BUFFER SOLUTIONS

Lesson 3

A buffer solution is a solution which resists changes in pH when a small amount of acid or base is added.

Typically a buffer solution is a mixture of a weak acid and a salt of its conjugated base

or a weak base and a salt of its conjugated acid

Biological Uses In biological systems (saliva, stomach, and blood) it is essential that the pH stays ‘constant’ in order for any processes to work properly. e.g. If the pH of blood varies by 0.5 it can lead to unconsciousness and coma Most enzymes work best at particular pH values. Other Uses Many household and cosmetic products need to control their pH values. Shampoo Buffer solutions counteract the alkalinity of the soap and prevent irritation Baby lotion Buffer solutions maintain a pH of about 6 to prevent bacteria multiplying Others Washing powder, eye drops, fizzy lemonade

USES OF BUFFERS

TYPES OF BUFFERS

Two types:

Acidic buffers (pH ‹ 7) – solution of a mixture of a weak acid and a salt of this weak acid with a strong base

Example: CH3COOH + CH3COONa acid salt

Alkaline buffers (pH > 7) – solution of a mixture of a weak base and a salt of this weak base with a strong acid

Example: NH3 + NH4Cl base salt

COMPOSITION NAME рН

HCOOH + HCOONa Formate 3,70

CH3COOH + CH3COONa Acetate 4,70

NH3 + NH4Cl Ammonium 9,20

Na2HPO4 + NaH2PO4 Phosphate 8

NaHCO3 + Na2CO3 Bicarbonate 9,7 - 10

EXAMPLES OF BUFFER SOLUTIONS

HOW BUFFERS WORK

Ex.: acidic buffer

HA H+ + A-

+

OH-

H2O

adding strong base

КA К+ + A-

+

H+

HА

adding strong acid

The resistive action is the result of equilibrium between weak acid (HA) and its conjugated base (A-) or vice versa.

It is essential to have a weak electrolyte for an equilibrium to be present so that ions can be removed and produced.

Adding acid Any H+ is removed by reacting with A¯ ions to form weak acid HA via the equilibrium. As a result c(H+) remains unchanged.

Adding alkali OH- react with the small concentration of H+ ions to produce water. Removal of H+ from the weak acid equilibrium means that more weak acid will dissociate to form ions to replace those being removed. As the added OH¯ ions remove the H+ from the weak acid system, the equilibrium moves to the right to produce more H+ ions. As a result c(H+) remains unchanged.

Conclusion: When hydrochloric acid is added to the acetate buffer, the salt reacts with the strong acid forming the weak acid, acetic acid.

Let’s consider as an example acetic buffer solution CH3COOH + CH3COONa

(рН = 4,70):

CH3COOH ↔ CH3COO- + Н+ (α << 1) CH3COONa ↔ CH3COO- + Na+ (α = 1)

As a result H+ will react with acetate anions to form weak acetic acid:

CH3COO- + Н+ ↔ CH3COOH CH3COONa + HCl ↔ CH3COOН + NaCl

с(Н+) ≈ const pH = const

а) Let us add a strong hydrochloric acid (HCl) to this buffer.

Let’s consider the previous example

CH3COOH ↔ CH3COO- + Н+ (α << 1) CH3COONa ↔ CH3COO- + Na+ (α = 1)

If too much H+ is added , the equilibrium is shifted all the way to the left , and there is no longer any more CH3COO- to “absorb” H+.

At this point the solution no longer resist the change in pH. It is useless as a buffer.

In order for a buffer to work well the concentration of the acid/base and its salt must be much higher than the strong acid/base added.

As a result a reaction of neutralization will occur:

CH3COOН + NaOH ↔ CH3COONa + Н2О

Н+ + ОН- → Н2О

Conclusion: when а base is added, the acid reacts with it forming salt and

water.

б) Let us add a strong base to acetic buffer solution (NaOH, КОН)

A buffer absorbs strong acid and base through the two reactions:

A- + H+ => HA HA + OH- => A- + H2O

The buffer will stop working when either one of its components (HA or A-) is exhausted, and therefore cannot neutralize any more strong acid or strong base. The most effective buffering solutions are those which have similar concentrations of HA and A- because then the buffer has the capacity to absorb both acid and base with the same effectiveness.

SUMMARY

The concentration of a buffer solution is important. If the concentration is too low, there won’t be enough HA and A¯ to cope with the ions added. For a buffer solution one needs: large c(HA) - for dissociating into H+ when alkali is added weak acid (equilibrium shifts to the right) large c(A¯) - for removing H+ as it is added conjugate base (equilibrium shifts to the left) This situation can’t exist if only acid is present; a mixture of the acid and salt is used. The weak acid provides the equilibrium and the large HA concentration. The salt provides the large A¯ concentration.

