Colourful Solutions > Proton transfer reactions > Buffer calculations

The Mad Science Lab

Higher-level only

Buffer solutions may be made directly by mixing weak acids/bases with their salts or by reaction between weak acids/bases and strong base/acid to produce the required mixture.

Syllabus ref: R3.1.17

Reactivity 3.1.17 - The pH of a buffer solution depends on both: (HL)

  • • the pKa or pKb of its acid or base
  • • the ratio of the concentration of acid or base to the concentration of the conjugate base or acid.
  • Solve problems involving the composition and pH of a buffer solution, using the equilibrium constant.

Guidance

  • Include explanation of the effect of dilution of a buffer.

Tools and links

  • Reactivity 2.3 - How does Le Chatelier’s principle enable us to interpret the behaviour of indicators and buffer solutions?

Preparation of an acidic buffer

The simplest way of preparing a buffer solution is to dissolve a known quantity of the salt of the weak acid (or base) in a solution of weak acid (or base) of known concentration.

Example: Sodium ethanoate is the salt of a weak acid - it has the formula CH3COONa (relative molecular mass = 82)

If 8.2 g of sodium ethanoate are dissolved in 100 cm3 of ethanoic acid then:

The concentration of the sodium ethanoate is equal to 0.1/0.1 = 1M

The concentration of the ethanoic acid = 1M

Therefore the buffer solution will have pKa = pH

Therefore the buffer has a pH value of 4.78

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Preparation of a basic buffer

A basic buffer is prepared using a weak base and its salt. The classic example is ammonia solution and ammonium chloride.

The pOH of the solution may be calculated using the pkb value of ammonia (4.75), or the Kb of ammonia (1.78 x 10-5). Once the pOH is found, the pH is calculated using the relationship pH + pOH = 14.

Example: Calculate the pH of a buffer solution consisting of a 0.2 mol dm-3 solution of ammonium chloride in 0.1 mol dm-3 ammonia solution (pKb = 4.75).

The base equilibrium expression:

Separate out the [OH-]

Take logs throughout

Now multiply through by minus (-1)

(Notice that ammonia and ammonium ion have now changed places)

The fraction [NH3]/[NH4+] = 1/2 = 0.5

Log 0.5 = -0.3

Therefore pKb = pOH - 0.3

pOH = 4.75 + 0.3 = 5.05

Therefore pH of the buffer solution = 14 - 5.05 = 8.95

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Preparation of a buffer by neutralisation

Another way to prepare a buffer solution (much favoured by examiners) is to neutralise an excess of weak acid (or weak base) with some strong base (or strong acid). The neutralisation produces the salt of the weak acid (or base) 'in situ' and, as the weak acid was in excess, there will still be some weak acid in the mixture. The resultant mixture contains both the salt of the weak acid and the weak acid itself.

Example: Find the pH of the solution formed by adding 50cm3 sodium hydroxide (1M) solution to 100 cm3 (1M) ethanoic acid.

Moles of sodium hydroxide added = 0.05 x 1 = 0.05 moles

This neutralises an equal number of moles of the ethanoic acid = 0.05 moles and produce exactly 0.05 moles of sodium ethanoate

CH3COOH + NaOH → CH3COONa + H2O

The concentration of the sodium ethanoate produced = 0.05 moles / 0.15 dm3 = 0.33M

The total initial acid moles = 1 x 0.1 = 0.1 moles

Moles of acid reacted with sodium hydroxide = 0.05 moles

Therefore moles of acid remaining = 0.1 - 0.05 = 0.05 moles

Therefore molarity of acid = 0.05 /0.15 = 0.33 M

The buffer solution pH can then be obtained from the buffer law:

pKa = pH + log[ethanoic acid]/[ethanoate ion]

pH = pKa - log[ethanoic acid]/[ethanoate ion]

pH = pKa - log 1

pH = pKa

In this case the concentration of the acid and the salt are equal and pH = pKa.

This represents a rapid shortcut in questions of this type. If it can be shown that the weak acid concentration (or the weak base concentration) equals the salt concentration then pH = pKa and NO calculation needs to be carried out!

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Buffers summary

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Worked examples

Q863-01 Which of the following combinations will produce a buffer solution?
  1. I.  20cm3 0.10 mol dm-3 CH3COOH and 10cm3 0.10 mol dm-3 CH3COONa
  2. II. 20cm3 0.10 mol dm-3 CH3COOH and 10cm3 0.10 mol dm-3 NaOH
  1. I only
  2. II only
  3. Both I and II
  4. Neither I nor II
Answer

I - contains a weak acid and its sodium salt and WILL have buffer action.

