10−3) = 2.52

pH = 14.00 − pOH = 14.00 − 2.52

pH = 11.48



Finding the pH of a Weak Acid

• There are also two sources of H3O+ in an aqueous solution

of a weak acid – the acid and the water

• However, finding the [H3O+] is complicated by the fact HA

the acid only undergoes partial ionization

• Calculating the [H3O+] requires the use of Ka for the

equilibrium :

HA + H2O A + H3O+



Acid Ionization Constant, Ka

• Acid strength measured by the size of the equilibrium constant when reacts with H2O

HA(aq) + H2O(ℓ) A−(aq) + H3O+(aq)

• The equilibrium constant for this reaction is called the acid ionization constant, Ka

larger Ka = stronger acid [A− ] [H3O+]

Ka = [HA ]



[HNO2] [NO2−] [H3O

+]

initial 0.200 0 0

change

equilibrium

Example 15.6: Find the pH of 0.200 M HNO2(aq) solution @

25 °C, given that Ka for HNO2 = 4.6 x 10−4

66

represent the change in

the concentrations in

terms of x

sum the columns to find

the equilibrium

concentrations in terms

of x

substitute into the

equilibrium constant

expression

+x +x x

0.200 x x x

HNO2 + H2O(l) NO2−

(aq) + H3O+

(aq)




25 °C, given that Ka for HNO2 = 4.6 x 10−4

using simplifying approximation method

67

because Ka is very small,

approximate the [HNO2]eq

= [HNO2]init and solve for

x

[HNO2] [NO2−] [H3O

+]

initial 0.200 0 ≈ 0

change −x +x +x

equilibrium 0.200 x x 0.200 x




25 °C

68

check if the

approximation is

valid by seeing if

x < 5% of [HNO2]init

Ka for HNO2 = 4.6 x 10−4

[HNO2] [NO2−] [H3O

+]

initial 0.200 0 ≈ 0

change −x +x +x

equilibrium 0.200 x x

the approximation is valid

x = 9.6 x 10−3



Example 15.6: Find the pH of 0.200 M HNO2(aq)

solution at 25 °C

69

substitute x into

the equilibrium

concentration

definitions and

solve

Ka for HNO2 = 4.6 x 10−4

[HNO2] [NO2−] [H3O

+]

initial 0.200 0 ≈ 0

cHAnge −x +x +x

equilibrium 0.200−x x x

x = 9.6 x 10−3

[HNO2] [NO2−] [H3O

+]

initial 0.200 0 ≈ 0

change −x +x +x

equilibrium 0.190 0.0096 0.0096



Example 15.6: Find the pH of 0.200 M HNO2(aq)

solution at 25 °C

70

substitute [H3O+]

into the formula for

pH and solve

[HNO2] [NO2−] [H3O

+]

initial 0.200 0 ≈ 0

change −x +x +x

equilibrium 0.190 0.0096 0.0096



Determining [H3O+] of a weak acid from Ka

using the direct approach

For a weak acid HA with [HA] = c mol/L

[A−] [H3O+]

Ka = [HA ]

Since [A−] = [H3O

+]

[HA ] x Ka = [H3O+]²

[H3O

+]² = c x Ka

[H3O

+] = c x Ka



Practice – What is the pH of a 0.012 M solution of nicotinic

acid, HC6H4NO2?

(Ka = 1.4 x 10−5 at 25 °C)

[HC6H4NO2 ] = 0.012 M = c

[H3O+] = c x Ka

[H3O+] = 0.012 x 1.4 x 10−5 = 4.1 x 10−4 M

pH = ‒ log(4.1 x 10−4) = 3.39



Example 15.8: What is the Ka of a weak acid if a

0.100 M solution has a pH of 4.25?

73

use the pH to find the

equilibrium [H3O+]

write the reaction for

the acid with water

Compute Ka

HA + H2O A + H3O+

Ka = [H3O+]² / c

= (5.6 x 10−5)²/0.100

= 3.1 x 10−8



Practice –

What is the Ka of nicotinic acid, HC6H4NO2, if

a 0.012 M solution of nicotinic acid has a pH

of 3.40?

