ACIDS AND BASESACIDS AND BASES

REVISEREVISE

pH = -log [H+]

pOH = -log [OH-]

pH + pOH = 14 at 25oC

Neutral: pH = 7 ([H+] = [OH-])

Acidic: pH < 7 ([H+] > [OH-])

Basic: pH > 7 ([H+] < [OH-])

• Brønsted-Lowry acids and basesBrønsted-Lowry acids and bases• Amphoteric substances• Conjugate acid base pairs• Neutralisation

Kw = [OH-][H+]

Kw = 14.0 at 25oC

Kw = KaKb

STRONG ACIDS AND BASESSTRONG ACIDS AND BASES

Common strong acids:

HCl, HBr, HI, H2SO4, HNO3, HClO4

(Why is HF not a strong acid?)

Common strong bases:

LiOH, NaOH, KOH, RbOH, CsOH, R4NOH

Strong acids and bases react nearly “completely” to produce H+ and OH-

equilibrium constants are large

e.g.: HCl H+ + Cl-

[HCl]

]][Cl[HK

Complete dissociation:

large

small

Example:

Calculate the pH of 0.1M LiOH.

Strong base

Dissociates completely

LiOH Li+ + OH-

Start:

Complete rxn:

0.1M 0 0

0 0.1M 0.1M

pOH = -log [OH-]

= -log (0.1) = 1

pH = 14 - 1 = 13

Problem:

What is the pH of 1x10-8 M KOH?

As before:

pOH = -log (1x10-8) = 8

pH = 14 – 8 = 6

BUT pH 6 acidic conditions and KOH is a strong base

IMPOSSIBLE!!!!!!

Since the concentration of KOH is so low (1x10-8 M), we need to take the ionisation of water into account.

We do this by systematic treatment of equilibrium.

Charge balance: [K+] + [H+] = [OH-]

Mass balance: [K+] = 1x10-8 M

Equilibria: [H+][OH-] = Kw = 1x10-14 M

3 equations + 3 unknowns solve simultaneously

Find: pH = 7.02

In pure water [OH-] = 1x10-7 M, which is greater than the concentration of OH- from KOH.

Hint: You end up with a quadratic equation which you solve using the formula.

Also note that:

• Only pure water produces 1x10-7 M H+ and OH-.

• If there is say 1x10-4 M HBr in solution, pH = 4 and [OH-] = 1x10-10 M

• But the only source of OH- is from the dissociation of water.

if water produces 1x10-10 M OH- it can only produce 1x10-10 M H+ due to the

dissociation of water.

pH in this case is due mainly to the dissociation of HBr and not the dissociation of water.

• It is thus important to look at the concentration of acid and bases present.

Some guidelines regarding the concentrations of acids and bases:

1) When conc > 1x10-6 M calculate pH as usual

2) When conc < 1x10-8 M pH = 7

(there is not enough acid or base to affect the pH of water)

3) When conc 1x10-8 - 1x10-6 M Effect of water ionisation and added acid and bases are comparable, thus:

use the systematic treatment of equilibrium approach.

WEAK ACIDS AND BASESWEAK ACIDS AND BASES

HA H+ + A-

Acid dissociation constant

[HA]

]][A[HKa

Partial dissociation

small

large

Weak acids and bases react only “partially” to produce H+ and OH-

equilibrium constants are small

Common weak acids:• carboxylic acids (e.g. acetic acid = CH3COOH)

• ammonium ions (e.g. RNH3

+, R2NH2+, R3NH+)

Common weak bases:• carboxylate anions (e.g. acetate = CH3COO-)

• amines (e.g. RNH2, R2NH, R3N)

Base hydrolysis:

B + H2O BH+ + OH-

[B]

]][OH[BHKb

base hydrolysis constant/ base “dissociation” constant

Weak base

partial dissociation

Kb small

NOTE:

pKa = -log Ka

pKb = -log Kb

As K increases, its p-function decreases and vice versa.

Problem:

Find the pH of a solution of formic acid given that the formal concentration is 2 M and Ka = 1.80x10-4.

