Topic 14 · page 14 of 18

pH of a buffer

The pH of a buffer depends on Ka and on the ratio of the weak acid to its conjugate base. Rearrange the Ka expression, put in the amounts, and take the log.

15 min read14.19

Read first: What a buffer is and how it works, pH of a weak acid

From Ka to the buffer equation 14.19

In a buffer both HA and A⁻ are present in large amounts, and the Ka expression still holds. Make [H⁺] the subject:

K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \qquad\Rightarrow\qquad [\ce{H+}] = K_\text{a} \times \frac{[\ce{HA}]}{[\ce{A-}]}

So [H⁺] depends on $K_\text{a}$ and on the ratio [HA] ÷ [A⁻], not on their sizes. Two assumptions let you put in the amounts you started with:

[HA] at equilibrium = the concentration of acid put in (the acid is only slightly dissociated, and the extra A⁻ from the salt holds it back even more).
[A⁻] at equilibrium = the concentration of salt put in (the salt is fully dissociated).

MethodpH of a buffer

Work out the amounts (moles) of the acid and of its conjugate base after any reaction.
Because both are in the same volume, the ratio of moles equals the ratio of concentrations.
[H⁺] = $K_\text{a}$ × (amount of acid ÷ amount of conjugate base).
pH = −lg [H⁺].

Worked examples 14.19

Worked exampleAcid and salt mixed

Calculate the pH of a buffer made from 0.200 mol dm⁻³ ethanoic acid and 0.100 mol dm⁻³ sodium ethanoate. $K_\text{a}$ = 1.74 × 10⁻⁵ mol dm⁻³.

[H⁺] = 1.74 × 10⁻⁵ × (0.200 ÷ 0.100) = 3.48 × 10⁻⁵ mol dm⁻³
pH = −lg (3.48 × 10⁻⁵) = 4.46

There is more acid than salt, so the pH is below the $\text{p}K_\text{a}$ of 4.76. Use this as a check.

Worked exampleMade by part-neutralising

50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid is mixed with 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide. Calculate the pH.

Moles of acid = 0.200 × 0.0500 = 1.00 × 10⁻² mol. Moles of NaOH = 0.100 × 0.0250 = 2.50 × 10⁻³ mol.
NaOH + CH₃COOH → CH₃COONa + H₂O. The NaOH is used up, so moles of CH₃COO⁻ = 2.50 × 10⁻³ mol and moles of CH₃COOH left = 1.00 × 10⁻² − 2.50 × 10⁻³ = 7.50 × 10⁻³ mol.
[H⁺] = 1.74 × 10⁻⁵ × (7.50 × 10⁻³ ÷ 2.50 × 10⁻³) = 1.74 × 10⁻⁵ × 3.00 = 5.22 × 10⁻⁵ mol dm⁻³
pH = 4.28

Worked exampleA basic buffer

Calculate the pH of a buffer containing 0.200 mol dm⁻³ ammonia and 0.100 mol dm⁻³ ammonium chloride. The $\text{p}K_\text{a}$ of NH₄⁺ is 9.25.

The acid of the pair is NH₄⁺, so use its $K_\text{a}$ : $K_\text{a}$ = $10^{-9.25}$ = 5.62 × 10⁻¹⁰ mol dm⁻³.
[H⁺] = $K_\text{a}$ × [NH₄⁺] ÷ [NH₃] = 5.62 × 10⁻¹⁰ × (0.100 ÷ 0.200) = 2.81 × 10⁻¹⁰ mol dm⁻³
pH = 9.55

If you are given $K_\text{b}$ for the base, convert it first: $K_\text{a}$ of the conjugate acid = $K_\text{w}$ ÷ $K_\text{b}$ .

A buffer in action 14.19

This is the calculation that shows a buffer working. A buffer has 0.100 mol each of ethanoic acid and sodium ethanoate in 1.00 dm³. Its pH is $\text{p}K_\text{a}$ = 4.76.

Worked exampleAdding acid to the buffer

0.0100 mol of HCl is added to 1.00 dm³ of this buffer. Calculate the new pH. Ignore the change in volume.

The H⁺ reacts with CH₃COO⁻: moles of CH₃COO⁻ = 0.100 − 0.0100 = 0.0900 mol. Moles of CH₃COOH = 0.100 + 0.0100 = 0.110 mol.
[H⁺] = 1.74 × 10⁻⁵ × (0.110 ÷ 0.0900) = 2.13 × 10⁻⁵ mol dm⁻³
pH = 4.67

The same 0.0100 mol of HCl in 1.00 dm³ of pure water would give a pH of 2.00. The buffer moved from 4.76 to 4.67, a change of only 0.09 units.