(Answer: 1.3 x 10−5 )



Weak Bases

75

NH3 + H2O NH4+ + OH−

• In weak bases, only a small fraction

of molecules accept H’s

weak electrolyte

most of the weak base molecules

do not take H from water

much less than 1% ionization in

water

• [HO–] << [weak base]

• Finding the pH of a weak base

solution is similar to finding the pH of

a weak acid



Base Ionization Constant, Kb

• Base strength measured by the size of the equilibrium

constant when a base B reacts with H2O

:B + H2O OH− + H:B+

• The equilibrium constant is called the base ionization

constant, Kb

larger Kb = stronger base



Structure of Amines



pK

• Another way of expressing the strength of a weak acid or

weak base is pK

• pKa = −log(Ka), Ka = 10−pKa

• The stronger the acid, the smaller the pKa

larger Ka = smaller pKa

because it is the –log

• pKb = −log(Kb), Kb = 10−pKb

• The stronger the base, the smaller the pKb

larger Kb = smaller pKb



Determine pKa

• a) HF has Ka = 3.5 x 10−4. Calculate the pKa.

• pKa = ‒logKa = ‒log(3.5 x 10−4) = 3.455

• b) The pKb of ethylamine, C2H5NH2 = 3.25. What is its Kb value?

• pKb = 3.25 Kb = 10−pKb = 10−3.25

• = 5.62 x 10−4



Example 15.12: Find the pH of 0.100 M NH3(aq)

81

determine the value of

Kb from Table 15.8

because Kb is very

small, approximate the

[NH3]eq = [NH3]init and

solve for x

Kb for NH3 = 1.76 x 10−5

[NH3] [NH4+] [OH]

initial 0.100 0 ≈ 0

change −x +x +x




Example 15.12: Find the pH of 0.100 M NH3(aq)

82

Example 15.12: To find pH of 0.100 M NH3(aq)

From previous calculation, obtained: x = 1.33 x 10−3



Practice – Find the pH of a 0.0015 M morphine solution, Kb =

1.6 x 10−6 using direct approximation method

B + H2O BH+ + OH

Taking [B] = c, and [BH+] = [OH]

[BH+][OH] [OH]²

Kb = =

[B] c

[OH] = (c x Kb) = (0.0015 x 1.6 x 10−6 )

= 4.9 x 10−5

pOH = ‒log(4.9 x 10−5) = 4.31

pH = 14 ‒ pOH = 14 ‒ 4.31

= 9.69



Acid–Base Properties of Salts: Salt Hydrolysis

• Salts are water-soluble ionic compounds

• Basic salts • Salts that contain the cation of a strong base and an anion

that is the conjugate base of a weak acid are basic NaF solutions are basic

Na+ is the cation of the strong base NaOH

F− is the conjugate base of the weak acid HF. F− sligthly hydrolyse in water to give OH−

F−(aq) + H2O(l) HF(aq) + OH−(aq)



SALTS

• Acidic Salts • Salts that contain cations that are the conjugate acid of a

weak base and an anion of a strong acid are acidic NH4Cl solutions are acidic

NH4+ is the conjugate acid of the weak base NH3

NH4+ under goes hydrolysis to produce slight excess of H+

NH4+ (aq) + H2O(l) NH4 OH(aq) + H+(aq)

Cl− is the anion of the strong acid HCl. No hydrolysis.

Neutral Salts

Salts that contain cations that are the conjugate acid of a strong base and an anion of a strong acid are neutral. Examples: NaBr, KNO3



Example 15.13: Use the table to determine if the given

anion is basic or neutral

a) NO3−

the conjugate base of a strong acid, HNO3, therefore neutral

b) NO2

−

the conjugate base of a weak acid, HNO2 therefore basic



Relationship between Ka of an Acid and Kb of Its

Conjugate Base

• Many reference books only give tables of Ka values

because Kb values can be found from them

when you add

equations, you

multiply the K’s


At 25 oC, Kw = 1 x 10 −14


Example 15.14: Find the pH of 0.100 M

NaCHO2(aq) solution

Na+ is the cation

of a strong base –

pH neutral. The

CHO2− is the

anion of a weak

acid – pH basic

write the reaction

for the anion with

water

construct an ICE

table for the

reaction

enter the initial

concentrations –

assuming the

[OH] from water

is ≈ 0

CHO2− + H2O HCHO2 + OH

[CHO2−] [HCHO2] [OH]

initial 0.100 0 ≈ 0

change

equilibrium



initial 0.100 0 ≈ 0

change

equilibrium 0.100 x

Example 15.14: Find the pH of 0.100 M

NaCHO2(aq) solution

91

represent the

change in the

concentrations in

terms of x

sum the columns to

find the equilibrium

concentrations in

terms of x

Calculate the value

of Kb from the value

of Ka of the weak

acid from Table 15.5

substitute into the

equilibrium constant

expression

+x +x x

x x



Example 15.14: Find the pH of 0.100 M NaCHO2(aq)

