Buffer solutions — resisting pH change — Chemistry

Before this

This leads to

Indicators and titration curves
Ka, Kb and pH calculations · coming soon

Two solutions with the same pH

Two beakers stand side by side, each holding 1.00 dm³ of solution at 25 °C. Both have a pH of 4.98. Beaker A is a very dilute solution of hydrochloric acid. Beaker B is a mixture of ethanoic acid and sodium ethanoate. A pH meter cannot tell them apart.

Now add the same small amount of strong acid, 0.010 mol of HCl, to each. Before you read on, predict what happens.

Predict — same pH, same acid added

Beakers A and B both start at pH 4.98. You add 0.010 mol of HCl to each 1.00 dm³. What happens to the two pH values? Choose one, then read the explanation.

This box gives the result of adding 0.010 mol of HCl to two solutions that both start at pH 4.98.

Beaker A, the dilute hydrochloric acid, falls to pH 2.00. Beaker B, the ethanoic acid and sodium ethanoate mixture, falls only to pH 4.96. The same pH at the start did not mean the same behaviour.

Beaker A falls by about 3 pH units. Beaker B falls by 0.02. The same starting pH did not mean the same solution. The pH tells you how much H⁺ is there now. It says nothing about what is waiting in the solution to react with added acid. Beaker B is a buffer solution, and this lecture explains how it works.

The last lecture ended with a promise. Your blood is held between pH 7.35 and 7.45, even though your cells keep producing acid, mostly as carbon dioxide. It said that this is done by mixtures that resist changes in pH, and that the next lecture would explain how such a mixture works. Here it is. At the end, you will see your blood's own buffer, with numbers.

Two reservoirs: one for acid, one for alkali

A buffer solution

A buffer solution resists changes in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid, HA, and its conjugate base, A⁻, both in large amounts.

The conjugate base usually comes from a salt of the acid. For ethanoic acid, CH₃COOH, the salt is sodium ethanoate, CH₃COONa, which gives ethanoate ions, CH₃COO⁻. The acid and the ion are a conjugate pair, as in the acid–base theories lecture: they differ by exactly one H⁺.

When acid is added, the conjugate base takes up the extra H⁺:

A^{-} (aq) + H^{+} (aq) \to HA (aq)

The added H⁺ does not stay free in the solution. It becomes part of a weak acid, which holds on to most of its protons. This is because a weak acid, by definition, gives away only a small fraction of its protons.

When alkali is added, the weak acid gives a proton to each OH⁻:

HA(aq) + OH⁻(aq) \to A⁻(aq) + H₂O(l)

The OH⁻ becomes water, and the weak acid becomes its conjugate base.

So a buffer holds two reservoirs. A⁻ is the reservoir that absorbs acid. HA is the reservoir that absorbs alkali. Each reaction only moves material from one reservoir to the other. Look at what happens when 0.10 mol of H⁺ is added to a buffer containing 0.30 mol of HA and 0.50 mol of A⁻:

Amounts in moles, in 1.00 dm³ of buffer made with ethanoic acid. Adding 0.10 mol of H⁺ moves 0.10 mol from the A⁻ reservoir to the HA reservoir. The ratio changes from 1.67 to 1.00, and the pH falls by only 0.22.

Why does it matter that the acid is weak? Try making a "buffer" from hydrochloric acid and sodium chloride. If you add more H⁺, the chloride ions do not take it up, because HCl is a strong acid: its conjugate base, Cl⁻, holds on to H⁺ so weakly that it does not react. So the added H⁺ stays free, and the pH falls just as it did in beaker A. A buffer needs a conjugate base that really reacts with H⁺, and that means the acid must be weak.

Ka: the equilibrium constant for an acid

To calculate a buffer's pH, you need one number for the acid: how strongly it gives away its protons. A weak acid sets up an equilibrium:

HA (aq) ⇌ H^{+} (aq) + A^{-} (aq)

You met the equilibrium constant $K_{c}$ in the equilibrium lectures: products over reactants, with water left out as the solvent. For an acid, this constant has its own name, the acid dissociation constant, $K_{a}$ :

K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}

Its unit is mol/dm³, because there are two concentrations on top and one on the bottom. A larger $K_{a}$ means a stronger acid. For ethanoic acid at 25 °C, $K_{a} = 1.74 \times 1 0^{- 5}$ mol/dm³. (Source: Chemguide.) That is small, so ethanoic acid is weak.

