A2 pH Revision Spr ‘18 0 __________________________________________________________________________________________________________________

A2 (yr 13) Chemistry Revision Workshop

- Acids, Bases & pH 20 March 2018

Preparing for your A-level Examinations

Participants Notes

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A2 TOPICS INDEX

1. Acids, Bases & pH p2

1.1 Bronsted-Lowry Theory p2

1.2 Strong Acids & Bases p2

1.3 Weak Acids & Bases p5

1.4 pH Curves p8

1.5 Indicators p12

1.6 Buffers p12

1. ACIDS, BASES & pH

1.1 BRONSTED-LOWRY THEORY

- applies to species in aqueous solution:

An acid is a proton donor.

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A base is a proton acceptor.

e.g. full equation: HCl + NaOH → NaCl + H2O

ionic equation: H+ + OH

- → H2O

HCl (a Bronsted-Lowry acid) donates a proton to the hydroxide ion (a Bronsted-Lowry base).

Also: HNO3 + H2SO4 → H2NO3+ + HSO4

-

Concentrated nitric acid accepts a proton from concentrated sulphuric acid, i.e. concentrated

nitric acid acts as a Bronsted-Lowry base in the presence of a stronger acid, sulphuric. This

reaction is important in the nitration of benzene.

All acid-base reactions involve the transfer of protons.

Note that water can act as both an acid and a base (this is called amphoteric).

As an acid: H2O + NH3 → NH4+ + OH

-

acid base acid base

As a base: H2O + HCl → H3O+ + Cl

-

base acid acid base

As both: H2O + H2O → H3O+ + OH

-

acid base acid base

1.2 STRONG ACIDS AND BASES

Are completely/totally/fully/100% dissociated into ions.

In aqueous solution, hydrochloric, nitric and sulphuric acids are strong, i.e. completely split

up into ions, e.g. HCl (aq) → H+ (aq) + Cl

- (aq)

0% 100%

Alkali metal hydroxides, e.g. NaOH or KOH are strong bases as the ionic lattice is completely

split up releasing all the hydroxide ions when the solids dissolve into a large volume of water.

pH is a logarithmic scale used to simplify the way the concentration of H+ is expressed.

pH = -log10[H+] or simply –log[H

+] [ ] means concentration in mol dm

-3

pH can be measured using Universal Indicator (a mixture of several indicators) or using a pH

meter which must first be calibrated using a buffer solution (see section 9.6 later).

In ALL calculations, pH MUST be given to 2 decimal places! (check your Exam

Board’s requirements!)

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a) Calculating pH from [H+]

e.g. (i) calculate the pH of a 0.200M solution of hydrochloric acid

The acid is fully ionised, so [H+] = 0.200M

Hence pH = -log[0.200] = 0.70 (note answer to 2dp!)

(ii) calculate the pH of a 1.00M solution of hydrochloric acid

[H+] = 1M hence pH = -log[1.00] = 0.00

(iii) calculate the pH of a 2.00M solution of hydrochloric acid

[H+] = 2M hence pH = -log[2.00] = -0.30 (again note answer to 2dp!)

Note that for solutions of monoprotic strong acids with concentrations greater than

1.00M, the pH will be negative.

b) Calculating [H+] from pH

e.g. calculate the concentration of hydrochloric acid when a solution has a pH of 3.20

pH = -log[H+] hence log[H

+] = -3.20 so [H

+] = 10

-3.20 = 6.31 x 10

-4 mol dm

-3

so [HCl] = 6.31 x 10-4

mol dm-3

(note default calculation accuracy of answer – 3 sig. figs.)

Ionic product of water, Kw

The equation: H2O + H2O H3O

+ + OH-

Can be simplified to: H2O H+ (aq) + OH-

(aq)

The equilibrium constant for this, Kc = [H+] [OH

-]

[H2O]

Since [H2O] is huge compared to the concentration of the ions, [H2O] is effectively a constant

in pure water and so can be incorporated into a new equilibrium constant, Kw, called the ionic

product of water:

Kw = [H+] [OH

-] and has the value 1.00 x 10

-14 mol

2 dm

-6 at 25

oC. LEARN THIS!!