92

because Kb is very

small, approximate

the [CHO2−]eq =

[CHO2−]init and

solve for x

Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

change −x +x +x





93

check if the

approximation is

valid by seeing if

x < 5% of

[CHO2−]init


x = 2.4 x 10−6

Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

change −x +x +x

equilibrium 0.100 x x




initial 0.100 0 ≈ 0

cHAnge −x +x +x

equilibrium 0.100 2.4E-6 2.4E-6


94

substitute x into

the equilibrium

concentration

definitions and

solve

x = 2.4 x 10−6

Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

−x +x +x

equilibrium




95

use the [OH−] to

find the [H3O+]

using Kw

substitute [H3O+]

into the formula

for pH and solve

Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

change −x +x +x





97

check by

substituting the

equilibrium

concentrations

back into the

equilibrium

constant

expression and

comparing the

calculated Kb to

the given Kb

though not exact,

the answer is

reasonably close

Kb for CHO2− = 5.6 x 10−11


initial 0.100 0 ≈ 0

cHAnge −x +x +x

equilibrium 0.100 2.4E−6 2.4E−6



If a 0.0015 M NaA solution HAs a pOH of

5.45, what is the Ka of HA?

Na+ is the cation of a

strong base – pOH

neutral. Because

pOH is < 7, the

solution is basic. A−

is basic.

write the reaction for

the anion with water

construct an ICE

table for the reaction

enter the initial

concentrations –

assuming the [OH]

from water is ≈ 0

A− + H2O HA + OH

[A−] [HA] [OH]

initial 0.100 0 ≈ 0

change

equilibrium

98


Practice – If a 0.15 M NaA solution HAs a pOH of 5.45,

what is the Ka of HA?

use the pOH to

find the [OH−]

use [OH−] to fill

in other items

[A−] [HA] [OH]

initial 0.15 0 ≈ 0

cHAnge −3.6E−6 +3.6E−6 +3.6E−6


[A−] [HA] [OH]

initial 0.15 0 ≈ 0

equilibrium



calculate the

value of Kb of A− [A−] [HA] [OH]

initial 0.15 0 ≈ 0

change −3.6E−6 +3.6E−6 +3.6E−6


100

Practice – If a 0.15 M NaA solution HAs a pOH of 5.45,

what is the Ka of HA?



use Kb of A− to

find Ka of HA

101

Practice – If a 0.15 M NaA solution HAs a pOH

of 5.45, what is the Ka of HA?

[A−] [HA] [OH]

initial 0.15 0 ≈ 0

change −3.6E−6 +3.6E−6 +3.6E−6


K

b 8.39 1011



Polyatomic Cations as

Weak Acids

• Some cations can be thought of as the conjugate acid of a weak base

others are the counter-ions of a strong base

• Therefore, some cations can potentially be acidic. They undergo hydrolysis

MH+(aq) + H2O(l) MOH(aq) + H3O+(aq)

• The stronger the base is, the weaker the conjugate acid is

a cation that is the counter-ion of a strong base is pH neutral

a cation that is the conjugate acid of a weak base is acidic

NH4+(aq) + H2O(l) NH3(aq) + H3O

+(aq)

because NH3 is a weak base, the position of this equilibrium favors the right



Metal Cations as Weak Acids

• Cations of small, highly charged metals are weakly acidic

alkali metal cations and alkali earth metal cations are pH neutral

cations are hydrated

Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)