Numbers like $1.74 \times 1 0^{- 5}$ are awkward, so chemists use the same trick as for pH:

p K_{a} = - lo g_{10} K_{a}

For ethanoic acid, $p K_{a} = - lo g_{10} (1.74 \times 1 0^{- 5}) \approx 4.76$ . Because of the minus sign, a smaller p $K_{a}$ means a stronger acid, just as a smaller pH means more H⁺. In this lecture, $K_{a}$ is used only for buffers. Finding the pH of a weak acid on its own is the work of the Pre-University lecture on $K_{a}$ .

The buffer equation

Now rearrange the $K_{a}$ expression to make [H⁺] the subject. Multiply both sides by [HA], and divide by [A⁻]:

[H^{+}] = K_{a} \times \frac{[ HA ]}{[ A ^{-} ]}

Take $- lo g_{10}$ of both sides. The log of a product is the sum of the logs, and turning the fraction upside down changes the sign of its log:

- lo g_{10} [H^{+}] = - lo g_{10} K_{a}

+ lo g_{10} \frac{[ A ^{-} ]}{[ HA ]}

The Henderson–Hasselbalch equation

$pH = p K_{a} + lo g_{10} \frac{[ A ^{-} ]}{[ HA ]}$

The pH of a buffer is set by the p $K_{a}$ of its acid and the ratio of conjugate base to acid.

In a buffer, you can use the amounts you put in for [HA] and [A⁻]. This is because the weak acid gives away only a tiny fraction of its protons, and the large amount of A⁻ already present pushes its equilibrium even further to the left. Both are in the same volume, so you can even use moles instead of concentrations: the volume cancels in the ratio.

Three results follow at once. When [A⁻] = [HA], the ratio is 1, $lo g_{10} 1 = 0$ , and pH = p $K_{a}$ . More base than acid gives a pH above p $K_{a}$ . More acid than base gives a pH below it. So to make a buffer at a chosen pH, choose an acid whose p $K_{a}$ is close to that pH, and then set the ratio.

That is it. That is the whole idea. The pH depends on a ratio. Adding a little acid or alkali changes the amounts a little, so it changes the ratio only a little, and the logarithm shrinks that change even more.

Worked examples

Worked Example 1 — The pH of a buffer

A buffer contains 0.30 mol/dm³ ethanoic acid and 0.50 mol/dm³ sodium ethanoate. $K_{a} = 1.74 \times 1 0^{- 5}$ mol/dm³. Find its pH.

Think first. HA is ethanoic acid and A⁻ comes from the salt. There is more base than acid, so the pH should be a little above the p $K_{a}$ of 4.76.

pH = 4.759 + lo g_{10} \frac{0.50}{0.30}

= 4.759 + 0.222 \approx 4.98

Check. 4.98 is a little above 4.76, as expected. ✓ Keep three decimal places in the p $K_{a}$ until the end. Rounding it to 4.76 first can change the last digit of the answer.

Worked Example 2 — Adding acid

0.10 mol of HCl is added to 1.00 dm³ of the buffer in Worked Example 1. Find the new pH. Compare it with adding the same acid to 1.00 dm³ of pure water.

Think first. The H⁺ reacts with A⁻ to make HA. So 0.10 mol of A⁻ is used up, and 0.10 mol of HA is made.

New amounts: HA $0.30 + 0.10 = 0.40$ mol; A⁻ $0.50 - 0.10 = 0.40$ mol.

pH = 4.759 + lo g_{10} \frac{0.40}{0.40}

= 4.759 + 0 \approx 4.76

The pH fell from 4.98 to 4.76, by 0.22. In pure water, the same acid gives [H⁺] = 0.10 mol/dm³, so the pH falls from 7.00 to 1.00, by 6 units.

Check. The ratio is now exactly 1, so the pH must equal the p $K_{a}$ . ✓

Worked Example 3 — Adding alkali

Instead, 0.10 mol of NaOH is added to 1.00 dm³ of the original buffer. Find the new pH.

