DUNCANRIG SECONDARY ADVANCED HIGHER CHEMISTRY

UNIT ONE

BOOKLET 5

pH

Acids have been described as substances that dissolve in water to form H+(aq)

ions, whilst bases are substances that react with acids.

However, a better, broader definition was produced independently by

Bronsted and Lowry.

any substance that is capable of donating hydrogen ions (protons)

to another substance – acids are proton donors.

any substance that is capable of accepting hydrogen ions (protons)

from another substance – bases are proton acceptors.

In general an acid, with the formula HA, will dissociate in water according to

This equation shows the hydronium ion, H3O+(aq).

For simplicity, this ion usually written in the shorthand way ---------- H+(aq)

The equation shown above can be written more simply as

In the forward reaction, HA loses a proton and so behaves as an acid. In the

reverse reaction, A- accepts a proton, i.e. behaves as a base. A- is said to be

the conjugate base of the acid HA. For every acid, there is a conjugate base

formed by loss of a proton {H+(aq) ion}.

In addition, every base has a conjugate acid.

cm3 0.10 mol l-1 NaOH added to 50.0 cm3 of 0.10 mol l-1 HCl

This idea can be summarised in a general equation for any acid/base pair.

To highlight this point more clearly consider what happens to ethanoic acid

when it is added to water.

Remember the Bronsted–Lowry definition of an acid: an acid is a proton donor.

In this example ethanoic acid donates its proton to H2O in the forward

reaction, resulting in the formation of the ethanoate ion. However, in the

reverse reaction, the ethanoate ion accepts a proton from H3O+ to revert to

ethanoic acid molecules.

CH3COOH + H2O CH3COO- + H3O+

From the Bronsted–Lowry definition, a base is a proton acceptor. Thus, the

ethanoate ion acts as a base in the reverse reaction, and is called the

conjugate base of ethanoic acid. Ethanoic acid and the ethanoate ion are known

as an acid–base conjugate pair. Interestingly enough, there is a second acid–

base conjugate pair in the reaction above. Notice how in the reverse reaction,

H3O+ donates a proton (making it an acid), while in the forward reaction, H2O

accepts a proton (making it a base). H2O and H3O+ are a conjugate acid–base

pair, with H3O+ being the conjugate acid and H2O being the conjugate base.

ethanoic acid hydronium ion water ethanoate ion

This reaction can be written more simply if we use the hydrogen ion, H+(aq),

instead of the hydronium ion.

CH3COOH(aq) CH3COO-(aq) + H+(aq)

Ethanoic acid is clearly the acid in this reaction – it has lost a proton.

The ethanoate ion is clearly the conjugate base of ethanoic acid as its formula

is identical to ethanoic acid minus the hydrogen ion.

1. Identify the acid-conjugate base pairs in the following equations.

The key to identifying the acid- conjugate base pairs is to spot the

movement of the hydrogen ion and apply the Bronsted-Lowry definition.

2. Write an equation showing how the H2PO4- ion can act as an acid.

ethanoic acid ethanoate ion –

conjugate base

of ethanoic acid

hydrogen ion

1. Write the formula for the species which is the conjugate base and for

the species which is the acid in the following reactions.

a. HCN + OH- CN- + H2O

b. NH2NH2 + H2O OH- + NH2NH3+

2. Each of the following species can act as an acid.

(i) HAsO42- (ii) HNO3 (iii) HVO4

2−

a. Write an equation for each of these species showing how they react with

water to form a conjugate base and the hydronium ion.

b. Write an equation for each of these species showing how they form a

conjugate base and a hydrogen ion.

3. Write equations to show how the following species can act as bases.

Circle the conjugate acid in each equation

(i) ClO- (ii) HN2- (iii) AsH3

4. Consider the reaction shown below.

SO32- + NH4

+ HSO3- + NH3

a. Which species is the acid in this reaction?

b. Explain your answer to part a.

c. Which species is the conjugate acid in this reaction?

5. Write the formula for the conjugate base of the following acids.

a. CH3CH2COOH b. CH3NH3+ c. H3O+ d. H2S

A compound is described as amphoteric if it has the ability to behave as an

acid and as a base. In certain conditions these substances will be proton

donors while in different conditions they can also be proton acceptors

Water has the ability to self-ionise.

When this happens one of the water molecules loses a proton (acts like an

acid) while the other gains a proton (acts like a base).

The amphoteric nature of water is also shown in the following reactions.

Acids are proton donors. Some acids are classified as WEAK and some are

classified as STRONG.

The STRENGTH of an acid depends on the DEGREE OF IONISATION - the

ability of the unionised acid molecule to produce hydrogen ions

The more an acid molecule tends to produce hydrogen ion, H+(aq), the stronger

it will be.

Strong acids, like hydrochloric acid, nitric acid and sulfuric acid totally

dissociate in water – all the acid molecules become ions. This is shown

by a one way arrow in the equation for the ionisation of the acid.

Strong acid HA H+ + A-

E. g. HCl H+ + Cl-

HNO3 H+ + NO3-

The strength of an acid is NOT RELATED to the CONCENTRATION of the

acid. The concentration of an acid is related to the volume of water that has

been added to the undiluted acid. More water will produce a more dilute acid

but it does not alter the STRENGTH of the acid.