In a neutral solution, [H+] = [OH

-] so Kw = [H

+]2 = 1.00 x 10

-14 at 25

oC

Hence [H+] = 1.00 x 10

-7 mol dm

-3 and thus pH = -log[1.00 x 10

-7] = 7.00

For the process: H2O H+ (aq) + OH-

(aq)

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H = +57.6 kJ mol-1

At higher temperatures, Le Chatelier’s Principle predicts that the equilibrium will move to the

right to act to oppose the change, so that [H+] rises and thus the pH falls, and the opposite if

the temperature is reduced.

This does NOT mean that pure water becomes acidic when it is heated since [OH-] also

increases by the same amount! It DOES however mean that the 0-14 scale of pH with 7

neutral only applies at 25oC!

For simplification of pH calculations, it will always be assumed that the solutions used are at

25oC.

Using Kw, we can calculate the pH of strong bases (alkalis)

a) calculating pH from known a concentrations of a strong alkali

(i) calculate the pH of a 0.200M solution of sodium hydroxide

The base is fully ionised, so [OH-] = 0.200M

[H+] = Kw / [OH

-] = 1.00 x 10

-14 / 0.200 = 5.00 x 10

-14 so pH = -log[5.00 x 10

-14] = 13.30

(ii) calculate the pH of a 1.00M solution of potassium hydroxide

[OH-] = 1.00M so [H

+] = Kw / [OH

-] = 1.00 x 10

-14 / 1.00 = 1.00 x 10

-14 so pH = 14.00

Note that for solutions more concentrated than 1.00M, the pH will be greater than 14.00!

b) calculating the concentration of a strong base given the pH of the solution

e.g. calculate the concentration of NaOH when a solution of it has a pH of 12.50

pH = 12.50 hence [H+] = 10-12.50 = 3.16 x 10

-13 mol dm

-3

so [OH-] = Kw / [H

+] = 1.00 x 10

-14 / 3.16 x 10

-13 = 0.0316 thus [NaOH] = 0.0316 mol dm

-3

Calculating pH of mixtures of strong acids and strong alkalis

e.g. (i) calculate the pH of the resulting solution produced by mixing 25.0cm3 of 0.100M

sodium hydroxide with 15.0cm3 0.120M hydrochloric acid

No. moles NaOH = MV / 1000 = 0.100 x 25.0 / 1000 = 2.50 x 10-3

= moles OH-

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No. moles HCl = MV / 1000 = 0.120 x 15.0 / 1000 = 1.80 x 10-3

= moles H+

Since H+ + OH

- → H2O

No. moles OH- in excess = 2.50 x 10

-3 – 1.80 x 10

-3 = 0.700 x 10

-3 = 7.00 x 10

-4 mol

These are present in 25.0 + 15.0 = 40.0cm3 of solution

So [OH-] = 1000n / V = 1000 x 7.00 x 10

-4 / 40 = 1.75 x 10

-2 mol dm

-3

[H+] = Kw / [OH

-] = 1.00 x 10

-14 / 1.75 x 10

-2 = 5.71 x 10

-13

so pH = -log[5.71 x 10-13

] = 12.24

e.g. (ii) calculate the pH of the resulting solution produced by mixing 25.0cm3 of 0.100M

sodium hydroxide with 35.0cm3 0.120M hydrochloric acid

No. moles NaOH = MV / 1000 = 0.100 x 25.0 / 1000 = 2.50 x 10-3

= moles OH-

No. moles HCl = MV / 1000 = 0.120 x 35.0 / 1000 = 4.20 x 10-3

= moles H+

Since H+ + OH

- → H2O

No. moles H+ in excess = 4.20 x 10

-3 – 2.50 x 10

-3 = 1.70 x 10

-3 mol

These are present in 25.0 + 35.0 = 60.0cm3 of solution

So [H+] = 1000n / V = 1000 x 1.70 x 10

-3 / 60.0 = 0.0280 mol dm

-3

pH = -log[0.0280] = 1.55

1.3 WEAK ACIDS AND BASES

Are not completely/fully/<100% dissociated into ions in aqueous solution – you can also say

partly/partially or incompletely but NOT “slightly”! (check your Exam Board’s

requirements – as some sadly accept this as a correct definition of weak acid/base! )

Note that the term weak in a chemical sense is not the same as weak used to describe diluting

fruit squash with water! It refers to the degree of dissociation of the parent molecules into ions

and has nothing to do with the concentration of the resulting solution.