Example 15.15: Determine if the given cation is

acidic or neutral

a) C5N5NH+

the conjugate acid of the weak base pyridine, C5N5N

therefore acidic

b) Ca2+

the counter-ion of the strong base Ca(OH)2, therefore

neutral

c) Cr3+

a highly charged metal ion, therefore acidic



Classifying Salt Solutions as

Acidic, Basic, or Neutral

• If the salt cation is the counter-ion of a strong base and the

anion is the conjugate base of a strong acid, it will form a

neutral solution

NaCl Ca(NO3)2 KBr

• If the salt cation is the counter-ion of a strong base

and the anion is the conjugate base of a weak

acid, it will form a basic solution

NaF Ca(C2H3O2)2 KNO2





• If the salt cation is the conjugate acid of a weak

base and the anion is the conjugate base of a

strong acid, it will form an acidic solution

NH4Cl

• If the salt cation is a highly charged metal ion and

the anion is the conjugate base of a strong acid, it

will form an acidic solution

Al(NO3)3





• If the salt cation is the conjugate acid of a weak

base and the anion is the conjugate base of a

weak acid, the pH of the solution depends on the

relative strengths of the acid and base

NH4F because HF is a stronger acid than NH4+, Ka of

NH4+ is larger than Kb of the F−; therefore the solution

will be acidic



Example 15.16: Determine whether a solution of the

following salts is acidic, basic, or neutral

a) SrCl2

Sr2+ is the counter-ion of a strong base, pH neutral

Cl− is the conjugate base of a strong acid, pH neutral

solution will be pH neutral

b) AlBr3

Al3+ is a small, highly charged metal ion, weak acid

Cl− is the conjugate base of a strong acid, pH neutral

solution will be acidic

c) CH3NH3NO3

CH3NH3+ is the conjugate acid of a weak base, acidic

NO3− is the conjugate base of a strong acid, pH neutral

solution will be acidic



Example 15.16: Determine whether a solution of the

following salts is acidic, basic, or neutral

d) NaCHO2

Na+ is the counter-ion of a strong base, pH neutral

CHO2− is the conjugate base of a weak acid, basic

solution will be basic

e) NH4F

NH4+ is the conjugate acid of a weak base, acidic

F− is the conjugate base of a weak acid, basic

Ka(NH4+) > Kb(F

−); solution will be acidic


Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach 110

Practice – Determine whether a solution of

the following salts is acidic, basic, or neutral

• KNO3

• CoCl3

• Ba(HCO3)2

• CH3NH3NO3


the solution is pH neutral

Co3+ is a highly charged cation, pH acidic

111

K+ is the counter-ion of a strong base, pH neutral

NO3− is the counter-ion of a strong acid, pH neutral

Cl− is the counter-ion of a strong acid, pH neutral

the solution is pH acidic

Ba2+ is the counter-ion of a strong base, pH neutral HCO3

− is the conjugate of a weak acid, pH basic

the solution is pH basic

CH3NH3+ is the conjugate of a weak base, pH acidic

NO3− is the counter-ion of a strong acid, pH neutral

the solution is pH acidic


• KNO3

• CoCl3

• Ba(HCO3)2

• CH3NH3NO3


Chapter 16

Aqueous Ionic

Equilibrium


Buffers

• Buffers are solutions that resist changes in pH when an acid

or base is added

• They act by neutralizing acid or base that is added to the

buffered solution

• Acidic buffers are made by mixing a solution of a weak acid

with a solution of a salt containing its conjugate base anion

Mixing acetic acid, HC2H3O2 with sodium acetate, NaC2H3O2

• Basic buffers are prepared by mixing a solution of a weak

base with a solution of a salt containing its conjugate acid cation

Mixing aqueous NH3 with ammonium chloride solution NH4Cl

113 Tro: Chemistry: A Molecular Approach, 2/e


Making an Acid Buffer



Basic Buffers

B:(aq) + H2O(l) H:B+(aq) + OH−

(aq)

• Buffers can also be made by mixing a weak base, (B:),

with a soluble salt of its conjugate acid, H:B+Cl−

H2O(l) + NH3 (aq) NH4+

(aq) + OH−(aq)



How Acid Buffers Work:

Addition of Base

HA(aq) + H2O(l) A−(aq) + H3O

+(aq)

• Buffers work by applying Le Châtelier’s Principle to weak

acid equilibrium

• Acid Buffer solutions contain significant amounts of the

weak acid molecules, HA

• These molecules neutralize the added base OH−(aq)

HA(aq) + OH−(aq) → A−(aq) + H2O(l)

you can also think of the H3O+ combining with the OH− to make

H2O; the H3O+ is then replaced by the shifting equilibrium



How Acid Buffers Work:

Addition of Acid


+(aq)

• The buffer solution also contains significant amounts of the conjugate base anion, A−

• These ions combine with added acid, H+(aq) to make more HA

H+(aq) + A−(aq) → HA(aq)



Common Ion Effect


+(aq)

• Adding a salt, NaA containing the anion, A−(aq) which is the

conjugate base of the acid (the common ion), shifts the

position of equilibrium to the left, lowering the H3O+ ion

concentration

• This causes the pH to be higher than the pH of the acid

solution



Example 16.1: What is the pH of a buffer that is

0.100 M HC2H3O2 and 0.100 M NaC2H3O2?

119

write the reaction

for the acid with

water

construct an ICE

table for the

reaction

enter the initial

concentrations –

assuming the

[H3O+] from

water is ≈ 0

HC2H3O2 + H2O C2H3O2 + H3O

+

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change

equilibrium

Tro: Chemistry: A Molecular Approach, 2/e


[HA] [A−] [H3O+]

initial 0.100 0.100 0

change

equilibrium

Example 16.1: What is the pH of a buffer that is 0.100 M

HC2H3O2 and 0.100 M NaC2H3O2?