Weak acid HA H+ + A-

E. g. CH3COOH H+ + CH3COO-

HCN H+ + CN-

A weak acid is an equilibrium mixture of unionised acid molecules, hydrogen

ions and the conjugate base of the acid. Equations for weak acids must always

be written with two-way arrows.

These arrows signify the only a proportion of the acid molecules IONISE.

Any acid, weak or strong, which is capable of donating only one hydrogen ion per

molecule is termed a monoprotic or monobasic acid. An acid like sulfuric acid

which can donate two hydrogen ions is called diprotic or dibasic.

Water molecules self-ionise according to the following equation

The equilibrium expression for this reaction is termed the ionic product of

water and has the form.

Kw = [H+(aq)] [OH-(aq)]

The value of Kw at 25oC is 1 x 10-14

In any aqueous solution this equilibrium reaction is responsible the pH of the

solution.

pH is measure of the concentration of hydrogen ions [H+(aq)] in an aqueous

solution. These quantities are related by the following formulae.

The table shows the values for pH, H+ and OH- at 25oC.

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

and

A neutral solution is one

which contains equal

concentrations of hydrogen

ions and hydroxide ions.

At 25oC the pH of any

neutral solution is 7.

At other temperatures this

value will alter as the value

of Kw is temperature

dependent.

The table shows how the value of Kw changes with temperature and the effect

this has on the pH of water.

The equation shows as water molecules ionise,

they produce one hydrogen ion for every

hydroxide ion.

Therefore, the equilibrium expression for the

ionic product of water becomes

Kw = [H+(aq)]2

This rearranges to give

[H+(aq)] = Kw

At 60oC, Kw = 9.61 x 10-14 [H+(aq)] = 9.61 x 10-14 = 3.1 x 10-7 mol l-1.

As pH = -log [H+(aq)] pH = -log 3.1 x 10-7 = 6.51.

It is very important to realise that although the pH of water is 6.51 it is still

neutral - the concentration of hydrogen ions and hydroxide ions are still equal.

At 60oC a pH greater than 6.51 would indicate an alkaline solution and a pH of

less than 6.51 would be acidic.

In addition, it is obvious from the table that Kw increases with increasing

temperature. This indicates that the ionisation of water is an endothermic

process.

The value of Kw is 51.3 x 10-14 at 100oC.

Calculate the pH of water at this temperature.

What does the very small value of Kw indicate

about the water equilibrium?

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

1. Calculate the pH of 0.030 mol l-1 hydrochloric acid, HCl.

pH = -log [H+(aq)] pH = -log [0.03] = 1.5

2. Calculate the concentration of hydrogen ions in solution of pH 8.5.

[H+(aq)] = 10-pH [H+(aq)] = 10-8.5 = 3.0 x 10-9 mol l-1.

3. Calculate the pH of a solution of 0.25 mol l-1 sodium hydroxide, NaOH.

For alkalis the ionic product of water must be used in the calculation.

Kw = [H+(aq)] [OH-(aq)] = 1 x 10-14

Kw = pH + pOH = 14

[H+] = 1 x 10-14 / [OH-] = 1 x 10-14 / 0.25 = 4 x 10-14

pH = -log[H+] = -log (4 x 10-14 ) = 13.4

OR

pOH = -log [OH-(aq)] pOH = -log [0.25] = 0.60

pH = 14 – pOH pH = 14 – 0.60 = 13.4

pOH = -log[OH-] pH = -log[H+]

4. 25.0 cm3 of a solution of a strong alkali contained 1.50 × 10–3 mol of

hydroxide ions.

Calculate the pH of this solution.

C of [OH-] = n/v = 1.50 x 10-3/0.025 = 0.06 mol l-1

[H+] = 1 x 10-14 / [OH-] = 1 x 10-14 / 0.06 = 1.67 x 10-13

pH = -log[H+] = -log (1.67 x 10-13 ) = 12.8

1. Calculate the pH of the following acids a. 0.16 mol l-1 HCl b. 0.45 mol l-1 HNO3

2. Calculate the pH of the following alkalis a. 0.32 mol l-1 NaOH b. 1.2 mol l-1 KOH

3. 7.3 g of HCl is dissolved in water and the solution made up to 500 cm3.

Calculate the pH of this solution.

4. 8.0 g of NaOH was dissolved in 100 cm3 of water and the solution made up to 250 cm3.

Calculate the pH of this solution.

5. If 25.6 cm3 of 0.0025mol l-1 of NaOH is added to 50.9 cm3 of 0.005 mol l-1 HCl.

Calculate the pH of the resulting solution.

NaOH + HCl NaCl + H2O

Hint:

This is an excess question – you must calculate how many moles of hydrogen ions or

hydroxide ions would be left at the end of the reaction. Calculate the concentration of

these excess ions. Do the pH calculation as in the examples above.

6. 250 cm3 of 0.1 mol l-1 of potassium hydroxide is added to 125 cm3 of 0.05mol l-1 of

hydrochloric acid.

Calculate the pH of the resulting solution.