Thus, a 5.0M solution of ethanoic acid is a concentrated solution of a weak acid (pH = -0.70),

whereas a 0.001M solution of hydrochloric acid is very dilute solution of a strong acid (pH =

3.00).

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i.e. pH is NOT a measure of the strength of the acid or base (in terms of strong or weak), just

the concentration of H+ ions in solution.

Weak acids

When a weak acid, HA, is added to water, the following equilibrium is established:

HA + H2O H3O+ + A-

Weak bases are also only partly dissociated; ammonia is the most common weak base:

NH3 + H2O NH4+ + OH-

For acids, if we write H+ in place of H3O

+, the equation simplifies to:

HA H+ + A-

The equilibrium constant for this is called the acid dissociation constant, Ka

Ka = [H+] [A-] The units of Ka are mol dm-3

[HA]

The larger the value of Ka, the more dissociated and thus the stronger the acid.

i.e. ethanoic acid CH3COOH Ka = 1.70 x 10-5

mol dm-3

citric acid H3C6H5O7 Ka = 7.40 x 10-3

mol dm-3

(i.e. stronger)

carbonic acid H2CO3 Ka = 4.20 x 10-7

mol dm-3

(i.e. weaker)

[note that this expression applies to all acids, thus for strong acids, [HA] = 0 and thus Ka is

infinitely large and incalculable!].

The value of Ka can be more conveniently expressed using a logarithmic scale:

pKa = -logKa

the larger the value of Ka, the lower the value of pKa, and thus the stronger the acid.

Calculations involving weak acids

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In the acid solution, we may assume that the concentration of H+ from the water negligible

compared with that from the acid, hence [H+] = [A

-], so when considering solutions of weak

acids only the expression for Ka simplifies to:

Ka = [H+]2

[HA]

Note that this expression is NOT the definition of Ka!

To calculate the pH of a solution of a weak acid

We need to know: 1. the value of Ka for the acid;

2. the equilibrium concentration of the undissociated acid.

Strictly, this concentration is not the same as the initial concentration of the acid, but as the

amount of dissociation is very small, to simplify calculations we can assume that the

equilibrium concentration of the undissociated acid is the same as the original acid.

Examples

(i) calculate the pH of a 0.250M solution of ethanoic acid, given that the pKa = 4.76

Ka = 10-4.76

= 1.74 x 10-5

mol dm-3

Rearranging the equation for Ka gives:

[H+]2 = Ka x [HA] = (1.74 x 10

-5)2 x 0.250 = 4.34 x 10

-6 hence [H

+] = 2.08 x 10

-3 mol dm

-3

and pH = -log[2.08 x 10-3

] = 2.68

(ii) the pH of a 0.100M solution of a weak acid is 2.88. Calculate Ka for this acid

If pH = 2.88 then [H+] = 10

-2.88 = 1.32 x 10

-3 mol dm

-3

Ka = [H+]2 / [HA] = (1.32 x 10

-3)2 / 0.100 = 1.74 x 10

-5 mol dm

-3

(iii) a solution of a weak acid, HA, with Ka = 6.08 x 10-4 mol dm-3 has a pH of 5.73. Calculate

the concentration of the acid, HA, in the solution

If pH = 5.73 then [H+] = 10

-5.73 = 1.86 x 10

-6 mol dm

-3

Ka = [H+]2 / [HA] so [HA] = [H

+]2 / Ka = (1.86 x 10

-6)2 / 6.08 x 10-4 = 5.69 x 10

-9 mol dm

-3

1.4 pH CURVES

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The shapes of the curves obtained when 0.1M solutions of alkalis are added to 0.1M solutions

of monobasic (monoprotic) acids are shown on pages 9 & 10 for all four strong and weak

combinations. For each one you should know:

(i) the pH at the start

(ii) the pH at equivalence (the end point) and

(iii) the pH after twice this volume of alkali has been added

Note the difference between equivalence point and end point:

The equivalence point is the pH at which the no. of moles of OH- added exactly equals the

no. moles of H+ originally present, i.e. the solutions have been exactly neutralised, and can

ONLY be determined from the pH curve.