Systematic treatment of equilibria:

Charge balance: [H+] = [HCOO-] + [OH-]

Mass balance: 2 M = [HCOOH] + [HCOO-]

Equilibria:

[HCOOH

]][HCOO[HKa

14w 101]][OH[HK

HCOOH H+ + HCOO-

H2O H+ + OH-

4 equations 4 unknowns

difficult to solve

Make an assumption:

[H+] due to acid dissociation [H+] due to water dissociation

Produces HCOO-

[HCOO-] large

Produces OH-

[OH-] small

[HCOO-] >> [OH-]

Charge balance: [H+] [HCOO-]

42

a 101.82[HCOOH]

K

x

xx.x

0103.6101.8 442 xx

0.019x

[H+] = [HCOO-] = 0.019 M

Charge balance: [H+] [HCOO-]

Mass balance: 2 M = [HCOOH] + [H+]

Equilibria:[HCOOH]

]][COOH[HKa

14w 101]][OH[HK

Let [H+] = [HCOO-] = x

Or x = -0.019

No negative conc’s

x

x

FK

2

a

pH = 1.7

OR since [HCOOH] > 1x10-6, we can calculate pH as usual

HCOOH H+ + HCOO-

Start:

Equilibrium:

2M 0 0

2-x x x

Weak acid

equilibrium conditions

[HCOOH]

]][COOH[HKa

42

a 101.8x2

x

x-2

x.xK

Solve as

before

FRACTION OF DISSOCIATION, FRACTION OF DISSOCIATION,

Fraction of acid in the form A-

][AF][A

][A

[HA]][A

][A

F

][A

For the above problem:

= [HCOO-]

F=

0.019 M

2 M= 0.0095

Acid is 0.95% dissociated at 2 M formal concentration

Weak electrolytes dissociate more as they are diluted.

WEAK BASE EQUILIBRIAWEAK BASE EQUILIBRIA

[B]

]][OH[BHKb

Charge balance: [BH+] = [OH-]

Mass balance: F = [B] + [BH+]

Equilibria:

Let [BH+] = [OH-] = x

x

xxx

F[B]K

2

b

.

B + H2O BH+ + OH-

F

][BH

FRACTION OF ASSOCIATIONFRACTION OF ASSOCIATION

Relationship between Ka and Kb for a conjugate acid- base pair:

Ka.Kb = Kw = 1x10-14 at 25oC

CONJUGATE ACIDS AND BASESCONJUGATE ACIDS AND BASES

If Ka is very large (strong acid)

Then Kb must be very small (weak conjugate base)

Base so weak it is not

a base at all in water

And vice versa

If Ka is very small, say 1x10-6 (weak acid)

Then Kb must be small, 1x10-8 (weak conjugate base)

Greater acid strength, weaker conjugate base strength, and vice versa.

Problem:

Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia.

NH3 + H2O NH4+ + OH-

acidbase

Kb

pKa = -log Ka

Ka = 5.70x10-10

= 1.75 x10-5Kb = Kw

Ka

=1x10-14

5 .70x10-10

Problem:

Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia.

= 1.75 x10-5

Kb = x2

F - x

=x2

0.1 - x

Solve for x using the quadratic equation

Where x = [OH-] = [NH+]

Find x = 1.31x10-3 M = [OH-]

pOH = -log [OH-] = 2.88

pH = 14 – 2.88 = 11.12

Negative value discarded

BUFFERSBUFFERS

Mixture of an acid and its conjugate base.

Buffer solution resists change in pH when acids or bases are added or when dilution occurs.

Mix:

A moles of weak acid + B moles of conjugate base

Find:• moles of acid remains close to A, and• moles of base remains close to B

Very little reaction

HA H+ + A- Le Chatelier’s principle

HENDERSON-HASSELBALCH EQUATIONHENDERSON-HASSELBALCH EQUATION

For acids:

For bases:

[HA]

][AlogpKpH a

][BH

[B]logpKpH a

B + H2O BH+ + OH-

acidbase acid base

Ka

Kb

pKa applies to this acid

When [A-] = [HA], pH = pKa

Why does a buffer resist change in pH when small amounts of strong acid or bases is added?

The acid or base is consumed by A- or HA respectively

A buffer has a maximum capacity to resist change to pH.

Buffer capacity, :

Measure of how well solution resists change in pH when strong acid/base is added.

pH

C

pH

C ab

d

d

d

d

?