The shortcut form 14.19

Take logs of the buffer equation. The result is the Henderson–Hasselbalch equation, which is a handy check:

\text{pH} = \text{p}K_\text{a} + \lg\frac{[\ce{A-}]}{[\ce{HA}]}

A graph of the pH of an ethanoic acid buffer against the logarithm of the ratio of ethanoate to ethanoic acid. It is a straight line of gradient 1 crossing pH 4.76 where the two are in equal amounts, and shading marks the region where the ratio lies between 1 to 10 and 10 to 1. — pH against lg ([A⁻] ÷ [HA]) for ethanoic acid. Equal amounts give pH = pKa, and a tenfold change in the ratio shifts the pH by 1.

Notice what the graph says.

When [A⁻] = [HA], the log is zero and pH = $\text{p}K_\text{a}$ .
More A⁻ than HA gives a pH above $\text{p}K_\text{a}$ . More HA gives a pH below.
A buffer works best when the ratio lies between 1 : 10 and 10 : 1, so the pH is within 1 unit of $\text{p}K_\text{a}$ . Choose a weak acid whose $\text{p}K_\text{a}$ is close to the pH you want.
Diluting a buffer changes both amounts equally, so the ratio and the pH do not change much. Its capacity to absorb acid or alkali does fall.

Check yourself

Calculate the pH of a buffer containing 0.150 mol dm⁻³ propanoic acid and 0.300 mol dm⁻³ sodium propanoate. $K_\text{a}$ of propanoic acid = 1.35 × 10⁻⁵ mol dm⁻³.
Calculate the pH of a buffer containing 0.250 mol dm⁻³ methanoic acid and 0.100 mol dm⁻³ sodium methanoate. $K_\text{a}$ of methanoic acid = 1.6 × 10⁻⁴ mol dm⁻³.
50.0 cm³ of 0.100 mol dm⁻³ ethanoic acid is mixed with 20.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide. Calculate the pH. $K_\text{a}$ = 1.74 × 10⁻⁵ mol dm⁻³.
Calculate the pH of a buffer containing 0.300 mol dm⁻³ ammonia and 0.200 mol dm⁻³ ammonium chloride. The $\text{p}K_\text{a}$ of NH₄⁺ is 9.25.
A buffer contains 0.100 mol each of ethanoic acid and sodium ethanoate in 1.00 dm³. Calculate its pH after 0.0200 mol of sodium hydroxide is added. Ignore the change in volume.
A buffer of 0.200 mol dm⁻³ ethanoic acid and 0.100 mol dm⁻³ sodium ethanoate is diluted ten times with water. What is the pH of the diluted buffer? Explain.

Answers cover these until you have tried the questions

[H⁺] = 1.35 × 10⁻⁵ × (0.150 ÷ 0.300) = 6.75 × 10⁻⁶ mol dm⁻³. pH = 5.17
[H⁺] = 1.6 × 10⁻⁴ × (0.250 ÷ 0.100) = 4.0 × 10⁻⁴ mol dm⁻³. pH = 3.40
Moles of acid = 0.100 × 0.0500 = 5.00 × 10⁻³ mol. Moles of NaOH = 0.100 × 0.0200 = 2.00 × 10⁻³ mol. Moles of CH₃COO⁻ formed = 2.00 × 10⁻³ mol; moles of CH₃COOH left = 3.00 × 10⁻³ mol. [H⁺] = 1.74 × 10⁻⁵ × (3.00 × 10⁻³ ÷ 2.00 × 10⁻³) = 2.61 × 10⁻⁵ mol dm⁻³. pH = 4.58
$K_\text{a}$ = $10^{-9.25}$ = 5.62 × 10⁻¹⁰. [H⁺] = 5.62 × 10⁻¹⁰ × (0.200 ÷ 0.300) = 3.75 × 10⁻¹⁰ mol dm⁻³. pH = 9.43
The OH⁻ reacts with CH₃COOH: moles of CH₃COOH = 0.100 − 0.0200 = 0.0800 mol; moles of CH₃COO⁻ = 0.100 + 0.0200 = 0.120 mol. [H⁺] = 1.74 × 10⁻⁵ × (0.0800 ÷ 0.120) = 1.16 × 10⁻⁵ mol dm⁻³. pH = 4.94
The pH is 4.46, the same as before (3.48 × 10⁻⁵ mol dm⁻³ of H⁺). Diluting reduces both [CH₃COOH] and [CH₃COO⁻] by the same factor, so their ratio, and [H⁺] = $K_\text{a}$ × ratio, does not change. (The buffer's capacity falls, though.)