7. A solution of a strong acid was found to have a pH of 0.5

a. Calculate the concentration of hydrogen ions in this solution

b. Calculate the volume of water which must be added to 25.0 cm3 of this solution to

increase its pH from 0.5 to 0.7.

Hint: You need to know the hydrogen ion concentration in both solutions.

Use dilution formula C1V1=C2V2.

8. At 25°C, Kw has the value 1.00 × 10–14.

Calculate the pH at 25 °C of

a. 0.150 mol l-1 solution of potassium hydroxide,

b. the solution formed when 35.0 cm3 of this solution of potassium hydroxide is mixed

with 40.0 cm3 of a 0.120 mol l-1 solution of hydrochloric acid.

KOH + HCl KCl + H2O

9. A 30.0 cm3 sample of a 0.480 mol l-1 solution of potassium hydroxide was partially

neutralised by the addition of 18.0 cm3 of a 0.350 mol l-1 solution of sulfuric acid.

2KOH + H2SO4 K2SO4 + 2H2O

a. Calculate the initial number of moles of potassium hydroxide.

b. Calculate the number of moles of sulfuric acid added.

c. Calculate the number of moles of potassium hydroxide remaining after mixing.

d. Calculate the concentration of hydroxide ions in the solution formed.

e. Calculate the pH of the solution formed.

10. Water molecules dissociate into hydrogen and hydroxide ions.

a. Write the expressions for the ionic product of water, Kw.

b. At 318 K, the value of Kw is 4.02 × 10–14 and hence the pH of pure water is 6.70

State why pure water is not acidic at 318 K.

c. Use the value of Kw given above and calculate the pH of the solution formed when

2.00 cm3 of 0.500 mol l-1 aqueous sodium hydroxide are added to 998 cm3 of pure

water at 318 K.

11 a. Calculate the change in pH when 0.250 mol l-1 hydrochloric acid is diluted with water to

produce 0.150 mol l-1 hydrochloric acid.

b. Calculate the volume of water which must be added to 30.0 cm3 of 0.250 mol l-1

hydrochloric acid in order to reduce its concentration to 0.150 mol l-1.

12. A mass, m, of solid KOH is added to 755 cm3 of 0.0120 mol l-1 HCl. The pH after this

addition is 11.60, measured at 25 °C.

The volume of the resulting solution is still 755 cm3.

KOH + HCl KCl + H2O

a. Calculate the number of moles of KOH needed to neutralise exactly the HCl present in

the 755 cm3 of 0.0120 mol l-1 HCl.

b. Calculate the number of moles of NaOH in excess when the pH is 11.60

c. Use these results to calculate the total number of moles of KOH added.

d. Deduce the value of m.

The Acid dissociation constant, Ka, is the equilibrium constant for the

reaction in which a weak acid is in equilibrium with its conjugate base in

aqueous solution.

Ethanoic acid dissociates according to the following equation

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

This can also be written as

CH3COOH(aq) CH3COO-(aq) + H+(aq)

This is you if you

get part d correct

The equilibrium expression for this reaction is

The value of Ka for ethanoic acid at 25oC is 1.7 x 10-5.

The size of Ka gives an indication of the degree of ionisation the acid ungergoes

and hence can be used as a measure of acid strength.

The larger the value of Ka,the stronger the acid

For most weak acids the Ka value is a fairly small value and so to give more

“readable numbers” Ka values are often expressed as pKa values.

Just as pH = -log[H+(aq)] pKa = -logKa

The larger the value of pKa,the weaker the acid

[CH3COO-(aq)] [H+(aq)]

[CH3COOH(aq)]

[H+(aq)]

Ka =

Hydrofluoric acid, HF(aq), is a weak acid with a Ka value of 6.3x10-4.

a. Write an equation which shows how hydrofluoric acid dissociates in water.

HF H+ + F-

b. Write the equilibrium expression for the dissociation constant Ka.

Ka =

[F-]

[H+]

[HF]

c. Calculate the value for pKa.

pKa = -logKa pKa = -log 6.3x10-4 = 3.2

d. State whether hydrofluoric acid is stronger or weaker than ethanoic acid.

Stronger(larger Ka)

The table shows the dissociation constants for several weak acids.

1. Which of the acids is the strongest?

2. Which of the acids is the weakest?

3. Write the formula for the conjugate base of hydrocyanic acid.

4. Write the equilibrium expression for the hydrogencarbonate ion.

5. The conjugate base of boric acid can also act as an acid.

Write an equation which shows how the conjugate base of boric acid dissociates in

water.

6. What is the relationship between Ka and pKa?

7. Hydrochloric acid is a strong acid.

a. Calculate the pH of 0.001 mol l-1 hydrochloric acid.

b. The pH of of 0.001 mol l-1 propanoic acid is 3.92.

Explain why the pH of 0.001 mol l-1 hydrochloric acid is different from the pH of

0.001 mol l-1 propanoic acid.

c. Would the pH of 0.001 mol l-1 phenol be higher or lower than 0.001 mol l-1

propanoic acid?

The pH of several acids of known concentration

were determined using a pH meter.

Complete the table with the results.