120

represent the

cHAnge in the

concentrations in

terms of x

sum the columns

to find the

equilibrium

concentrations in

terms of x

substitute into the

equilibrium

constant

expression

+x +x x

0.100 x 0.100 + x x


+



Example 16.1: What is the pH of a buffer that is

0.100 M HC2H3O2 and 0.100 M NaC2H3O2?

121

determine the

value of Ka

because Ka is very

small, approximate

the [HA]eq = [HA]init

and [A−]eq = [A−]init ,

then solve for x

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 x 0.100 x 0.100 +x

Ka for HC2H3O2 = 1.8 x 10−5





122

check if the

approximation

is valid by

seeing if x <

5% of

[HC2H3O2]init

Ka for HC2H3O2 = 1.8 x 10−5


x = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

cHAnge −x +x +x

equilibrium 0.100 0.100 x





123

substitute x

into the

equilibrium

concentration

definitions

and solve x = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 1.8E-5 0.100 + x x 0.100 x





124

substitute [H3O+]

into the formula

for pH and solve

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 1.8E−5



Practice − What is the pH of a buffer that is 0.14 M

HF (pKa = 3.15) and 0.071 M KF?

125

write the

reaction for the

acid with water

construct an

ICE table for

the reaction

enter the initial

concentrations

– assuming the

[H3O+] from

water is ≈ 0

HF + H2O F + H3O+

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change

equilibrium



[HA] [A−] [H3O+]

initial 0.14 0.071 0

cHAnge

equilibrium

Practice – What is the pH of a buffer that is 0.14 M HF (pKa =

3.15) and 0.071 M KF?

126

represent the

cHAnge in the

concentrations in

terms of x

sum the columns

to find the

equilibrium

concentrations in

terms of x

substitute into

the equilibrium

constant

expression

+x +x x

0.14 x 0.071 + x x

HF + H2O F + H3O+



pKa for HF = 3.15 Ka for HF = 7.0 x 10−4

Practice – What is the pH of a buffer that is 0.14 M and 0.071

M KF?

127

determine the

value of Ka

because Ka is

very small,

approximate the

[HA]eq = [HA]init

and [A−]eq = [A−]init

solve for x

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

cHAnge −x +x +x

equilibrium 0.012 0.100 x 0.14 x 0.071 +x



Practice – What is the pH of a buffer that is 0.14 M

HF (pKa = 3.15) and 0.071 M KF?

128

check if the

approximation

is valid by

seeing if

x < 5% of

[HC2H3O2]init

Ka for HF = 7.0 x 10−4


x = 1.4 x 10−3

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.14 0.071 x




HF (pKa = 3.15) and 0.071 M KF?

129

substitute x into

the equilibrium

concentration

definitions and

solve

x = 1.4 x 10−3

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.14 0.072 1.4E-3 0.071 + x x 0.14 x



Practice – WHAt is the pH of a buffer tHAt is

0.14 M HF (pKa = 3.15) and 0.071 M KF?

130

substitute

[H3O+] into the

formula for pH

and solve

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.14 0.072 1.4E−3



Henderson-Hasselbalch Equation

• Calculating the pH of a buffer solution can be simplified by

using an equation derived from the Ka expression called

the Henderson-Hasselbalch Equation

• The equation calculates the pH of a buffer from the pKa

and initial concentrations of the weak acid and salt of the

conjugate base

as long as the ―x is small‖ approximation is valid



Deriving the Henderson-Hasselbalch Equation





133

assume the [HA] and [A−]

equilibrium

concentrations are the

same as the initial

substitute into the

Henderson-Hasselbalch

Equation

check the ―x is small‖

approximation


+

Ka for HC7H5O2 = 6.5 x 10−5




HF (pKa = 3.15) and 0.071 M KF?

134

find the pKa from the

given Ka

assume the [HA] and

[A−] equilibrium


same as the initial

substitute into the

Henderson-

Hasselbalch equation


approximation

HF + H2O F + H3O+



• The Henderson-Hasselbalch equation is written for a

chemical reaction with a weak acid reactant and its

conjugate base as a product

• The chemical equation of a basic buffer is written with a

weak base as a reactant and its conjugate acid as a

product

B: + H2O H:B+ + OH−

• To apply the Henderson-Hasselbalch equation, the

chemical equation of the basic buffer must be looked at like

an acid reaction

H:B+ + H2O B: + H3O+

this does not affect the concentrations, just the way we are looking

at the reaction

Henderson-Hasselbalch Equation for Basic Buffers



Relationship between pKa and pKb

• Just as there is a relationship between the Ka of a weak

acid and Kb of its conjugate base, there is also a

relationship between the pKa of a weak acid and the pKb of

its conjugate base

Ka Kb = Kw = 1.0 x 10−14

−log(Ka Kb) = −log(Kw) = 14

−log(Ka) + −log(Kb) = 14

pKa + pKb = 14




NH3 (pKb = 4.75) and 0.20 M NH4Cl?