The end point is the volume of alkali (in these cases) which is needed to reach the

equivalence point – this can be determined either from the pH curve or by using a suitable

acid-base indicator.

The equivalence point is NOT always at pH = 7 – indeed, it is only this value for the strong

acid-strong base combination.

The salt of a strong acid and a weak base is acidic and the equivalence point is at pH < 7.

The salt of a weak acid and a strong base is alkaline and the equivalence point is at pH > 7.

Finally, note that a weak acid – weak base combination does not have an easily measureable

equivalence point and thus this type of titration is not carried out.

The pH at half-equivalence for a weak acid, HA, is the point at which [HA] = [A-].

Thus Ka = [H+] [A

-] / [HA] simplifies to Ka = [H

+] or pKa = pH.

This provides an experimental method to measure pKa and hence Ka for a weak acid:

1. titrate HA solution with strong base (e.g. NaOH) and determine equivalence point

from pH curve – thus the volume (V) of NaOH added to reach equivalence point;

2. determine V/2 value and read off the pH for this volume – this pH = pKa

pH curve for weak acid (HA) titrated against a strong base (NaOH) – Ka determination

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Volume of 0.100M sodium hydroxide added (cm3)

[Image edited from https://chemistry.stackexchange.com/questions/27800/whats-happening-at-the-beginning-of-a-weak-acid-titration]

1.4a pH curves

pH

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1.5 INDICATORS

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An acid-base indicator is simply a weak acid where the colours of the acid (HIn) itself and its

anion (In-) are noticeably different.

e.g. HIn H+ + In-

colour 1 colour 2

At low pH, the unionised form HIn predominates and thus the solution appears colour 1.

At high pH, the ionised form In- predominates and thus the solution appears colour 2.

Complete colour change from one colour to the other usually takes place over around 1 - 2 pH

units.

Hence it is very important to choose the correct indicator for each combination of acid-alkali

titration.

The indicators colour change pH range MUST lie completely on the steep part of the

titration curve (i.e. around the equivalence point) so one drop from the burette will change

the pH sufficiently to change the indicator colour completely.

The two most common indicators are methyl orange and phenolphthalein:

Indicator Colour 1: acid (HIn) pH range Colour 2: base (In-)

Methyl orange Red 3.2 – 4.4 Yellow

Phenolphthalein Colourless 8.2 – 10.0 Pink

Now look back at the pH curves and decide which is the best indicator from these two to use

for each of the four types of acid-base titration?

1.6 BUFFER SOLUTIONS

A buffer solution resists a change in pH when small amounts of acid or alkali are added or

when it is diluted.

An acid buffer maintains a pH below 7. Acid buffers are prepared by mixing equimolar

quantities of a weak acid and its salt, e.g. ethanoic acid and sodium ethanoate (an alternative

way of achieving this is to half-neutralise some ethanoic acid with a strong alkali).

CH3COOH H+ + CH3COO-

LARGE small small

CH3COONa Na+ + CH3COO-

small LARGE

When a small amount of H+ is added to this buffer mixture, the equilibrium moves slightly to

the left – the large [CH3COOH] rises slightly and the large [CH3COO-] falls slightly.

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Ka = [H+] [CH3COO

-] thus [H

+] = Ka x [CH3COOH]

[CH3COOH] [CH3COO-]

Both [CH3COOH] and [CH3COO-] are large relative to H

+ and so their ratio remains almost

constant – as Ka is a constant and so [H+] and thus the pH remains almost constant.

If the buffer solution is simply diluted with water, the ratio of [CH3COOH] / [CH3COO-]

again remains constant, so [H+] and thus pH remain constant.