The formula pH = -log[H+(aq)] CANNOT be used to calculate the pH of weak

acids. In a monoprotic strong acid there will be 0.1 mo l-1 of hydrogen ions in a

0.1 mol l-1 solution because the acid totally ionises in solution.

In a weak acid only a small proportion of the molecules ionise and so in a

0.1 mol l-1 solution of a monoprotic weak acid there will be much less than

0.1 mol l-1 of hydrogen ions.

When calculating the pH of a weak acid the Ka or pKa value must be included

because, together with the concentration of the acid, this gives a measure of

the concentration of hydrogen ions in solution.

The formula used to calculate the pH of a weak acid is

Acid Concentration/

mol l-1

Ka pKa pH

pH meters

Use this formula to calculate the theoretical pH

of the acid solutions used in the experiment.

Do your answers match the experimental values?

Calculate the pH of a 0.36 mol l-1 solution of ethanoic acid.

Use

From data booklet pKa = 4.76

pH = ½ (4.76) - ½ log (0.36)

pH = 2.38 – ½ (-0.44)

pH = 2.38 + 0.22 = 2.60

Note that a strong acid of concentration 0.36 mol l-1 would have a pH of 0.44.

1. The pH of a 0.15 M solution of a weak acid, HA, is 2.82.

a.. Write an expression for the acid dissociation constant, Ka, of HA, and determine the

value of Ka for this acid.

b. The dissociation of HA into its ions in aqueous solution is an endothermic process. How

would its pH change if the temperature were increased? Explain your answer.

2. Phenol is a weak acid. The dissociation of phenol in aqueous solution is represented by

the following equation:

C6H5OH(aq) + H2O(l) H3O+(aq) + C6H5O–(aq)

a. Write an expression for the acid dissociation constant, Ka, for phenol.

b. The value of the acid dissociation constant, Ka, for phenol is 1 × 10–10 mol l-1.

Calculate the pKa value of phenol.

c. Calculate the pH of 0.0025 mol l-1 aqueous solution of phenol.

3. The acid dissociation constant, Ka, for propanoic acid has the value of 1.35 × 10–5.

a. Calculate the pH of a 0.117 mol l-1 aqueous solution of propanoic acid.

b. Write the formula for the conjugate base of propanoic acid.

c. Use the data booklet to find an acid which could described to be weaker than

propanoic acid.

4. An acid, HA, has a pKa = 4.20.

a. Calculate the value of the dissociation constant, Ka, for this acid.

b. Calculate the pH of a 0.40 mol l-1 aqueous solution of this acid.

5. The hydrogen halides all react with water to form acids. Hydrogen fluoride forms a

weak acid while the others all form strong acids.

a. Write equations to show the reactions that occur when hydrogen fluoride and

hydrogen chloride are dissolved in water.

b. Calculate the pH of an aqueous solution of hydrofluoric acid of concentration

0.050 mol l-1, given that Ka = 5.6 × 10–4.

c. When hydrogen fluoride is dissolved in pure nitric acid, a reaction takes place that can

be represented by the equation:

HNO3 + HF H2NO3+ + F–

d. State, with a reason, which reactant acts as a Bronsted-Lowry acid in this reaction and

give the formula of its conjugate base.

6. Fizzy drinks contain carbon dioxide dissolved in water which dissociates, as shown, to

produce carbonic acid.

a. What is the Bronsted-Lowry definition of an acid?

b. Write the expression for the dissociation constant, Ka, for carbonic acid.

c. Calculate the pH of a 0.10 mol l-1 solution of carbonic acid.

7. When an ant bites, it injects methanoic acid (HCOOH)

a. Methanoic acid is a weak acid

(i) What is the conjugate base of methanoic acid?

(ii) Write the expression for the dissociation constant, Ka, for methanoic acid.

b. In a typical bite, an ant injects 3.6 x 10-3 g of

methanoic acid. Assuming that the methanoic

acid dissolves in 1 cm3 of water in the body,

calculate the pH of this methanoic acid solution.

What are they?

Buffer solutions have a fixed pH. A buffer solution

resists changes to its pH when small volumes of acid

or alkali are added to it.

The pH of a buffer solution will not change when it is

diluted with water.

Different buffer solutions will have different pH values

What are the ingredients of a buffer solution?

Buffer solutions are formed by mixing (in the proper concentration ratio) two

different chemicals.

Acidic Buffer

These are made by mixing a weak acid and its salt.

E.G. ethanoic acid/sodium ethanoate or benzoic acid/potassium benzoate.

Basic Buffer

These are made by mixing a weak base and its salt.

E.G. ammonia/ammonium chloride or ammonia/ammonium sulphate.

Buffer solutions can be thought of as either a weak acid and it conjugate base

or a weak base and its conjugate acid.

Buffer solutions cannot be made with strong acids or alkalis.

What are buffers used for?

Their resistance to changes in pH gives buffers many uses in biochemcal

systems and chemical reactions. The blood in our bodies is buffered at a pH

value of 7.36-7.42 due to bicarbonate - carbonic acid buffer. A mere change

of 0.2 pH units can cause death. Certain enzymes get activated only at certain

definite pH values. Industrially, buffer solutions are used in fermentation

processes and in setting the correct conditions for dyes used in colouring

fabrics. They are also used in chemical analysis (e.g. EDTA titration) and in the

calibration of pH meters.