Larger more resistance to pH change

A buffer is most effective in resisting changes in pH when:

pH = pKa

i.e.: [HA] = [A-]

Choose buffer whose pKa is as close as possible to the desired pH.

pKa 1 pH unit

Problem:

Calculate the pH of a solution containing 0.200 M NH3 and 0.300 M NH4Cl given that the acid dissociation constant for NH4

+ is 5.7x10-10.

NH3 + H2O NH4+ + OH-

acidbaseKa

][BH

[B]logpKpH a pKa applies

to this acid

pKa = 9.244

pH = 9.244 + log (0.200)

(0.300)

pH = 9.07

POLYPROTICPOLYPROTIC ACIDS AND BASES ACIDS AND BASES

Can donate or accept more than one proton.

Relationships between Ka’s and Kb’s:

Ka1. Kb2 = Kw

Ka2. Kb1 = Kw

Diprotic acid:

H2L HL- + H+ Ka1 K1

HL- L2- + H+ Ka2 K2

Diprotic base:

L2- + H2O HL- + OH- Kb1

HL- + H2O H2L + OH- Kb2

In general:

Using pKa values and mass balance equations, the fraction of each species can be determined at a given pH.

ACID-BASE TITRATIONSACID-BASE TITRATIONS

We will construct graphs to see how pH changes as titrant is added.

Start by:

• writing chemical reaction between titrant and analyte

• using the reaction to calculate the composition and pH after each addition of titrant

TITRATION OF STRONG BASE WITH TITRATION OF STRONG BASE WITH STRONG ACIDSTRONG ACID

Example:

Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr.

HBr + KOH KBr + H2O

What is of interest to us in an acid-base titration:

H+ + OH- H2O

Mix strong acid and strong base

reaction goes to completion

H+ + OH- H2O

* Calculate volume of HBr needed to reach the equivalence point, Veq:

C1V1 C2V2

n1 n2

= But n1 = n2 = 1

CHBrVeq = CKOHVKOH

(0.1000 M)Veq = (0.02000 M)(50.00 ml)

Veq = 10.00 ml

Example:

Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr.

There are 3 parts to the titration curve:

1) Before reaching the equivalence point

excess OH- present

2) At the equivalence point

[H+] = [OH-]

3) After reaching the equivalence point

excess H+ present

1

2

3

1) Before reaching the equivalence point excess OH- present

Say 2.00 ml HBr has been added.

COH- =nunreacted

Vtotal

Starting nOH-

= (0.02 M)(0.050 L)

= 1x10-3 mol

nH+ added

= (0.1 M)(0.002 L)

= 2x10-4 mol

nOH- unreacted

= 8x10-4 mol

Vtotal = 50 + 2 mL

= 52 mL

= 0.052 L

COH- = 0.01538 M

HBr + KOH KBr + H2O

Kw = [H+][OH-]

1x10-14 = [H+](0.01538 M)

[H+] = 6.500x10-13 M pH = 12.19

2) At the equivalence point

nH+ = nOH-

pH is determined by dissociation of H2O:

H2O H+ + OH-

x x

Kw = [H+][OH-]

1x10-14 = x2

x = 1x10-7 M [H+] = 1x10-7 M

pH = 7

pH = 7 at the equivalence point ONLY for strong acid – strong base titrations!!

3) After reaching the equivalence point

excess H+ present

Say 10.10 ml HBr has been added.

CH+ =nexcess

Vtotal

Starting nOH-

= 1x10-3 mol

nH+ added

= (0.1 M)(0.0101 L)

= 1.010x10-3 mol

nH+ excess

= 1x10-5 molVtotal = 50 + 10.1 mL

= 60.1 mL

= 0.0601 L

CH+ = 1.664x10-4 M

pH = 3.78

HBr + KOH KBr + H2O

Note:

A rapid change in pH near the equivalence point occurs.

Equivalence point where:

• slope is greatest

aV

pHslope

d

d

• second derivative is zero (point of inflection)

0V

pH2

a

2

d

d

Calculate titration curve by calculating pH values after a number of additions of HBr.

TITRATION OF WEAK ACID WITH TITRATION OF WEAK ACID WITH STRONG BASESTRONG BASE

Example:

Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH.