137

find the pKa of the

conjugate acid (NH4+)

from the given Kb

assume the [B] and

[HB+] equilibrium


same as the initial

substitute into the

Henderson-

HAsselbalch equation


approximation




• The Henderson-HAsselbalch equation is written for a

chemical reaction with a weak acid reactant and its

conjugate base as a product

• The chemical equation of a basic buffer is written with a

weak base as a reactant and its conjugate acid as a

product

B: + H2O H:B+ + OH−

• We can rewrite the Henderson-HAsselbalch equation for

the chemical equation of the basic buffer in terms of pOH

Henderson-HAsselbalch Equation for Basic Buffers




NH3 (pKb = 4.75) and 0.20 M NH4Cl?

139

find the pKb if given Kb

assume the [B] and

[HB+] equilibrium


same as the initial

substitute into the

Henderson-HAsselbalch

equation base form,

find pOH


approximation

calculate pH from pOH




Buffering Effectiveness

• A good buffer should be able to neutralize moderate amounts of added acid or base

• However, there is a limit to how much can be added before the pH changes significantly

• The buffering capacity is the amount of acid or base a buffer can neutralize

• The buffering range is the pH range the buffer can be effective

• The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base



Buffering Capacity

a concentrated

buffer can

neutralize

more added

acid or base

tHAn a dilute

buffer



Effectiveness of Buffers

• A buffer will be most effective when the

[base]:[acid] = 1

equal concentrations of acid and base

• A buffer will be effective when 0.1 <

[base]:[acid] < 10

• A buffer will be most effective when the [acid] and

the [base] are large



Buffering Range

• We have said that a buffer will be effective when

0.1 < [base]:[acid] < 10

• Substituting into the Henderson-Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective

Lowest pH Highest pH

Therefore, the effective pH range of a buffer is pKa ± 1

When choosing an acid to make a buffer, choose one whose

is pKa closest to the pH of the buffer



Formic acid, HCHO2 pKa = 3.74

Example 16.5a: Which of the following acids would be the

best choice to combine with its sodium salt to make a

buffer with pH 4.25?

Chlorous acid, HClO2 pKa = 1.95

Nitrous acid, HNO2 pKa = 3.34

Hypochlorous acid, HClO pKa = 7.54

The pKa of HCHO2 is closest to the desired pH of the

buffer, so it would give the most effective buffering

range



Example 16.5b: What ratio of NaCHO2 : HCHO2 would be

required to make a buffer with pH 4.25?

145

Formic acid, HCHO2, pKa = 3.74

To make a buffer with pH

4.25, you would use 3.24

times as much NaCHO2 as

HCHO2



Practice – What ratio of NaC7H5O2 : HC7H5O2 would be required

to make a buffer with pH 3.75?

Benzoic acid, HC7H5O2, pKa = 4.19

to make a buffer with

pH 3.75, you would use

0.363 times as much

NaC7H5O2 as HC7H5O2 146 Tro: Chemistry: A Molecular Approach, 2/e


Buffering Capacity

• Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH

• The buffering capacity increases with increasing absolute concentration of the buffer components

• As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves

• Buffers that need to work mainly with added acid generally have [base] > [acid]

• Buffers that need to work mainly with added base generally have [acid] > [base]



Titration

• In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete

when the reaction is complete we have reached the endpoint of the titration

• An indicator may be added to determine the endpoint

an indicator is a chemical that changes color when the pH changes

• When the moles of H3O+ = moles of OH−, the titration has

reached its equivalence point



Titration



Titration Curve

• A plot of pH vs. amount of added titrant

• The inflection point of the curve is the equivalence point of the titration

• Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH

• The pH of the equivalence point depends on the pH of the salt solution

equivalence point of neutral salt, pH = 7

equivalence point of acidic salt, pH < 7

equivalence point of basic salt, pH > 7

Rapid pH change from about pH = 3 to pH = 11



Titration Curve:

Unknown Strong Base Added to Strong Acid



Before Equivalence

(excess acid)

After Equivalence

(excess base)