How do buffers work?

If a buffer is to stabilise pH, it must have a component which is able to react

with any extra hydrogen ions added to the buffer and a component which is

able to react with any extra hydroxide ions which are added to the buffer.

The two components ( weak acid and the salt) dissociate as follows

1 HA(aq) H+(aq) + A-(aq) Weak acid

2 NaA(s) Na+(aq) + A-(aq) Salt

In the buffer there will be large concentration of undissociated HA(aq)

molecules from reaction 1.

There will also be a large concentration of the conjugate base A-(aq) from

reaction 2.

Add acid,H+(aq), to the buffer

Any additional hydrogen ions added to the buffer will react with the large

concentration of A-(aq) present due to the complete ionisation of the salt.

This will cause reaction 1 to shift to left removing the extra hydrogen ions

and maintaining the buffer pH.

Add alkali, OH-(aq), to the bufferAny additional hydroxide ions added to

the buffer will react with hydrogen ions present. This will cause reaction 1 to

shift to right maintaining the buffer pH.

The pH of a buffer solution.

The pH of a buffer is determined by two factors;

1. The equilibrium constant Ka of the weak acid and

2. The ratio of the concentration of the weak acid [Acid] to the

concentration of the conjugate base [Salt] in the solution.

Remember - the salt and the conjugate base are the same thing.

These facts are formalised in the Henderson-Hasselbalch equation:

In practice to obtain a buffer of a certain pH, a weak acid with a pKa value

around the desired pH would be selected. The concentrations of both the acid

and the base can then be manipulated to “fine tune” the buffer solution to the

desired pH.

Salt

We can deduce several facts from this equation:

1. The equation allows calculation of pH of an acid buffer from its

composition and acid dissociation constant, or calculation of

composition from the other two values. Values for Ka and pKa are

available in the data booklet.

2. If the [ACID] = [SALT] then pH = pKa (since the log of 1 is 0)

3. If the buffer mixture is diluted with water, the ratio [ACID]/[SALT]

will not change as both components will be diluted equally. This means

that diluting the buffer will NOT alter its pH value.

Calculate the pH of a buffer solution made with 0.10 mol l-1 ethanoic acid

(pKa = 4.8) and 0.30 mol l-1 sodium ethanoate.

pH = 4.8 - log [0.1]/[0.3]

pH = 4.8 - log 0.333

pH = 4.8 - (-0.477)

pH = 5.28

Calculate the concentration of ethanoic acid and sodium ethanoate

required to make a buffer solution with pH 4.9 (pKa = 4.8)

4.9 = 4.8 - log[Acid]/[Base]

log[Acid]/[Base] = - 0.1

[Acid]/[Base] = 10- 0.1

[Acid]/ = = 0.79

[Salt]

[Salt]

[Salt]

[Salt]

[Salt]

Therefore the [Acid] should be 0.79 mol l-1 and the [Salt] should be 1 mol l-1.

This buffer solution could be made by mixing any volume of 0.79 mol l-1

ethanoic acid with any volume of 1 mol l-1 sodium ethanoate.

Alternatively any volume of 0.79 mol l-1 ethanoic acid could be measured out in

a beaker and the appropriate mass of sodium ethanoate could be mixed with

the acid.

For example, if 500 cm3 of the buffer was required, 0.5 mol of sodium

ethanoate would be added to 500 cm3 of 0.79 mol l-1 ethanoic acid.

1. Calculate the pH of a buffer solution containing 0.1 mol l-1 ethanoic acid and 0.50 mol l-1

sodium ethanoate.

2. Calculate the concentration of methanoic acid and sodium methanoate required to

make a buffer solution with a pH of 4.0.

3. 25 cm3 of 0.10 mol l-1 of a weak acid and 25 cm3 of 0.40 mol l-1 of the sodium salt of

the weak acid are mixed together. The resulting buffer solution has a pH 5.35.

Calculate the dissociation constant, Ka, of the weak acid.

4. Calculate the pH of a buffer solution which contains 7.20g of sodium benzoate,

C6H5COONa, dissolved in one litre of 0.02 mol l-1 benzoic acid.

5. Buffer solutions are important in Chemistry.

a. State what is meant by the term buffer solution.

b. Identify a reagent which could be added to a solution of ammonia in order to form a

buffer solution.

c. An acidic buffer solution is obtained when sodium ethanoate is dissolved in aqueous

ethanoic acid.

(i) Explain how this buffer can resist a decrease in pH when a small amount of acid is

added to it.

(ii) Explain why this pH of this buffer is unaffected when the buffer is diluted with

water.

(iii) Calculate the pH of the buffer solution formed at when 0.125 mol of sodium ethanoate

is dissolved in 250 cm3 of a 1.00 mol l-1 solution of ethanoic acid.

6. A mixture of the acid HA and the sodium salt of this acid, NaA, can be used to prepare

a buffer solution.

a. Write an ionic equation for the reaction which occurs when a small volume of a dilute

acid is added to this buffer solution.

b. The concentration of HA in a buffer solution is 0.250 mol l-1 .

