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Buffer calculations

The one relationship you need

The booklet gives you the $K_a$ expression. Rearrange it and every buffer calculation follows:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \quad \Longrightarrow \quad [\text{H}_3\text{O}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]}$

The acid is on top and the conjugate base underneath. Inverting this is the commonest error, and it puts the pH on the wrong side of the $\text{p}K_a$ .

Moles or concentrations — both work

The acid and its conjugate base are in the same solution, so they share the same volume.
In the ratio $[\text{HA}]/[\text{A}^-]$ that volume cancels, so you may use moles directly.
This saves a step and avoids errors, especially when a solid is added to a solution.

When the ratio is 1

If $[\text{HA}] = [\text{A}^-]$ , the ratio is 1 and $[\text{H}_3\text{O}^+] = K_a$ , so

$\text{pH} = \text{p}K_a$

This is the buffer's most effective composition, because it has equal capacity against added acid and added base.
It is also how the $\text{p}K_a$ is read off a titration curve at the half-equivalence point.

Choosing a buffer

To make a buffer of a given pH, choose a weak acid whose $\text{p}K_a$ is close to that pH, then adjust the ratio.
A buffer works well within about one pH unit of its $\text{p}K_a$ — beyond that the ratio becomes lopsided and one component runs low.

The three question types

Find the pH from given amounts of acid and base. Substitute the ratio and take the log.
Find the ratio or an amount needed for a target pH. Work backwards: find $[\text{H}_3\text{O}^+] = 10^{-\text{pH}}$ , then rearrange for the unknown.
Find the new pH after adding acid or base. Adjust the moles of each component first, then substitute.

Adjusting the amounts

When a strong acid or base is added, work out the new amounts before touching the $K_a$ expression:

Added	Reacts with	$\text{HA}$	$\text{A}^-$
$\text{H}_3\text{O}^+$	the conjugate base	increases	decreases
$\text{OH}^-$	the weak acid	decreases	increases

Both change by the same number of moles as the amount added.

Tips and Tricks

Acid on top. $[\text{H}_3\text{O}^+] = K_a \times [\text{HA}]/[\text{A}^-]$ . A quick sanity check: more acid should mean a lower pH, so if increasing $[\text{HA}]$ raises your pH, you have inverted the fraction.

Use moles, not concentrations, when the two components are in the same solution. The volume cancels, and this is explicitly allowed.

When a solid is added to a solution, its volume change is usually stated to be negligible — read the question, and say you are assuming it.

For an "adding acid" question, write out the before / change / after amounts in a small table before you substitute. Doing the arithmetic in your head is where the sign errors come from.

Check your answer against the $\text{p}K_a$ . If the buffer has more acid than base the pH must be below the $\text{p}K_a$ ; if more base than acid, above it.

Worked ExampleMaking a buffer of a specified pH

A chemist needs a buffer of pH $4.50$ . They start with $500$ mL of $0.200$ mol L⁻¹ ethanoic acid and add solid sodium ethanoate.

Calculate the mass of sodium ethanoate required. Assume the volume does not change when the solid dissolves.

$K_a(\text{CH}_3\text{COOH}) = 1.74 \times 10^{-5}$ , $M(\text{CH}_3\text{COONa}) = 82.0$ g mol⁻¹

Step 1 — Find the hydronium concentration needed

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-4.50} = 3.162 \times 10^{-5} \text{ mol L}^{-1}$

Step 2 — Rearrange the buffer relationship for $[\text{A}^-]$

$[\text{H}_3\text{O}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} \quad \Longrightarrow \quad [\text{A}^-] = \frac{K_a \times [\text{HA}]}{[\text{H}_3\text{O}^+]}$

Step 3 — Substitute

$[\text{A}^-] = \frac{1.74 \times 10^{-5} \times 0.200}{3.162 \times 10^{-5}} = \frac{3.48 \times 10^{-6}}{3.162 \times 10^{-5}}$

$[\text{CH}_3\text{COO}^-] = 0.1100 \text{ mol L}^{-1}$

Step 4 — Convert to moles, then to mass

The volume is $500$ mL $= 0.500$ L:

$n = cV = 0.1100 \times 0.500 = 0.05502 \text{ mol}$

$m = nM = 0.05502 \times 82.0 = 4.512 \text{ g}$

Step 5 — Check the answer is sensible

$\text{p}K_a = -\log(1.74 \times 10^{-5}) = 4.76$

The target pH of 4.50 is below the $\text{p}K_a$ , so the buffer must contain more acid than conjugate base.

Our answer gives $[\text{HA}] = 0.200$ and $[\text{A}^-] = 0.110$ — the acid is in excess, by a ratio of about 1.8 : 1 ✓

Buffer calculations

Covers AS91392

The one relationship you need

The booklet gives you the $K_a$ expression. Rearrange it and every buffer calculation follows:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \quad \Longrightarrow \quad [\text{H}_3\text{O}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]}$

The acid is on top and the conjugate base underneath. Inverting this is the commonest error, and it puts the pH on the wrong side of the $\text{p}K_a$ .