II - contains an excess of a weak acid and sodium hydroxide. The sodium hydroxide reacts with some of the weak acid making the sodium salt of ethanoic acid and leaving the excess ethanoic acid. It is also now a buffer mixture.

Correct response I and II = C

Q863-02 Citric acid (C6H8O7) is a triprotic acid whose 1st ionisation has a pKa of 3.13. What mass of sodium dihydrogen citrate (C6H7O7Na, Mr=214) must be added to 100cm3 of a 0.1 mol dm-3 solution of citric acid to produce a buffer of pH = 4.0?
Answer

The acid dissociation constant of a weak acid: Ka = [H+][A-]/[HA]

Derive the buffer law expression (rearrange to isolate [H+], take logs, multiply through by -1, replace terms):

pH = pKa - log([HA]/[A-])

First, calculate the molarity of the sodium dihydrogen citrate needed:

4.0 = 3.13 - log([HA]/[A-])

log([HA]/[A-]) = -0.87

[HA]/[A-] = 0.135

[A-] = 0.1/0.135 = 0.741 mol dm-3

But only 100cm3 are needed therefore moles required = 0.741/10 = 0.0741 moles

Therefore mass of sodium dihydrogen citrate (Mr=214) needed = 0.0741 x 214 = 15.9 g

Q863-03 A buffer solution can be prepared by adding which of the following to 50cm3 of 0.10 mol dm-3 CH3COOH(aq)?
  1. I.  50cm3 of a 0.10 mol dm-3 CH3COONa(aq)
  2. II.  25cm3 of a 0.10 mol dm-3 NaOH(aq)
  3. III. 50cm3 of a 0.10 mol dm-3 NaOH(aq)
  1. I only
  2. I and II only
  3. II and III only
  4. I, II and III
Answer

50cm3 of 0.10 mol dm-3 CH3COOH(aq) contains 0.05 x 0.1 moles = 0.005 moles of ethanoic acid

I - 50cm3 of a 0.10 mol dm-3 CH3COONa(aq) creates a buffer mixture of a weak acid + the salt of a weak acid.

II - Addition of 25cm3 of a 0.10 mol dm-3 NaOH(aq) = 0.0025 moles of strong base. This will all react with some of the ethanoic acid, leaving a mixture of ethanoic acid and its sodium salt. It is a buffer solution.

III - In this case the sodoim hydroxide = 0.005 moles and will neutralise all of the ethanoic acid. The result is just a solution containing only sodium ethanoate. It has no buffer action.

Correct response I and II only = B

Q863-04 Which of the following combinations produce a buffer solution when equal volumes are mixed?
  1. I.   0.1 M HCl and 0.1 M NH4Cl
  2. II.  0.1 M HCl and 0.2 M NH3
  3. III. 0.1 M NH3 and 0.1 M NH4Cl
  1. I only
  2. II only
  3. II and III only
  4. I, II and III
Answer

I - The strong acid is not affected by the ammonium salt in any way; it does not produce a buffer solution.

II - The excess ammonia reacts will all of the strong acid making a salt, ammonium chloride, and leaving the excess weak base. This is now a buffer solution consisting of a weak base and its salt.

III - This is a direct mixture of a weak base and its salt; it is a buffer.

Correct response II and III only = C

Q863-05 A buffer solution is made from a weak acid HA and its sodium salt. If the pH of the buffer is 5.4 and the solution was made by dissolving 0.2 moles of the sodium salt in 100 cm3 of 0.1 mol dm-3 acid, calculate the acid dissociation constant, Ka, of the weak acid. (assume that the volume of the solution does not change on addition of the salt).
Answer

The acid dissociation constant Ka = [H+][A-]/[HA]

Rearrange to isolate [H+] and take logs: log[H+] = logKa + log([HA]/[A-])

Multiply through by -1: pH = pKa - log([HA]/[A-])

  • Molarity of the sodium salt = 0.2/0.1 = 1.0 mol dm-3
  • Molarity of the acid [HA] = 0.1 mol dm-3

Substitute values: 5.4 = pKa - log(0.1/2)

Therefore pKa = 5.4 - 1.3 = 4.1

Therefore Ka of the weak acid = 7.94 x 10-5
Q863-06 If the molarity of the carbonic acid (Ka = 4.26 x 10-7) in blood is 0.001 mol dm-3, calculate the molarity of sodium hydrogen carbonate needed to give a pH of 7.45.
Answer

The acid dissociation constant of a weak acid: Ka = [H+][A-]/[HA]