The dissociation constant, Ka, for this acid is 1.45 × 10–4.

Calculate the concentration of the conjugate base, A–, in this buffer solution when the

pH is 3.59.

Acid/base indicators (or simply indicators) are

themselves weak acids which change colour

depending on the pH of the solution.

HIn can be used as a general formula for an

indicator and its dissociation can be

represented by the equation:

In a good indicator, the undissociated acid, HIn, will have a distinctly

different colour from its conjugate base, In-.

For example; the indicator litmus, HIn is red and In- is blue.

As this is an equilibrium reaction, adding ACID (low pH) will shift the reaction

to the left and the indicator colour will be that of HIn. Adding ALKALI (high

pH) will have the opposite effect and the colour will be that of In-

The acid dissociation constant for an indicator HIn is given the symbol KIn and

is represented by:

When the concentrations of In- and HIn are equal the indicator will change

from one colour to the other. At this point pH = pKIn which means the

indicator changes colour at the pH which is equivalent to its pKIn.

The pH of the solution is determined by the pKIn of the indicator and the ratio

of [In-] to [HIn]. Since these are different colours, the ratio of [In-] to [HIn]

determines the overall colour of the solution. For a given indicator, the overall

colour is dependent on the pH of the solution.

There are a huge range of pH indicators. The table below shows some data for

three common indicators.

Indicator pKIn Colour at low pH Colour at high pH pH range

Methyl Orange 3.7 Red Yellow 3.1 – 4.4

Bromothymol Blue 7.0 Yellow Blue 6.0 - 7.6

Phenolphthalein 9.3 Colourless Pink 8.3 – 10.0

The pH range of indicators

Indicators don't change colour sharply at

one particular pH (given by their pKind).

Instead, they change over a narrow range

of pH.

Assume the indicator equilibrium is firmly

to one side, but now you add something to

start to shift it. As the equilibrium shifts,

you will start to get more and more of the

second colour formed, and at some point

the eye will start to detect it.

For example, suppose you had methyl

orange in an alkaline solution so that the

dominant colour was yellow. Now start to

add acid so that the indicator equilibrium

begins to shift.

At some point there will be enough of the

red form of the methyl orange present

that the solution will begin to take on an

orange tint. As you go on adding more acid,

the red will eventually become so dominant

that you can no longer see any yellow.

There is a gradual smooth change from

one colour to the other, taking place over

a range of pH. As a rough "rule of thumb",

the visible change takes place about 1 pH

unit either side of the pKind value.

Some more indicators and their pH ranges are shown above. Universal indicator

is a mixture of different indicators each with its own pKa value and colour

change. A universal indicator is typically composed of water, propan-1-ol,

phenolphthalein , sodium hydroxide, methyl red, bromothymol blue, and thymol

blue.

In general when an acid reacts with an alkali a salt and water are produced.

acid + alkali salt + water

E.g. hydrochloric acid + potassium hydroxide potassium chloride + water

HCl + KOH KCl + H2O

ethanoic acid + sodium hydroxide sodium ethanoate + water

CH3COOH + NaOH CH3COONa + H2O

When enough acid is added to completely react with an alkali the EQUIVALENCE

POINT has been reached.

Suppose 25 cm3 of a 0.1 mol l-1 solution of a monoprotic acid {HA} is titrated

with 0.1 mol l-1 sodium hydroxide {NaOH}.

HA + NaOH KCl + H2O

The equivalence point would occur when 25 cm3 of

NaOH had been added. This equivalence point would be

the same for any monoprotic acid regardless of its

strength.

However the pH at the equivalence point is

dependent on the strength of the acid and

base used in the reaction.

The pH at the equivalence point depends on the nature of the salt formed

during the reaction. This is summarised in the table below.

If the acid used was CH3COOH and the

base was NaOH, the salt formed would be

CH3COONa. When dissolved in water this

salt will form free CH3COO- ions. These

ions will react with the hydrogen ions

present in the water, leaving excess

hydroxide ions and hence the solution is

alkaline. If the base used was NH3 and the acid was HCl the salt formed would be NH4Cl.

Acid Base equivalence point pH

strong strong neutral (= 7)

strong weak acidic (< 7)

weak strong alkaline (> 7)

HA(aq)

Burette containing

When dissolved in water this salt forms free NH4+ ions. These can react with the hydroxide

ions in the water leaving excess hydrogen ions and hence the solution is acidic.

If the pH is monitored when an acid is added to a base, an acid base titration curve can be

plotted.

Strong acid v Strong base

The graph shows the pH curve for adding a strong acid to a strong base.

For example; adding 0.1 mol l-1 hydrochloric acid to 25 cm3 of 0.1 mol l-1 sodium hydroxide

would produce this curve – the equivalence point would be when 25 cm3 of hydrochloric acid

had been added.

Superimposed on the graph are the pH ranges for methyl orange and phenolphthalein

indicators

3.

25 cm3

1.

The equivalence point occurs half way down the steepest part of the curve.

2.

As this titration involved a strong acid and a strong base the equivalence point

occurs at pH = 7.