Let’s consider another example ammonia buffer solution NH3·H2O + NH4Cl

рН = 9,20:

NH3 ·H2O ↔ NH4+ + OH- (α ≤ 1)

NH4Cl ↔ NH4+ + Cl- (α = 1)

а) Let us add a strong hydrochloric acid (HCl) to this buffer.

As a result neutralization occurs NH3 ·H2O + HCl ↔ NH4Cl + Н2О

Conclusion: Strong acid is replaces by water

(рН = const).

Conclusion: Strong electrolyte KOH is replaced by weak base, weak electrolyte

б) Let us add a strong base to ammonia buffer solution (КОН)

NH4Cl + NaOH ↔ NH3 ·H2O + NaCl

NH4+ + OH- ↔ NH3 ·H2O

As a result ammonia cations will react with ОН- ions:

Calculating the pH of buffer solutions

)(

)()(

3

3

COOHCHc

HcCOOCHc

aK)(

)(

3

3)(COOCHc

COOHCHc

aKHc

)(

)()(

saltc

acidc

aKHc

1. Calculating the pH of an acidic buffer solution (ex. CH3COOH + CH3COONa)

if с(CH3COOH) ≈ с(acid) с(CH3COO-) ≈ с(salt)

CH3COOH ↔ CH3COO- + Н+ (α << 1) CH3COONa ↔ CH3COO- + Na+ (α = 1)

)(

)(lg

baseс

acidc

apKpH

Let’s take the –log on both parts :

or

)(

)(lg

basen

acidn

apKpH

)(

)()(

saltc

acidc

aKHc

This relationship known as the Henderson-Hasselbalch equation, that is often used to perform the calculations required in preparation of buffers for use in the laboratory, or other applications.

Calculate the pH and c(Н+) of a buffer whose c(CH3COOH) is

0,01 mol/l and c(CH3COONa) is 0,01 mol/l (К = 1,75∙10-5).

lmolKHcCOOCHc

COOHCHc

a /1075.11075.1)( 5

01.001.05

)(

)(

3

3

рН = -lgc(H+) = 4.75

Solution:

Problem to consider

)(

)()(

23

4

OHNHc

OHcNHc

bK )(

)(

4

23)(NHc

OHNHc

bKOHc

)(

)()(

saltc

basec

bKOHc

NH3 ·H2O ↔ NH4+ + OH- (α << 1)

NH4Cl ↔ NH4+ + Cl- (α = 1)

с(NH3 ·H2O) ≈ с(base) с(NH4Cl) ≈ с(salt)

2. Calculating the pH of an alkali buffer solution (ex. NH3 ·H2O + NH4Cl)

)(

)(

)(

)(lglg

saltn

basen

bsaltс

basec

b pKpKpOH

)(

)(lg14

saltс

basec

bpKpH

)(

)(lg14

saltn

basen

bpKpH

the pH of a buffer depends on the ratio c(HA)/c(A-), it means that the рН of buffer solution practically does not change at dilution! it depends also on the value of pK: The lower the value of pK, the lower is the pH of the solution.

Calculate the pH of the solution formed, when 20 mL of 0,1

M solution of NH3 ∙H2O is mixed with 40 mL of 0,2 M solution

of NH4Cl (Кb = 1,76∙10-5).

Solution:

Problem

Diluting a solution reduces the concentration, the Molarity

ti VcVc 21

To take the dilution in consideration, let's apply the following expression:

)()()()( 332331 NHVNHcNHVNHc ti

lmolNHcNHV

NHVNHc

t

i /033,060

201,0)(

)(

)()(32

3

331

64.8lg76.414lg14133.0033.0

)(

)(

saltс

basec

bpKpH

)()()()( 442441 ClNHVClNHcClNHVClNHc ti

lmolClNHcClNHV

ClNHVClNHc

t

i /133,060

402,0)(

)(

)()(42

4

441

BUFFER CAPACITY

It is defined as a numbers of moles of an acid or a base required to be added to one liter of a buffer solution so as to change its pH by one. The buffering capacity of а buffer is, defined as the ability of the buffer to resist changes in pH when an acid or base is added.

BUFFER RANGE

The buffer range is the pH range over which the buffer is effective. Buffer range is related to the ratio of buffer component concentrations. The closer c(HA)/c(A) is to one, the more effective the buffer. A buffer works most effectively at pH values that are + 1 pH unit from the pKa

Buffer’s characteristics

Factors in choosing a buffer

Be sure it covers the pH range you need

Generally: pKa of acid ± 1 pH unit

Consult tables for ranges or pKa values

Be sure it is not toxic.

Be sure it would not confound the experiment.