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

Equivalence Point

equal moles of

HCl and NaOH

pH = 7.00

Because the

solutions are equal

concentration, and

1:1 stoichiometry, the

equivalence point is

at equal volumes



HCl NaCl NaOH

mols before 2.50E-3 0 5.0E-4

mols cHAnge −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0

HCl NaCl NaOH


mols cHAnge

mols end

molarity, new

HCl NaCl NaOH


mols cHAnge −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new

5.0 x 10−4 mole NaOH added


• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

• Initial pH = −log(0.100) = 1.00

• Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence point added 5.0 mL NaOH




• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• To reach equivalence, the added moles NaOH = initial moles

of HCl = 2.50 x 10−3 moles

• At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH

left over

• Because the NaCl is a neutral salt, the pH at equivalence =

7.00



Practice – Calculate the pH of the solution that results when

10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M

HNO3



HNO3 NaNO3 NaOH


mols cHAnge −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new

HNO3 NaNO3 NaOH


mols cHAnge −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new 0.018 0.025 0



HNO3

• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

• Initial pH = −log(0.250) = 0.60

• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• Before equivalence point added 10.0 mL NaOH



HNO3 NaNO3 NaOH


mols cHAnge −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new



HNO3

• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• After equivalence point

added 100.0 mL NaOH


Titration of a Strong Base with a Strong Acid

• If the titration is run so

that the acid is in the

burette and the base

is in the flask, the

titration curve will be

the reflection of the

one just shown



Titration of a Weak Acid with a Strong Base

• Titrating a weak acid with a strong base results in

differences in the titration curve at the equivalence point and

excess acid region

• The initial pH is determined using the Ka of the weak acid

• The pH in the excess acid region is determined as you

would determine the pH of a buffer

• The pH at the equivalence point is determined using the Kb

of the conjugate base of the weak acid

• The pH after equivalence is dominated by the excess strong

base

the basicity from the conjugate base anion is negligible



Titrating Weak Acid with a Strong Base

• The initial pH is that of the weak acid solution

calculate like a weak acid equilibrium problem

e.g., 15.5 and 15.6

• Before the equivalence point, the solution becomes a buffer

• Half-neutralization pH = pKa

• At equivalence point, pH ≈ 9

and rapid change of pH from about pH = 7 to pH = 11



Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH

• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)

• Initial pH

[HCHO2] [CHO2−] [H3O

+]

initial 0.100 0.000 ≈ 0

change −x +x +x

equilibrium 0.100 − x x x

Ka = 1.8 x 10−4




• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)

• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence added 5.0 mL NaOH

HA A− OH−

mols before 2.50E-3 0 0

mols added – – 5.0E-4

mols after 2.00E-3 5.0E-4 ≈ 0



HA A− OH−

mols before 2.50E-3 0 0

mols added – – 2.50E-3

mols after 0 2.50E-3 ≈ 0

[HCHO2] [CHO2−] [OH−]

initial 0 0.0500 ≈ 0

cHAnge +x −x +x

equilibrium x 5.00E-2-x x


added 25.0 mL NaOH CHO2

−(aq) + H2O(l) HCHO2(aq) + OH−

(aq)

Kb = 5.6 x 10−11 [OH−] = 1.7 x 10−6 M

• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)

• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• At equivalence


added 35.0 mL NaOH

0.00100 mol NaOH xs

pH = 12.22

initial HCHO2 solution

0.00250 mol HCHO2

pH = 2.37

added 5.0 mL NaOH

0.00200 mol HCHO2

pH = 3.14

added 10.0 mL NaOH

0.00150 mol HCHO2

pH = 3.56

added 25.0 mL NaOH

equivalence point

0.00250 mol CHO2−

[CHO2−]init = 0.0500 M

[OH−]eq = 1.7 x 10−6

pH = 8.23

added 30.0 mL NaOH

0.00050 mol NaOH xs

pH = 11.96

added 20.0 mL NaOH

0.00050 mol HCHO2

pH = 4.34

added 15.0 mL NaOH

0.00100 mol HCHO2

pH = 3.92

added 12.5 mL NaOH

0.00125 mol HCHO2

pH = 3.74 = pKa

HAlf-neutralization

Adding NaOH to HCHO2

added 40.0 mL NaOH

0.00150 mol NaOH xs

pH = 12.36

added 50.0 mL NaOH

0.00250 mol NaOH xs

pH = 12.52



Titrating Weak Acid with a Strong Base

• At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established

mol A− = original mole HA

calculate the volume of added base as you did in Example 4.8

[A−]init = mol A−/total liters

calculate like a weak base equilibrium problem

e.g., 15.14

• Beyond equivalence point, the OH is in excess

[OH−] = mol MOH xs/total liters

[H3O+][OH−]=1 x 10−14



Example 16.7a: A 40.0 mL sample of 0.100 M HNO2 is

titrated with 0.200 M KOH. Calculate the volume of KOH at

the equivalence point.