Think first. The OH⁻ reacts with HA, so HA falls by 0.10 mol and A⁻ rises by 0.10 mol.

pH = 4.759 + lo g_{10} \frac{0.60}{0.20}

= 4.759 + 0.477 \approx 5.24

Check. Alkali was added, so the pH should rise. It rose by 0.26. ✓ In pure water, the same alkali would raise the pH from 7.00 to 13.00.

Worked Example 4 — Making a buffer at pH 5.00

You have 0.50 mol of ethanoic acid in 1.00 dm³ of water. How much solid NaOH must you add to make a buffer with a pH of 5.00?

Think first. Adding NaOH turns some of the acid into ethanoate. So you make the A⁻ from the acid itself. First find the ratio you need.

lo g_{10} \frac{[ A ^{-} ]}{[ HA ]} = 5.00 - 4.759

= 0.241

\frac{[ A ^{-} ]}{[ HA ]} = 1 0^{0.241} \approx 1.74

If you add $x$ mol of NaOH, then A⁻ = $x$ and HA = $0.50 - x$ . So $x = 1.74 (0.50 - x)$ , which gives $2.74 x = 0.870$ and $x \approx 0.32$ mol.

Check. With 0.32 mol A⁻ and 0.18 mol HA, $4.759 + lo g_{10} (0.3175 \div 0.1825) \approx 5.00$ . ✓ This way of making a buffer, part-neutralising a weak acid, is how many laboratory buffers are made, and it returns in the titration curves lecture.

How much can a buffer take?

A buffer resists change only while both reservoirs last. Add more acid than there is A⁻, and the extra H⁺ has nothing left to react with. The pH then falls as fast as it would in water. Try it below. You can also make the buffer ten times more dilute, with the same ratio.

Explore — add acid or alkali to a buffer

Add HCl (move right) or NaOH (move left) to 1.00 dm³ of the ethanoic acid buffer, and to 1.00 dm³ of pure water. Watch both pH values. Then switch to the diluted buffer and try again.

Try:

Added: nothing

This table shows the pH of 1.00 dm³ of ethanoic acid buffer, and of 1.00 dm³ of pure water, after adding HCl or NaOH. The diluted buffer has the same ratio but one tenth of the amounts. Values are calculated from the full equilibrium, not only from the buffer equation.

Added	Buffer (0.30 + 0.50)	Diluted buffer	Pure water
nothing	4.98	4.98	7.00
0.10 mol HCl	4.76	1.30	1.00
0.10 mol NaOH	5.24	12.85	13.00
0.30 mol HCl	4.28	0.60	0.52
0.45 mol HCl	3.59	0.40	0.35
0.55 mol HCl	1.30	0.30	0.26

The full buffer holds its pH until almost all 0.50 mol of A⁻ is used up. The diluted buffer starts at the same pH but is overwhelmed by 0.10 mol of acid.

You should have seen two things. The full buffer holds its pH between about 4.3 and 5.9, while you add up to 0.30 mol of acid or 0.25 mol of alkali. Then it fails suddenly, when one reservoir runs out. The diluted buffer starts at exactly the same pH, but 0.10 mol of acid overwhelms it.

Stop and think

A student says: "Diluting a buffer with water must change its pH, because the concentrations fall." Is the student right? And does dilution change anything at all?

Show the answer

The pH stays the same. Water lowers [HA] and [A⁻] by the same factor, so their ratio does not change, and the pH depends only on the ratio. But dilution does change something important: the amount the buffer can absorb. The ability of a buffer to absorb added acid or alkali is called its buffer capacity. It depends on the actual amounts of HA and A⁻, not on their ratio. A buffer ten times more dilute has the same pH and one tenth of the capacity. It is also best when [HA] and [A⁻] are similar: a buffer works well only within about one pH unit of its p $K_{a}$ , where neither reservoir is small.

The buffer in your blood

Your blood's main buffer is made from carbon dioxide. Dissolved CO₂ reacts with water to make carbonic acid, H₂CO₃, which gives up a proton to form the hydrogencarbonate ion, HCO₃⁻ (also called bicarbonate):

CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻

Here the weak acid is the dissolved CO₂ (with its carbonic acid), and the conjugate base is HCO₃⁻. Blood plasma normally contains about 24 mmol/dm³ of HCO₃⁻ and about 1.2 mmol/dm³ of dissolved CO₂, a ratio of 20 to 1. At body temperature, the effective p $K_{a}$ of this pair is 6.1. (Source: WikiLectures, "Bicarbonate buffer", which gives these values and the pK_a of 6.1.) So:

pH = 6.1 + lo g_{10} 20

\approx 6.1 + 1.30 = 7.40

That is exactly normal blood pH. But notice something strange. The best buffers have a pH within about one unit of their p $K_{a}$ , with similar amounts of acid and base. Blood sits 1.3 units above, with 20 times more base than acid. On its own, this would be a poor buffer for acid.