HCO2H + NaOH HCO2Na + H2O

HA A-

OR HCO2H + OH- HCO2- + H2O

pKa = 3.745

Ka = 1.80x10-4

Kb = 5.56x10-11

10

b

101.80K

1K

Equilibrium constant so large reaction “goes to completion” after each addition of OH-

Strong and weak react completely

* Calculate volume of NaOH needed to reach the equivalence point, Veq:

C1V1 C2V2

n1 n2

= But n1 = n2 = 1

CNaOHVeq = CFAVFA

(0.1000 M)Veq = (0.02000 M)(50.00 ml)

Veq = 10.00 ml

Example:

Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH.

HCO2H + OH- HCO2- + H2O

There are 4 parts to the titration curve:

2) From first addition of NaOH to immediately before equivalence point mixture of unreacted HA and A-

HA + OH- A- + H2O

BUFFER!! use Henderson-Hasselbalch eqn for pH

1) Before base is added HA and H2O present. HA weak acid, pH determined by equilibrium:

HA H+ + A-Ka

1

2

4) Beyond the equivalence point

excess OH- added to A-.

Good approx: pH determined by strong base (neglect small effect from A-)

3) At the equivalence point

all HA converted to A-. A- is a weak base whose pH is determined by reaction:

A- + H2O HA + OH-Kb 3

4

1) Before base is added

HA and H2O present. HA = weak acid.

[HA]

]][A[HKa

x2 + 1.80x10-4x – 3.60x10-6 = 0

x = 1.81x10-3

[H+] = 1.81x10-3 pH = 2.47

Ka = 1.80x10-4HA H+ + A-

F- x x x

1.80x10-4 = x2

0.02 - x

2) From first addition of NaOH to immediately before equivalence point mixture of unreacted HA and A- . BUFFER!!

[HA]

][AlogpKpH a

HA

Aa n

nlogpKpH

pH = 3.745 + log 2x10-4

8x10-4

Say 2.00 ml NaOH has been added.

HA + OH- A- + H2O

pH = 3.14

Starting nHA

= (0.02 M)(0.05 L)

= 1x10-3 mol

nOH- added

= (0.1 M)(0.002 L)

= 2x10-4 mol

1x10-3 2x10-4

8x10-4 2x10-4

Start -

-End

HCO2H + OH- HCO2- + H2O

When volume of titrant = ½ Veq

pH = pKa

Special condition:

HA

Aa n

nlogpKpH

nHA = nA-

Since:

3) At the equivalence point

all HA converted to A-. A- = weak base. (nHA = nNaOH)

Starting nHA

= 1x10-3 mol

Solution contains just A- a solution of weak base

HA + OH- A- + H2O1x10-3

1x10-3

1x10-3Start -

-End - nOH-

= 1x10-3 mol

A- + H2O HA + OH-

F- x x xKb = 5.56x10-11

= 0.0167 M

Vtotal = 50 + 10 mL

= 60 mL = 0.060 L

][A

][HA][OHK

-b

x2 + 5.56x10-11x – 9.27x10-13 = 0

x = 9.63x10-7 pH = 7.98

[OH-] = 9.63x10-7 M

pOH = 6.02

pH is slightly basic at equivalence point for strong base-weak acid titrations

FA- = nA-

V=

1x10-3 mol

0.060 L

5.56x10-11 = x2

0.0167 - x

Vol NaOH pH0 2.471 2.792 3.143 3.384 3.575 3.756 3.927 4.118 4.359 4.7

9.5 5.029.8 5.4410 7.98

10.2 10.5210.5 10.9211 11.2112 11.5113 11.6814 11.815 11.89

0

2

4

6

8

10

12

14

0 5 10 15

Vol NaOH / ml

pH

CALCULATED TITRATION CURVE

Titration curve depends on Ka of HA.

As HA becomes a weaker acid the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect

not practical to titrate an acid or base that is too weak.

Titration curve depends on extent of dilution of HA.

As HA becomes a more dilute the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect

not practical to titrate a very dilute acid or base.

TITRATION OF WEAK BASE WITH TITRATION OF WEAK BASE WITH STRONG ACIDSTRONG ACID

This is the reverse of the titration of weak base with strong acid.

The titration reaction is:

B + H+ BH+

Recall:

Strong and weak react completely

There are 4 parts to the titration curve:

1) Before acid is added

B and H2O present.

