pH of Acidic and Alkaline Solutions

Lessons

Senior 6 · Chemistry

pH of Acidic and Alkaline Solutions

5 lessons

Lesson 5 of 5

Buffer Solutions

25 min

How does blood stay at pH 7.4?

Your muscles pour lactic acid and carbon dioxide into your blood every time you run, yet the pH of your blood stays between 7.35 and 7.45. Below 7.35 (acidaemia) or above 7.45 (alkalaemia), enzymes and organs stop working properly. Add the same amount of acid to pure water and its pH would crash by several units. Blood resists the change because it contains a buffer: a weak acid, carbonic acid, together with its conjugate base, the hydrogencarbonate ion.

Chemists rely on buffers too. Every pH meter is calibrated with buffer solutions of known pH, and many medicines, foods and fermentations are kept at a fixed pH by a buffer. In this lesson you will see how a buffer works and calculate its pH.

Definition

Buffer solution: a solution whose pH changes very little when a small amount of acid or base is added, or when it is diluted. Example: a mixture of ethanoic acid and sodium ethanoate.

Acidic buffer: a weak acid and a salt of its conjugate base, with pH below 7 (e.g. $\mathrm{CH_3COOH/CH_3COONa}$ ). Basic buffer: a weak base and a salt of its conjugate acid, with pH above 7 (e.g. $\mathrm{NH_3/NH_4Cl}$ ).

Henderson–Hasselbalch equation: $\mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}$ , giving the pH of a buffer from the p $K_{a}$ of its acid and the ratio of base to acid.

Buffer capacity: the amount of strong acid or base a buffer can absorb before its pH changes sharply. It is largest when the concentrations are high and $[\mathrm{A^{-}}] \approx [\mathrm{HA}]$ .

How a buffer works

Consider a solution that is 0.100 mol dm⁻³ in both ethanoic acid and sodium ethanoate. It contains large "reservoirs" of both members of a conjugate pair, in equilibrium:

$\mathrm{CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^{-}(aq) + H_3O^{+}(aq)}$

Add a little acid: the added $\mathrm{H_3O^{+}}$ is removed by the large reservoir of ethanoate ions, $\mathrm{CH_3COO^{-} + H_3O^{+} \rightarrow CH_3COOH + H_2O}$ .
Add a little alkali: the added $\mathrm{OH^{-}}$ is removed by the large reservoir of ethanoic acid, $\mathrm{CH_3COOH + OH^{-} \rightarrow CH_3COO^{-} + H_2O}$ .

Either way the ratio $\frac{[\mathrm{CH_3COO^{-}}]}{[\mathrm{CH_3COOH}]}$ changes only a little, so the pH changes only a little (Figure 5.1). Once one reservoir is used up, the buffer fails and the pH changes sharply.

Figure 5.1: Adding 1.00 mol dm⁻³ HCl to 100 cm³ of pure water and to 100 cm³ of a buffer (0.100 mol dm⁻³ in both ethanoic acid and sodium ethanoate), calculated by Ganzaa.

The Henderson–Hasselbalch equation

Rearrange the $K_{a}$ expression for the weak acid and take $-\log_{10}$ of both sides:

$[\mathrm{H_3O^{+}}] = K_a \times \frac{[\mathrm{HA}]}{[\mathrm{A^{-}}]} \quad\Rightarrow\quad \mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}$

In a buffer the acid is hardly ionised (the $\mathrm{A^{-}}$ present pushes its equilibrium to the left) and the anion hardly hydrolyses, so $[\mathrm{HA}]$ and $[\mathrm{A^{-}}]$ can be taken as the amounts mixed in. Four consequences follow:

When $[\mathrm{A^{-}}] = [\mathrm{HA}]$ , $\mathrm{pH} = \mathrm{p}K_a$ . To make a buffer of a given pH, choose a weak acid whose p $K_{a}$ is within about 1 unit of it.
Only the ratio matters, so the amounts in moles can be used directly, since both are in the same volume.
Dilution leaves the ratio, and so the pH, almost unchanged; but it lowers the buffer capacity.
For a basic buffer, use the $K_{a}$ of the conjugate acid: $\mathrm{pH} = \mathrm{p}K_a(\mathrm{NH_4^{+}}) + \log_{10}\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^{+}}]}$ .