An alkaline buffer maintains a pH above 7. This consists of an equimolar mixture of a weak

base together with its salt, e.g. ammonia and ammonium chloride:

NH3 + H2O NH4+ + OH-

LARGE small small

NH4Cl NH4+ + Cl-

small LARGE

When H+ is added, the equilibrium moves to the right; [NH4

+] rises slightly and [NH3] falls

slightly, but [OH-] remains almost constant and so pH remains almost constant.

Buffers are very important in the control of pH both industrially in the manufacture of

medicines, dyes, shampoos, etc. They are also found in biological systems, where they ensure

that the pH of aqueous solutions in living creatures is maintained within narrow ranges, e.g.

the blood in our bodies is kept within the pH range 7.35 – 7.45 to ensure that enzymes within

our bodies work correctly and most efficiently.

Quantitative treatment of acid buffers

A buffer was 0.100M with respect to both HA and A- (Ka = 1.80 x 10

-5 mol dm

-3)

(i) calculate the pH of this buffer

Ka = [H+] [A

-] / [HA] so Ka = [H

+] hence pH = -log[1.80 x 10

-5] = 4.74

This is known as the half-equivalence point, and can be used experimentally to find the

value of Ka (since Ka = [H+] at this point, pH = pKa, which can be read off the pH curve – see

also the pH curves section 1.4, p8).

(ii) calculate the change in pH when 10.0cm3 of 1.00M HCl are added to 1000cm

3 of this

buffer

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No. moles H+ added = MV / 1000 = 1.00 x 10.0 / 1000 = 0.0100 mol

The extra H+ reacts with A

- to make more HA, so:

No. moles HA present after addition = original + created = 0.100 + 0.010 = 0.110 mol

New concentration of HA = 0.110 / V

No. moles A- present after addition = original – removed = 0.100 – 0.010 = 0.090 mol

New concentration of A- = 0.090 / V

Ka = [H+] [A

-] / [HA] = 1.80 x 10

-5

[H+] = Ka x [HA] / [A

-] = 1.80 x 10

-5 x 0.110 / 0.090 (the volumes V cancel out)

[H+] = 2.20 x 10

-6 so pH = 4.66

Note pH has only decreased by a very small amount = 4.74 – 4.66 = -0.08 units

(iii) calculate the change in pH when 10.0cm3 of 1.00M NaOH are added to 1000cm

3 of this

buffer

No. moles OH- added = MV / 1000 = 1.00 x 10.0 / 1000 = 0.010 mol

The extra OH- reacts with HA to make more A

-, so:

No. moles HA present after addition = original - removed = 0.100 – 0.010 = 0.090 mol

New concentration of HA = 0.090 / V

No. moles A- present after addition = original + created = 0.100 + 0.010 = 0.110 mol

New concentration of A- = 0.110 / V

Ka = [H+] [A

-] / [HA] = 1.80 x 10

-5

[H+] = Ka x [HA] / [A

-] = 1.80 x 10

-5 x 0.090 / 0.110 (again, the volumes V cancel out)

[H+] = 1.47 x 10-5

so pH = 4.83

Note pH has only increased by a very small amount = 4.74 – 4.83 = +0.09 units

(iv) calculate pH of the buffer formed when 25.0cm3 of 0.100M NaOH are added to 75.0cm

3

of 0.200M HA (for which Ka = 1.80 x 10-5

mol dm-3

)

Reaction which occurs is: HA + OH- -----> A

- + H2O

No. moles HA used = 75.0 x 0.200 / 1000 = 0.0150

No. moles OH- added = 25.0 x 0.100 / 1000 = 0.00250

No. moles A- formed = no. moles OH

- used = 0.00250

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Concentration of A- in the buffer = 0.00250 / V

No. moles HA removed = 0.00250

No. moles HA in the buffer = 0.0150 – 0.00250 = 0.01250

Concentration of HA = 0.01250 / V

Ka = [H+] [A

-] / [HA] = 1.80 x 10

-5

[H+] = Ka x [HA] / [A

-] = 1.80 x 10

-5 x 0.0125 / 0.00250 (again, the volumes V cancel out)

[H+] = 9.00 x 10

-5 so pH = 4.05