B weak base pH determined by equilibrium:

B + H2O BH+ + OH-Kb

F-x x x

2) From first addition of acid to immediately before equivalence point

mixture of unreacted B and BH+

B + H+ BH+

BUFFER!!

use Henderson-Hasselbalch equation for pH

][BH

[B]logpKpH a pKa applies

to this acid

3) At the equivalence point

all B converted to BH+.

BH+ is a weak acid determined pH by reaction:

BH+ B + H+Ka

F-x x x

FBH+ = n

Vtotal

Take dilution into account

pH is slightly acidic (pH below 7) for strong acid-weak base titrations

4) Beyond the equivalence point

excess H+ added to BH+.

Good approx: pH determined by strong acid (neglect small effect from BH+)

Example:

50.00 ml of 0.05 M NaCN is titrated with 0.1 M HCl. Ka for NaCN = 6.20x1010

Draw the titration curve by calculating pH at various volumes of HCl.

Vol HCl pH0 10.95

2.5 10.165 9.81

7.5 9.5810 9.39

12.5 9.2115 9.03

17.5 8.8420 8.61

22.5 8.2623.5 8.0224.5 7.5225 5.34

25.5 3.1826.5 2.7127.5 2.4930 2.2

32.5 2.0435 1.93

37.5 1.8540 1.78

0

2

4

6

8

10

12

0 10 20 30 40

Volume HCl/ml

pH

TITRATION CURVE OF WEAK BASE WITH STRONG ACID

TITRATIONS IN DIPROTIC SYSTEMSTITRATIONS IN DIPROTIC SYSTEMS

Example - a base that is dibasic:

B + H+ BH+

BH+ + H+ BH22+

pKb1 = 4.00

pKb2 = 9.00

With corresponding reactions:

Two end points will be observed.

FINDING END POINTS WITH A pH FINDING END POINTS WITH A pH ELECTRODEELECTRODE

After each small addition of titrant the pH is recorded and a titration curve is plotted.

2 ways of determining end points from this:

• using derivatives

• using a Gran plot

Setup

But there are autotitrators!

Titrando from Metrohm

USING DERIVATIVES

End point is taken where the slope is greatest

V

pH

d

d

Or where the 2nd derivative is zero

0V

pH2

2

d

d

USING A GRAN PLOT

A problem with using derivatives

titration data is most to obtain near the end point

Example – titration of a weak acid, HA

HA H+ + A- Ka =[H+]H+[A-] A-

[HA] HA

NOTE:

pH electrode responds to hydrogen ion ACTIVITY, not concentration

Say we titrated Va ml of HA (formal conc = Fa) with Vb ml of NaOH (formal conc = Fb) : HA + OH- A- + H2O

[A-] = nOH(titrated)

VTotal

=VbFb

Va + Vb

[HA] = nHA(initial) – nOH(titrated)

VTotal

=VaFa- VbFb

Va + Vb

Substitute into the equilibrium constant:

Ka =[H+]H+[A-] A-

[HA] HA

HAba

bbaa

Aba

bbH

a

VVFVFV

VVFV

][H

K

HAbbaa

AbbHa FVFV

FV][HK

Rearrange:

b

bbaa

HA

AaHb F

FVFVK][HV

[H+]H+ = 10-pH = Ve - Vb

VaFa

Fb

- Vb

beHA

Aa

pHb VVK10V

HAbbaa

AbbHa FVFV

FV][HK

beHA

Aa

pHb VVK10V

Gran plot equation:

Gran plot Graph of Vb10-pH vs Vb

If is constant, then:A

-HA

Slope = -Ka

A

-HA

and x-intercept = Ve

• Use data taken before end point to find end point

• Can determine Ka from slope

Use only linear portion of graph

Extrapolate graph to get Ve

FINDING END POINTS WITH FINDING END POINTS WITH INDICATORSINDICATORS

Acid-base indicator acid or base itself

Various protonated species have different colours

HIn H+ + In-

Choose indicator whose colour change is as close as possible to the pH of the end point

Indicators transition range overlaps the steepest part of the titration curve

Indicator error: difference between the observed end point (colour change) and the true equivalence point.

Systematic error

Random error

Visual uncertainty associated with distinguishing the colour of the indicator reproducibly

Why do we only add a few drops of indicator?

Indicator is an acid/base itself will react with analyte/titrant

Few drops neglible relative to amount of analyte