166

write an equation for

the reaction for B

with HA

use stoichiometry to

determine the volume

of added B

HNO2 + KOH NO2 + H2O



Example 16.7b: A 40.0 mL sample of 0.100 M HNO2 is

titrated with 0.200 M KOH. Calculate the pH after adding

5.00 mL KOH.

167

write an

equation for the

reaction for B

with HA

determine the

moles of HAbefore

& moles of

added B

make a

stoichiometry

table and

determine the

moles of HA in

excess and

moles A made


HNO2 NO2− OH−

mols before 0.00400 0 ≈ 0

mols added – – 0.00100

mols after ≈ 0 0.00300 0.00100




titrated with 0.200 M KOH. Calculate the pH after adding

5.00 mL KOH.

168

write an

equation for the

reaction of HA

with H2O

determine Ka

and pKa for HA

use the

Henderson-

Hasselbalch

equation to

determine the

pH

HNO2 + H2O NO2 + H3O

+

HNO2 NO2− OH−



mols after 0.00300 0.00100 ≈ 0

Table 15.5 Ka = 4.6 x 10−4



Example 16.7b: A 40.0 mL sample of 0.100 M HNO2

is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

169

write an

equation for the

reaction for B

with HA

determine the

moles of HAbefore

& moles of added

B

make a

stoichiometry

table and

determine the

moles of HA in

excess and

moles A made


HNO2 NO2− OH−



mols after ≈ 0 0.00200 0.00200

at HAlf-equivalence, moles KOH = ½ mole HNO2




titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

170

write an equation

for the reaction of

HA with H2O

determine Ka and

pKa for HA

use the

Henderson-

Hasselbalch

equation to

determine the pH

HNO2 + H2O NO2 + H3O

+

HNO2 NO2− OH−



mols after 0.00200 0.00200 ≈ 0

Table 15.5 Ka = 4.6 x 10-4



At equivalence point, pH ≈ 5

Rapid change of pH range between pH = 3 to pH = 7


Titrating Weak Base with a Strong Acid


Titration Curve of a Weak Base with a Strong Acid



Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M

HCl. Calculate the initial pH of the NH3(aq)

173

• NH3(aq) + HCl(aq) NH4Cl(aq)

• Initial: NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

[HCl] [NH4+] [NH3]

initial 0 0 0.10

cHAnge +x +x −x

equilibrium x x 0.10−x

pKb = 4.75

Kb = 10−4.75 = 1.8 x 10−5



Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75)

with 0.10 M HCl. Calculate the pH of the solution at

equivalence.





equivalence.

175

• NH3(aq) + HCl(aq) NH4Cl(aq)

• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• At equivalence mol NH3 = mol HCl = 2.50 x 10−3

added 25.0 mL HCl NH3 NH4Cl HCl


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new 0 0.050 0





equivalence.

176

NH3(aq) + HCl(aq) NH4Cl(aq)

at equivalence [NH4Cl] = 0.050 M

[NH3] [NH4+] [H3O

+]

initial 0 0.050 ≈ 0

change +x −x +x

equilibrium x 0.050−x x

NH4+

(aq) + H2O(l) NH3(aq) + H3O+

(aq)



Indicators • Many dyes change color depending on the pH of the

solution

• These dyes are weak acids, establishing an equilibrium

with the H2O and H3O+ in the solution

HInd(aq) + H2O(l) Ind(aq) + H3O

+(aq)

• The color of the solution depends on the relative

concentrations of Ind:HInd

when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and

HInd

when Ind:HInd > 10, the color will be mix of the colors of Ind

when Ind:HInd < 0.1, the color will be mix of the colors of HInd



Phenolphthalein



Methyl Red

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)

2N N N N

H

NaOOC

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)

2N N N N

NaOOC

H3O+ OH-



Monitoring a Titration with an Indicator

• For most titrations, the titration curve shows a very

large change in pH for very small additions of

titrant near the equivalence point

• An indicator can therefore be used to determine

the endpoint of the titration if:

pKa of indicator ≈ pH at equivalence point

it changes color within the same range as the rapid

change in pH



Acid-Base Indicators


chapter 3 acids & bases

Documents

strong electrolyte

acid hydrogens

base molecules

titrating

lowry acidbase

form basic

excess acid

initial concentrations