The buffer in your blood

In blood at 37 °C, counting dissolved carbon dioxide as carbonic acid, the effective p $K_{a}$ is 6.1 (higher than the 25 °C value in Table 3.1 because the conditions are different). To hold the pH at 7.4, the ratio must be

$7.4 = 6.1 + \log_{10}\frac{[\mathrm{HCO_3^{-}}]}{[\mathrm{H_2CO_3}]} \quad\Rightarrow\quad \frac{[\mathrm{HCO_3^{-}}]}{[\mathrm{H_2CO_3}]} = 10^{1.3} \approx 20$

A 20 : 1 ratio is far from the ideal 1 : 1, yet the system works because it is open: the lungs breathe out excess carbon dioxide (removing $\mathrm{H_2CO_3}$ ) and the kidneys adjust the hydrogencarbonate level.

Worked example

Problem. A buffer is 0.100 mol dm⁻³ in ethanoic acid and 0.150 mol dm⁻³ in sodium ethanoate (p $K_{a}$ = 4.76). (a) Calculate its pH. (b) Calculate the pH after 1.00 cm³ of 1.00 mol dm⁻³ HCl is added to 100 cm³ of it, and compare with adding the same acid to 100 cm³ of pure water.

Solution.

(a) $\mathrm{pH} = 4.76 + \log_{10}\frac{0.150}{0.100} = 4.76 + 0.18 = 4.94$

(b) In 100 cm³: $n(\mathrm{CH_3COOH}) = 0.0100\,\mathrm{mol}$ , $n(\mathrm{CH_3COO^{-}}) = 0.0150\,\mathrm{mol}$ . The acid adds $1.00 \times 10^{-3}\,\mathrm{mol}$ of $\mathrm{H_3O^{+}}$ , which converts that much ethanoate into ethanoic acid:

	$\mathrm{CH_3COOH}$ / mol	$\mathrm{CH_3COO^{-}}$ / mol
before	0.0100	0.0150
after	0.0110	0.0140

$\mathrm{pH} = 4.76 + \log_{10}\frac{0.0140}{0.0110} = 4.76 + 0.10 = 4.86$

In pure water, $[\mathrm{H_3O^{+}}] = \frac{1.00 \times 10^{-3}}{0.101} = 9.9 \times 10^{-3}$ , pH 2.00.

Answer: (a) pH 4.94; (b) the buffer's pH falls by less than 0.1, to 4.86, while pure water falls by 5 units, from 7.00 to 2.00.

Worked example

Problem. What mass of sodium ethanoate ( $M = 82.03$ ) must be dissolved in 500 cm³ of 0.200 mol dm⁻³ ethanoic acid to make a buffer of pH 5.00? Assume no volume change.

Solution.

$5.00 = 4.76 + \log_{10}\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]} \quad\Rightarrow\quad \frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]} = 10^{0.24} = 1.74$

$n(\mathrm{HA}) = 0.500 \times 0.200 = 0.100\,\mathrm{mol}$ , so $n(\mathrm{A^{-}}) = 0.174\,\mathrm{mol}$ .

$m = 0.174\,\mathrm{mol} \times 82.03\,\mathrm{g\,mol^{-1}} = 14.3\,\mathrm{g}$

Answer: 14.3 g. The same buffer could be made without the salt by adding 0.174 mol of NaOH to a larger amount of ethanoic acid, so that the leftover acid and the ethanoate formed are in the ratio 1 : 1.74.

Warning

Common mistakes with buffers

Putting the ratio upside down: it is base over acid, $\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}$ . Check: more base must give a higher pH.
Forgetting to react the added acid or alkali with the buffer before using the equation.
Calling a strong acid and its salt (HCl + NaCl) a buffer. $\mathrm{Cl^{-}}$ cannot remove added $\mathrm{H_3O^{+}}$ , so there is no buffering.
Thinking a buffer keeps the pH perfectly constant. It only resists change, and only until one component runs out.

Tip

Buffer questions follow three steps: (1) react any added strong acid or base with the buffer, in moles; (2) find the new amounts of HA and $\mathrm{A^{-}}$ ; (3) substitute the ratio into $\mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}$ . If the question gives $K_{a}$ instead, use $[\mathrm{H_3O^{+}}] = K_a\frac{[\mathrm{HA}]}{[\mathrm{A^{-}}]}$ : it is the same equation.