Here is a simple model. Suppose exercise adds 2 mmol of H⁺ to each dm³ of plasma. HCO₃⁻ falls to 22 mmol/dm³ and CO₂ rises to 3.2. If the CO₂ stayed in the blood, the pH would be $6.1 + lo g_{10} (22 \div 3.2) \approx 6.94$ : dangerously low. But blood flows through the lungs, and you breathe out CO₂. Breathing faster keeps the CO₂ near 1.2 mmol/dm³, and then the pH is $6.1 + lo g_{10} (22 \div 1.2) \approx 7.36$ , still inside the normal range. The kidneys can also adjust the amount of HCO₃⁻, more slowly. So your blood's buffer is not a closed beaker. It is a buffer whose acid reservoir is emptied by your lungs. (This model leaves out the blood's other buffers, such as proteins and haemoglobin, which also absorb some of the acid.)

Three traps

Trap 1 — "A buffer keeps the pH constant"

A buffer resists changes in pH. The pH still changes a little every time, as Worked Example 2 showed (0.22 units). And once one reservoir is used up, the buffer stops working altogether. Write "resists changes in pH", never "keeps the pH constant".

Trap 2 — A strong acid and its salt

HCl with NaCl, or HNO₃ with NaNO₃, is not a buffer. The conjugate base of a strong acid does not take up H⁺, so added acid stays free. A buffer needs a weak acid and its conjugate base (or a weak base and its conjugate acid, such as NH₃ with NH₄Cl).

Trap 3 — The ratio upside down

The buffer equation has base over acid: $lo g_{10} ([A^{-}] \div [HA])$ . Putting acid on top changes the sign of the log. In Worked Example 1 that gives 4.54 instead of 4.98. Check every answer: more base than acid must give a pH above the p $K_{a}$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. For ethanoic acid, use $K_{a} = 1.74 \times 1 0^{- 5}$ mol/dm³ (p $K_{a}$ = 4.759).

(a) Find the pH of a buffer containing 0.10 mol/dm³ ethanoic acid and 0.10 mol/dm³ sodium ethanoate. (b) The sodium ethanoate is doubled to 0.20 mol/dm³. Find the new pH.

Hint 1
In (a), the ratio is 1.

Hint 2
In (b), $pH = 4.759 + lo g_{10} (0.20 \div 0.10)$ .

Full answer
(a) The ratio is 1 and $lo g_{10} 1 = 0$ , so pH = p $K_{a}$ ≈ 4.76. (b) $4.759 + lo g_{10} 2 = 4.759 + 0.301 \approx 5.06$ . Check: more base than acid, so the pH rose above the p $K_{a}$ . ✓ Doubling the base raised the pH by only 0.30, because the logarithm of 2 is 0.30.
Which of these mixtures are buffers? Give a reason for each. (a) HCl and NaCl. (b) CH₃COOH and CH₃COONa. (c) NH₃ and NH₄Cl. (d) 0.20 mol CH₃COOH with 0.10 mol NaOH. (e) 0.10 mol CH₃COOH with 0.20 mol NaOH.

Hint 1
A buffer needs a weak acid and its conjugate base, both present in large amounts after any reaction has finished.

Hint 2
For (d) and (e), first let the NaOH react with the acid. What is left?

Full answer
(a) No: HCl is a strong acid, and Cl⁻ does not take up H⁺. (b) Yes: a weak acid and its conjugate base. (c) Yes: NH₄⁺ is a weak acid and NH₃ is its conjugate base; this buffer works near pH 9. (d) Yes: 0.10 mol NaOH turns 0.10 mol of the acid into ethanoate, leaving 0.10 mol CH₃COOH and 0.10 mol CH₃COO⁻, so pH = p $K_{a}$ . (e) No: all 0.10 mol of acid is used up, leaving ethanoate and 0.10 mol of excess NaOH. With no HA left, nothing can absorb added alkali. Check: in every "yes", both members of a conjugate pair are left. ✓
A 0.500 dm³ buffer contains 0.25 mol of ethanoic acid and 0.25 mol of sodium ethanoate. 0.050 mol of HCl is added. Find the pH before and after.

