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How do buffer solutions resist changes in pH?

Explain how a buffer maintains pH using equilibrium and calculate buffer pH.

How a weak acid and its conjugate base resist pH change, the action of buffers when acid or base is added, buffer capacity, and calculating buffer pH with the Henderson-Hasselbalch relationship, with worked TASC-style examples.

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What this dot point is asking

TASC expects you to explain buffer action through equilibrium and Le Chatelier's principle, write the neutralising equations, and calculate buffer pH.

What a buffer is

A common example is ethanoic acid mixed with sodium ethanoate: the solution contains plenty of the weak acid CH3COOH\text{CH}_3\text{COOH} and plenty of its conjugate base CH3COO\text{CH}_3\text{COO}^-. A buffer can be made either by mixing the weak acid with a salt of its conjugate base, or by partially neutralising a weak acid with a strong base.

How a buffer works

The buffer relies on the equilibrium CH3COOH(aq)H+(aq)+CH3COO(aq)\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^+(aq) + \text{CH}_3\text{COO}^-(aq), with a large reservoir of both partners.

  • When acid (extra H+\text{H}^+) is added, it is consumed by the conjugate base: CH3COO(aq)+H+(aq)CH3COOH(aq)\text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{CH}_3\text{COOH}(aq). The added hydrogen ions are mopped up, so the pH barely changes.
  • When base (extra OH\text{OH}^-) is added, it is neutralised by the weak acid: CH3COOH(aq)+OH(aq)CH3COO(aq)+H2O(l)\text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq) \rightarrow \text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l). The added hydroxide is removed, so the pH barely changes.

Because both partners are present in large amounts, the buffer absorbs additions of either acid or base.

Buffer capacity

Buffer capacity is the amount of acid or base a buffer can absorb before the pH changes sharply. It is greatest when the two components are present in roughly equal, high concentrations. Once one partner is largely used up, the buffer fails and the pH swings rapidly.

Choosing a buffer for a target pH

To make a buffer at a particular pH, choose a weak acid whose pKapK_a is close to the desired pH, then adjust the ratio of acid to conjugate base. Because pH=pKa+log10[base][acid]\text{pH} = pK_a + \log_{10}\dfrac{[\text{base}]}{[\text{acid}]}, equal amounts of the two give pH=pKa\text{pH} = pK_a, and shifting the ratio up or down moves the pH by at most about one unit before the buffer becomes ineffective. This is why each buffer system has a useful working range of roughly pKa±1pK_a \pm 1.

Biological and natural buffers

Buffers are essential in living systems. Human blood is held near pH 7.47.4 mainly by the carbonic acid and hydrogencarbonate system, H2CO3H++HCO3\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-, with the lungs and kidneys removing or retaining CO2\text{CO}_2 and HCO3\text{HCO}_3^- to keep the ratio steady. Even small departures from this pH are dangerous, which shows how tightly biological buffers must work. Ocean water is similarly buffered by the carbonate system, which is why rising atmospheric CO2\text{CO}_2 gradually lowers ocean pH as the buffer is progressively consumed.

In the exam, name both components of the buffer, write the neutralising equations for added acid and added base, link the resistance to Le Chatelier's principle, and use KaK_a with the concentration ratio to calculate the pH.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20225 marksA buffer is prepared containing 0.20 mol L10.20\ \text{mol L}^{-1} ethanoic acid, CH3COOH\text{CH}_3\text{COOH}, and 0.30 mol L10.30\ \text{mol L}^{-1} sodium ethanoate, CH3COONa\text{CH}_3\text{COONa}. (Ka(CH3COOH)=1.8×105K_a(\text{CH}_3\text{COOH}) = 1.8 \times 10^{-5}.) (a) Write equations showing how this buffer responds to added H+\text{H}^+ and added OH\text{OH}^-. (b) Calculate the pH of the buffer.
Show worked answer →

(a) Added acid is removed by the conjugate base: CH3COO(aq)+H+(aq)CH3COOH(aq)\text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{CH}_3\text{COOH}(aq). Added base is removed by the weak acid: CH3COOH(aq)+OH(aq)CH3COO(aq)+H2O(l)\text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq) \rightarrow \text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l). (2 marks)

(b) [H+]=Ka×[acid][base]=1.8×105×0.200.30=1.2×105 mol L1[\text{H}^+] = K_a \times \dfrac{[\text{acid}]}{[\text{base}]} = 1.8 \times 10^{-5} \times \dfrac{0.20}{0.30} = 1.2 \times 10^{-5}\ \text{mol L}^{-1}. (2 marks)

pH=log10(1.2×105)=4.92\text{pH} = -\log_{10}(1.2 \times 10^{-5}) = 4.92. (1 mark)

TCE 20213 marksCarbonic acid and hydrogencarbonate form the main buffer of human blood: H2CO3(aq)H+(aq)+HCO3(aq)\text{H}_2\text{CO}_3(aq) \rightleftharpoons \text{H}^+(aq) + \text{HCO}_3^-(aq). Explain, using this equilibrium and Le Chatelier's principle, how the blood buffer responds when a small amount of lactic acid (a source of H+\text{H}^+) is produced during exercise.
Show worked answer →

Adding lactic acid increases [H+][\text{H}^+]. By Le Chatelier's principle the equilibrium shifts left to oppose this increase, consuming the added H+\text{H}^+. (1 mark)

The added H+\text{H}^+ reacts with the conjugate base: HCO3(aq)+H+(aq)H2CO3(aq)\text{HCO}_3^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{CO}_3(aq). (1 mark)

Because most of the extra H+\text{H}^+ is converted to H2CO3\text{H}_2\text{CO}_3 (which can be removed as CO2\text{CO}_2 by the lungs), the blood pH stays close to its normal value of about 7.47.4 rather than dropping sharply. (1 mark)

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