Bale's Chemistry: How do buffers work (AQA A level Chemistry). Watch how acidic and basic buffers remove small amounts of added acid or alkali, and which equilibrium shifts in each case.

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Practice

Explain, with equations, how a mixture of ammonia and ammonium chloride resists a change in pH when a little acid or a little alkali is added.
Calculate the pH of a buffer that is 0.200 mol dm⁻³ in methanoic acid and 0.100 mol dm⁻³ in sodium methanoate (p $K_{a}$ = 3.75).
Calculate the pH of a solution that is 0.10 mol dm⁻³ in ammonia and 0.20 mol dm⁻³ in ammonium chloride (p $K_a(\mathrm{NH_4^{+}})$ = 9.25).
25.0 cm³ of 0.200 mol dm⁻³ ethanoic acid is mixed with 10.0 cm³ of 0.200 mol dm⁻³ NaOH. Show that the mixture is a buffer and calculate its pH.
0.50 cm³ of 1.0 mol dm⁻³ NaOH is added to 100 cm³ of the buffer in the first worked example (pH 4.94). Calculate the new pH.
Explain why diluting a buffer tenfold hardly changes its pH, but makes it worse at resisting added acid.
A patient's blood has $[\mathrm{HCO_3^{-}}] = 24\,\mathrm{mmol\,dm^{-3}}$ and $[\mathrm{H_2CO_3}] = 2.4\,\mathrm{mmol\,dm^{-3}}$ . Calculate the pH (p $K_{a}$ = 6.1), say whether it is normal, and evaluate the claim that "carbonic acid is a poor choice of buffer for pH 7.4".

Show answersHide answers

Added acid: $\mathrm{NH_3 + H_3O^{+} \rightarrow NH_4^{+} + H_2O}$ . Added alkali: $\mathrm{NH_4^{+} + OH^{-} \rightarrow NH_3 + H_2O}$ . Both reservoirs are large, so the ratio $\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^{+}}]}$ , and the pH, change only slightly.
$\mathrm{pH} = 3.75 + \log_{10}\frac{0.100}{0.200} = 3.75 - 0.30 = 3.45$ .
$\mathrm{pH} = 9.25 + \log_{10}\frac{0.10}{0.20} = 8.95$ .
$n(\mathrm{CH_3COOH}) = 5.00 \times 10^{-3}$ , $n(\mathrm{OH^{-}}) = 2.00 \times 10^{-3}$ . After reaction: $3.00 \times 10^{-3}\,\mathrm{mol}$ acid and $2.00 \times 10^{-3}\,\mathrm{mol}$ ethanoate, both present, so it is a buffer. $\mathrm{pH} = 4.76 + \log_{10}\frac{2.00}{3.00} = 4.58$ .
$n(\mathrm{OH^{-}}) = 5.0 \times 10^{-4}$ : acid 0.0100 → 0.0095 mol, ethanoate 0.0150 → 0.0155 mol. $\mathrm{pH} = 4.76 + \log_{10}\frac{0.0155}{0.0095} = 4.97$ .
Both concentrations fall by the same factor, so their ratio, and so the pH, is unchanged. But each reservoir now holds a tenth as many moles, so a given amount of added acid uses up a larger fraction of the ethanoate and changes the ratio more: the buffer capacity is lower.
$\mathrm{pH} = 6.1 + \log_{10}\frac{24}{2.4} = 7.1$ : below 7.35, so acidaemia, not normal. The claim is half right: 7.4 is 1.3 units above the p $K_{a}$ , outside the usual "p $K_{a}$ ± 1" range, so as a closed buffer it would be weak against added acid. But the body controls both components, removing $\mathrm{CO_2}$ through the lungs and adjusting $\mathrm{HCO_3^{-}}$ through the kidneys, so as an open system it is very effective.

1How does blood stay at pH 7.4?

2How a buffer works

3The Henderson–Hasselbalch equation

4The buffer in your blood

How does blood stay at pH 7.4?

How a buffer works

The Henderson–Hasselbalch equation

The buffer in your blood