Moles or concentrations — both work

The acid and its conjugate base are in the same solution, so they share the same volume.
In the ratio $[\text{HA}]/[\text{A}^-]$ that volume cancels, so you may use moles directly.
This saves a step and avoids errors, especially when a solid is added to a solution.

When the ratio is 1

If $[\text{HA}] = [\text{A}^-]$ , the ratio is 1 and $[\text{H}_3\text{O}^+] = K_a$ , so

$\text{pH} = \text{p}K_a$

This is the buffer's most effective composition, because it has equal capacity against added acid and added base.
It is also how the $\text{p}K_a$ is read off a titration curve at the half-equivalence point.

Choosing a buffer

To make a buffer of a given pH, choose a weak acid whose $\text{p}K_a$ is close to that pH, then adjust the ratio.
A buffer works well within about one pH unit of its $\text{p}K_a$ — beyond that the ratio becomes lopsided and one component runs low.

The three question types

Find the pH from given amounts of acid and base. Substitute the ratio and take the log.
Find the ratio or an amount needed for a target pH. Work backwards: find $[\text{H}_3\text{O}^+] = 10^{-\text{pH}}$ , then rearrange for the unknown.
Find the new pH after adding acid or base. Adjust the moles of each component first, then substitute.

Adjusting the amounts

When a strong acid or base is added, work out the new amounts before touching the $K_a$ expression:

Added	Reacts with	$\text{HA}$	$\text{A}^-$
$\text{H}_3\text{O}^+$	the conjugate base	increases	decreases
$\text{OH}^-$	the weak acid	decreases	increases

Both change by the same number of moles as the amount added.

Tips and Tricks

Use moles, not concentrations, when the two components are in the same solution. The volume cancels, and this is explicitly allowed.

When a solid is added to a solution, its volume change is usually stated to be negligible — read the question, and say you are assuming it.

For an "adding acid" question, write out the before / change / after amounts in a small table before you substitute. Doing the arithmetic in your head is where the sign errors come from.

Check your answer against the $\text{p}K_a$ . If the buffer has more acid than base the pH must be below the $\text{p}K_a$ ; if more base than acid, above it.

Worked ExampleMaking a buffer of a specified pH

A chemist needs a buffer of pH $4.50$ . They start with $500$ mL of $0.200$ mol L⁻¹ ethanoic acid and add solid sodium ethanoate.

Calculate the mass of sodium ethanoate required. Assume the volume does not change when the solid dissolves.

$K_a(\text{CH}_3\text{COOH}) = 1.74 \times 10^{-5}$ , $M(\text{CH}_3\text{COONa}) = 82.0$ g mol⁻¹

Step 1 — Find the hydronium concentration needed

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-4.50} = 3.162 \times 10^{-5} \text{ mol L}^{-1}$

Step 2 — Rearrange the buffer relationship for $[\text{A}^-]$

$[\text{H}_3\text{O}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} \quad \Longrightarrow \quad [\text{A}^-] = \frac{K_a \times [\text{HA}]}{[\text{H}_3\text{O}^+]}$

Step 3 — Substitute

$[\text{A}^-] = \frac{1.74 \times 10^{-5} \times 0.200}{3.162 \times 10^{-5}} = \frac{3.48 \times 10^{-6}}{3.162 \times 10^{-5}}$

$[\text{CH}_3\text{COO}^-] = 0.1100 \text{ mol L}^{-1}$

Step 4 — Convert to moles, then to mass

The volume is $500$ mL $= 0.500$ L:

$n = cV = 0.1100 \times 0.500 = 0.05502 \text{ mol}$

$m = nM = 0.05502 \times 82.0 = 4.512 \text{ g}$

Step 5 — Check the answer is sensible

$\text{p}K_a = -\log(1.74 \times 10^{-5}) = 4.76$

The target pH of 4.50 is below the $\text{p}K_a$ , so the buffer must contain more acid than conjugate base.

Our answer gives $[\text{HA}] = 0.200$ and $[\text{A}^-] = 0.110$ — the acid is in excess, by a ratio of about 1.8 : 1 ✓

Buffer calculations

The one relationship you need

Moles or concentrations — both work

When the ratio is 1

Choosing a buffer

The three question types

Adjusting the amounts

Step 1 — Find the hydronium concentration needed

Step 2 — Rearrange the buffer relationship for [A−][\text{A}^-][A−]

Step 3 — Substitute

Step 4 — Convert to moles, then to mass

Step 5 — Check the answer is sensible

Buffer calculations

The one relationship you need

Moles or concentrations — both work

When the ratio is 1

Choosing a buffer

The three question types

Adjusting the amounts

Step 1 — Find the hydronium concentration needed

Step 2 — Rearrange the buffer relationship for [A−][\text{A}^-][A−]

Step 3 — Substitute

Step 4 — Convert to moles, then to mass

Step 5 — Check the answer is sensible

Step 2 — Rearrange the buffer relationship for $[\text{A}^-]$

Step 2 — Rearrange the buffer relationship for $[\text{A}^-]$