Hint 1
Work in moles. Both substances are in the same volume, so the volume cancels in the ratio.

Hint 2
After the HCl: HA = 0.30 mol and A⁻ = 0.20 mol.

Full answer
Before: equal amounts, so pH = 4.76. After: HA $0.25 + 0.050 = 0.30$ mol and A⁻ $0.25 - 0.050 = 0.20$ mol. $pH = 4.759 + lo g_{10} (0.20 \div 0.30) = 4.759 - 0.176 \approx 4.58$ . Check: acid was added, so the pH fell, by 0.18. ✓ Using concentrations instead (0.60 and 0.40 mol/dm³) gives the same ratio and the same answer. ✓
A stretch. In a patient with a kidney problem, the HCO₃⁻ in plasma falls to 12 mmol/dm³, while the dissolved CO₂ stays at 1.2 mmol/dm³. Use p $K_{a}$ = 6.1. (a) Find the blood pH. (b) The body responds by breathing faster. What concentration of dissolved CO₂ would bring the pH back up to 7.30? (c) Explain in words why breathing faster helps.

Hint 1
(a) The ratio is 12 ÷ 1.2.

Hint 2
(b) For pH 7.30, $lo g_{10} (ratio) = 7.30 - 6.1 = 1.20$ , so the ratio is $1 0^{1.20}$ .

Full answer
(a) Ratio $= 12 \div 1.2 = 10$ , so $pH = 6.1 + 1.0 = 7.1$ , well below the normal range. (b) The ratio needed is $1 0^{1.20} \approx 15.8$ , so the CO₂ must fall to $12 \div 15.8 \approx 0.76$ mmol/dm³. (c) Breathing faster removes CO₂, the acid half of the buffer, from the blood. A smaller acid reservoir raises the base-to-acid ratio, and so raises the pH. Check: $6.1 + lo g_{10} (12 \div 0.757) \approx 7.30$ . ✓ This is only partial: doctors call it "compensation", and the kidneys must still restore the HCO₃⁻.

The bigger idea

The buffer equation began with blood. In 1908, the American physiologist Lawrence Henderson wrote an equation for the hydrogen ion concentration of the carbon dioxide–hydrogencarbonate buffer. The next year, Sørensen introduced the pH scale, which you met in the last lecture. A few years later, the Danish physician Karl Hasselbalch rewrote Henderson's equation in terms of pH, and the result carries both their names. (Sources: L. J. Henderson, American Journal of Physiology, 1908; OpenStax Chemistry 2e, section 14.6; Wikipedia, "Henderson–Hasselbalch equation".)

So the equation in this lecture is not just a laboratory formula. Doctors use it every day to read blood test results: they measure the CO₂ and the HCO₃⁻, and the ratio tells them whether the lungs or the kidneys are the problem. The same idea runs through chemistry. Wherever a pH must stay steady, in a cell, a shampoo or a chemical reaction, there is a weak acid and its conjugate base doing the work. The next lecture uses buffers again: the flat middle part of a titration curve is a buffer forming in the flask.

Key concepts in this lesson

A buffer solution resists changes in pH because it contains a weak acid to react with added alkali and its conjugate base to react with added acid.
The acid dissociation constant $K_{a} = [H^{+}] [A^{-}] \div [HA]$ is the equilibrium constant for a weak acid, and $p K_{a} = - lo g_{10} K_{a}$ .
The pH of a buffer is $p K_{a} + lo g_{10} ([A^{-}] \div [HA])$ , so it depends on the ratio of base to acid, and equals the p $K_{a}$ when they are equal.
Adding a little acid or alkali changes the ratio only a little, which is why the pH changes so little.
Buffer capacity depends on the actual amounts of acid and base, so a diluted buffer has the same pH but absorbs less.
Blood is buffered by dissolved carbon dioxide and hydrogencarbonate at a ratio of about 1 to 20, and breathing out carbon dioxide keeps that buffer working.