Deriving the buffer law expression (rearrange to isolate [H+], take logs, multiply through by -1, replace terms):

pH = pKa - log([HA]/[A-])

7.45 = 6.37 - log([HA]/[A-])

log([HA]/[A-]) = -1.08

[HA]/[A-] = 0.083

[A-] = [HA]/0.083 = 0.012 mol dm-3
Q863-07 A student carries out a titration by adding 0.10 mol dm-3 hydrochloric acid dropwise to 50 cm3 of 0.10 mol dm-3 ammonia solution (pKb = 4.76). Calculate the pH of the solution when she has added exactly 25cm3 of ammonia solution.
Answer

Ammonia solution reacts wth hydrochloric acid according to the equation: NH3 + HCl → NH4Cl

When 25cm3 of hydrochloric acid have been added, exactly half the ammonia has been neutralised and the moles of salt produced exactly equals the moles of ammonia remaining.

By the base dissociation equation: Kb = [BH+][OH-]/[B]

When [BH+] = [B], Kb = [OH-] therefore pKb = pOH and pOH = 4.76

As pOH + pH = 14, then pH = 14 - 4.76 = 9.24

Q863-08 A student carries out a titration by adding 0.10 mol dm-3 sodium hydroxide dropwise to 50 cm3 of 0.10 mol dm-3 hydrofluoric acid (Ka = 6.7 x 10-4). Calculate the pH of the solution when she has added exactly 25cm3 of sodium hydroxide.
Answer

Sodium hydroxide reacts with hydrofluoric acid according to the equation: NaOH + HF → NaF + H2O

When 25cm3 of sodium hydroxide have been added, the moles of sodium fluoride produced exactly equals the moles of hydrofluoric acid remaining. Therefore the concentration of the weak acid and the salt are equal.

Inspection of the acid dissociation: Ka = [H+][A-]/[HA] shows us that when [A-]=[HA] the hydrogen ion concentration = Ka

Therefore pH = pKa and in this case Ka = 6.7 x 10-4

Therefore pH = 3.17
Q863-09 Calculate the mass of sodium ethanoate (Mr = 82) that must be added to 100cm3 of 0.1 mol dm-3 ethanoic acid to produce a buffer of pH 5.85. Ka of ethanoic acid = 1.78 x 10-5
Answer

The acid dissociation constant for ethanoic acid: Ka = [H+][CH3COO-]/[CH3COOH]

Rearrange to separate [CH3COO-] (we are asked for the sodium ethanoate): [CH3COO-] = Ka[CH3COOH]/[H+]

Assume that the concentration of ethanoic acid is approximately the same as that given and substitute values:

[CH3COO-] = (1.78 x 10-5 x 0.1)/[H+]

The hydrogen ion concentration can be found from the pH = -log[H+], pH = 5.85,
therefore [H+] = 10-5.85 = 1.41 x 10-6 mol dm-3

Therefore [CH3COO-] = (1.78 x 10-5 x 0.1)/1.41 x 10-6 = 1.26 mol dm-3

This is the concentration that must be achieved in 100cm3 so the number of moles that must be added = 1.26/10 = 0.126

Mass of sodium ethanoate = 82, therefore mass of sodium ethanoate needed = 0.126 x 82 = 10.33g

Q863-10 Calculate the mass of sodium hydroxide that must be added to 100cm3 of ethanoic acid solution (0.1 mol dm-3) needed to produce a solution with a pH = 4.5. (Ka of ethanoic acid = 1.78 x 10-5). Assume that there is no change in volume on addition of NaOH.
Answer

The acid dissociation constant of a weak acid: Ka = [H+][A-]/[HA]

Therefore:

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

If pH = 4.5, [H+] = 10-4.5

[H+] = 3.16 x 10-5

the molarity ratio = [A-]/[HA] = 0.56 (as the total volume is common for both substances the mole ratio must be the same)

From the equation for the reaction:

CH3COOH + NaOH → CH3COONa + H2O

The initial moles of ethanoic acid = 0.10 x 0.10 = 0.01 moles

We can see that 1 mole of CH3COOH produces 1 mole of CH3COONa

Therefore if x moles of CH3COONa are made (0.01 - x) moles of CH3COOH remain.

If the ratio [A-]/[HA] = 0.56, then x/(0.01 - x) = 0.56

(0.01 - x)0.56 = x

0.0056 = 1.56x

x = 0.00359

So the moles of acid that have to be neutralised = 0.00359 and this requires an equivalent number of moles of NaOH (Mr=40)

Therefore mass of NaOH required = 0.00359 x 40 = 0.144 g
  Now test yourself

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