The pH changes from around 12.5 to 1.5 when a very small volume of acid is added.

Therefore at the equivalence point there is a large change in the concentration of hydrogen

ions.

This is ideal if an indicator is to be used to determine the equivalence point.

HIn H+ + In-

A large change in [H+] will cause the equilibrium to move greatly one way or the other and

hence a rapid change in the colour of the indicator will occur.

Weak acid v Strong base

6.

5.

4.

Using PHENOLPHTHALEIN, the indicator would start off PINK and you would titrate until

it just becomes COLOURLESS (at pH 8.3) because that is as close as you can get to the

equivalence point.

On the other hand, using METHYL ORANGE, the indicator would start off YELLOW and you

would titrate until there is the very first trace of ORANGE (a mixture of yellow and red

make orange) in the solution – this would happen at pH 4.4. If the solution becomes red, you

are getting further from the equivalence point.

Ideally the indicator colour change should happen as close to the equivalence point as possible.

In this case the indicator should have a pKIn of 7 as the equivalence point is at pH 7

You can see that neither indicator changes colour at the equivalence point. However, the graph

is so steep at that point that there will be virtually no difference in the volume of acid added

whichever indicator you choose. For this reason

both indicators would be suitable in this titration.

To be effective the indicator must change colour on the steep part of the curve as it is here

that the hydrogen ion concentration is changing rapidly. The indicator equilibrium will swing

from one side to the other resulting in a rapid colour change.

The PHENOLPHTHALEIN changes colour

(pink to colourless) exactly where you want

it to – the ideal indicator.

This time, the METHYL ORANGE is

hopeless! As the pH never gets low enough

for the indicator to change colour.

The equivalence point is at pH 9.2 because

the titration is between a strong base and

a weak acid.

Strong acid v Weak base

Weak acid v Weak base

This time it is obvious that PHENOLPHTHALEIN

would be completely useless as it would change

from pink to colourless well before the equivalence

point.

METHYL ORANGE starts to change

from yellow to orange very close to the

equivalence point. The pKIn for methyl

orange is 4.4. Ideally the indicator for

this titration should have a pKIn of 4.8.

Methyl orange is suitable here as it

changes colour on the steep section of

the curve.

This curve is for a case where the acid and base

are both equally weak (their dissociation

constants are the same) - for example,

ethanoic acid and ammonia solution – this gives

an equivalence pH of seven In other cases, the

equivalence point will be at some other pH.

You can see that neither indicator is any use.

Phenolphthalein will have finished changing well

before the equivalence point, and methyl orange

falls off the graph altogether.

Notice that the pH change at the equivalence point here is much more gradual than in any of

the other titrations. This means that the concentration of hydrogen ions will also be changing

gradually. This will lead to a very slow colour change in the indicator, often over several

millilitres of added acid even if the indicator had a pKIn exactly on the equivalance point.

Therefore there is no indicator which can be used to accurately detect the equivalence point

in the titration of a weak acid with a weak base.

Summary Table

There will be a range of suitable indicators for acid/base titrations – the important idea to

remember is that the pKIn of the indicator should match, as closely as possible, the

equivalence point pH of the titration.

All the titration curves we have looked at so far have been for MONOPROTIC

acids (acids which yield only one hydrogen ion) like HCl, HNO3 and CH3COOH.

Ethanedioic acid was also known as oxalic acid. It is a diprotic acid, it can

donate 2 protons (hydrogen ions) to a base.

The reaction of sodium hydroxide with oxalic takes place in two stages

because one of the hydrogen ions is easier to remove than the other. The two

successive reactions are:

Titration Equivalence

point pH

Suitable

indicator

pKIn

Strong acid/Strong Base 7 Several 3 - 11

Weak acid/Strong Base More than 7 Phenolphthalein 8 - 10

Strong acid/Weak Base Less than 7 Methyl Orange 3 - 6

Weak acid/Weak Base Depends on Ka of

both reactants None None

then

The pH curve for the titration of 25 cm3 of 0.1 mol l-1 oxalic acid with

0.1 mol l-1 sodium hydroxide is shown below.

This titration may need a mixture of two separate indicators to detect both

equivalence points – one with a pKIn of 2.5, the other with a pKIn of 8.5.

1. The graph below shows how the pH changes when 0.12 mol l-1 NaOH is added to

25.0 cm3 of a solution of a weak monoprotic acid, HA.

The equation for this reaction is NaOH + HA NaA + H2O

14

13

12

11

10

9

8

7

6

5

4

3

2

1

0

0 63 9 131 74 10 142 8 125 11

Volume of 0.12 M NaOH/cm3

pH

volume of 0.12 mol l-1 NaOH/cm3

c. c. Calculate the pH of the propanoic acid before

any potassium hydroxide was added.

a. What volume of alkali was required to reach the equivalence point of the reaction?

b. Use the graph to determine the pH at the equivalence point of the reaction.

c. Use your answer to part a to calculate the initial concentration of the weak acid HA.

d. Use the graph and your answer to part c to calculate the value of the dissociation

constant, Ka, of the weak acid HA.

2. The pH indicator phenolphthalein is a weak acid which can be represented by the

formula HIn. It dissociates in solution and has a pKa value 9.3

HIn H+ + In-

colourless red

a. State the colour of the indicator at the following pH values

(i) 6.3 (ii) 7.9 (iii) 9.8

b. Explain why phenolphthalein is unsuitable for use in a titration between a strong acid

and a weak base.

c. Explain why phenolphthalein is unsuitable for use in a titration between a weak acid

and a weak base.

d. Calculate the value of the dissociation constant Ka for phenolphthalein.

3. A sample of 0.005 mol l-1 was propanoic acid was titrated with potassium hydroxide.

a. Sketch the pH graph that would be obtained for this titration. Your graph should

clearly show the pH at the equivalence point.

b. From the list below select the best indicator for this titration and justify your choice.

Indicator pH range

bromophenol blue 3.0-4.6

methyl red 4.2-6.3

bromothymol blue 6.0-7.6

thymol blue 8.0-9.6

14

12

10

8

6

4

2

0

pH

0 10 20 30 40 50

Volume/cm3

A

14

12

10

8

6

4

2

0

pH

0 10 20 30 40 50

Volume/cm3

B

14

12

10

8

6

4

2

0

pH

0 10 20 30 40 50

Volume/cm3

C

14

12

10

8

6

4

2

0

pH

0 10 20 30 40 50

Volume/cm3

D

4. The pH indicator thymol blue has different structures at different pH values.

a. The formula of the unionised molecule responsible for the red colour of the indicator

can be shown as HIn.

Use this information to write an equation for the dissociation which takes place when

the indicator changes colour from yellow to blue.

b. Explain why this indicator would be suitable for both the titration of a strong acid

with a weak base and the titration of a weak acid with a strong base.

5. Titration curves labelled A,B, C and D for combinations of different acids and bases

are shown below. All solutions have a concentration of 0.1 mol l-1.

a. Select from A,B,C or D the curve produced by the addition of

(i) aqueous ammonia to 25 cm3 of hydrochloric acid

(ii) benzoic acid to 25 cm3 of potassium hydroxide

(ii) sodium hydroxide to 25 cm3 of nitric acid.

red yellow blue

b. A table of acid–base indicators and the pH ranges over which they change colour is

shown.

Select from the table an indicator which could be used in the titration which produces

curve A but not in the titration which produces curve B.

7. The table gives some information about three different pH indicators.

a. At what pH would the colour change be expected to take place in each of the

indicators.

b. Which indicator would be most suitable in a titration of ammonia solution and

hydrochloric acid?

Indicator pH range

thymol blue 1.2 – 2.8

bromophenol blue 3.0 – 4.6

methyl red 4.2 – 6.3

cresolphthalein 8.2 – 9.8

thymolphthalein 9.3 – 10.5

Indicator Kindicator Colour change (low pH to high pH)

Methyl yellow 5.1 x 10–4 Red to yellow

Bromothymol blue 1.0 x 10–7 Yellow to blue

Thymol blue 1.3 x 10–9 Yellow to blue

A list of “learning outcomes” for the topic is shown below. When the

topic is complete you should review each learning outcome.

Your teacher will collect your completed notes, mark them,

and then decide if any revision work is necessary.

State the Bronsted - Lowry definition of an acid and a base.

Define the terms strong and weak acid or base.

Write equations to represent the dissociation of weak acids.

Identify the conjugate base of an acid and the conjugate acid of a base.

State that water is amphoteric and that an amphoteric substance can act as an

acid or a base.

Write an equation which represents the ionic product of water.

State that the ionic product of water, Kw, is an equilibrium constant and that its

value is temperature dependent.

State that water and any solution which contains [H+] =[OH-] is neutral but

the pH of a neutral solution is only 7 when the temperature is 25oC.

State that pH = -log[H+] and use this expression along with the ionic product of

water to calculate the pH of any strong acid or strong base solution.

State that weak acids have a dissociation constant, Ka, and that this can be

converted to a more readable pKa value using pKa = -logKa.

State that as the value of Ka increases the strength of the acid increases .

Use the expression to calculate the pH of a weak acid.

State that a buffer solution is one which resists pH changes when small volumes

of acid or base are added to it and to state some uses of buffer solutions.

State that a buffer solution is prepared from a weak acid and its salt or a weak

base and its salt.

Explain how a buffer solution is able to resist pH changes.

Explain why the pH of a buffer solution is unaffected by dilution with water.

Need Help

Understand

Revise

Be able to use the equation to calculate either the

pH of a buffer of the composition of a buffer

State that pH indicators are weak acids.

State that the pKa value of a particular indicator is equal to the pH at which

the indicator will change colour.

State the approximate equivalence point pH in a titration based on knowledge

of the acid and base used in the titration.

Be able to select the most suitable indicator for an acid/base titration from a

selection of indicators based on the relative strengths of the acid and base.

State that pH indicators are of little use in the titration of a weak acid with a

weak base as at the equivalence point the pH does not change rapidly enough.

I have discussed the learning outcomes with my teacher.

Teacher comments.

Date. __________________________________

Pupil Signature. __________________________

Teacher Signature. _______________________

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