Buffers

What you'll learn

What a buffer solution is, and why it resists pH change.
How weak acid/conjugate base pairs such as ethanoic acid/ethanoate work.
How to calculate buffer pH using $K_a$ and mole ratios.
How buffers appear in weak acid–strong base titration curves.

Prerequisites: pH and weak acid equilibria

Before buffers, you need the idea of a weak acid. A weak acid only partially dissociates in water, so an equilibrium is set up:

HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA(aq)} \rightleftharpoons \mathrm{H}^{+}\mathrm{(aq)} + \mathrm{A}^{-}\mathrm{(aq)}

For this equilibrium:

Ka=[H+][A−][HA]K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}

The pH is linked to hydrogen ion concentration by:

pH=−log⁡[H+]\mathrm{pH} = -\log[\mathrm{H}^{+}]

Definition

Conjugate acid-base pair

A conjugate acid-base pair differs by one proton, $H+\mathrm{H}^{+}$ . In $HA/A−\mathrm{HA}/\mathrm{A}^{-}$ , $HA\mathrm{HA}$ is the acid and $A−\mathrm{A}^{-}$ is its conjugate base.

Definition

pKa

$pKa=−log⁡Ka\mathrm{p}K_a = -\log K_a$ . A lower $pKa\mathrm{p}K_a$ means a stronger acid. Buffers work best when the desired pH is close to the acid’s $pKa\mathrm{p}K_a$ .

What is a buffer?

A buffer solution resists changes in pH when small amounts of acid or alkali are added.

Definition

Buffer solution

A buffer solution contains a mixture that can remove added $H+\mathrm{H}^{+}$ and added $OH−\mathrm{OH}^{-}$ , so the hydrogen ion concentration changes only slightly.

The most common A-Level buffer is an acidic buffer, made from:

a weak acid, $HA\mathrm{HA}$
a salt containing its conjugate base, $A−\mathrm{A}^{-}$

For example, an ethanoic acid/ethanoate buffer contains:

ethanoic acid, $CH3COOH\mathrm{CH_3COOH}$
sodium ethanoate, $CH3COONa\mathrm{CH_3COONa}$ , which supplies $CH3COO−\mathrm{CH_3COO}^{-}$ ions

The sodium ions are spectator ions; they do not control the pH.

Common Mistake

A weak acid alone is not a good buffer

A weak acid solution contains a little conjugate base, but not usually enough to resist added acid effectively. A proper acidic buffer needs significant amounts of both $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ .

How an acidic buffer works

In an acidic buffer, the weak acid and conjugate base act as two “reservoirs”.

If acid is added, the conjugate base removes the added $H+\mathrm{H}^{+}$ :

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

If alkali is added, the weak acid removes the added $OH−\mathrm{OH}^{-}$ :

HA(aq)+OH−(aq)→A−(aq)+H2O(l)\mathrm{HA(aq)} + \mathrm{OH}^{-}\mathrm{(aq)} \to \mathrm{A}^{-}\mathrm{(aq)} + \mathrm{H_2O(l)}

The key point is that the ratio of $HA\mathrm{HA}$ to $A−\mathrm{A}^{-}$ changes only slightly, so $[H+][\mathrm{H}^{+}]$ and pH change only slightly.

Diagram showing how an acidic buffer removes added hydrogen ions and hydroxide ions

Key Idea

The ratio controls the pH

For an acidic buffer, the pH mainly depends on the ratio of weak acid to conjugate base, not just the total amount present.

The buffer pH equation

Start with the weak acid equilibrium:

Ka=[H+][A−][HA]K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}

Rearrange for hydrogen ion concentration:

[H+]=Ka[HA][A−][\mathrm{H}^{+}] = K_a \frac{[\mathrm{HA}]}{[\mathrm{A}^{-}]}

Then calculate pH using:

pH=−log⁡[H+]\mathrm{pH} = -\log[\mathrm{H}^{+}]

You may also see the same idea written as:

pH=pKa+log⁡([A−][HA])\mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)

This is the Henderson–Hasselbalch form. Edexcel questions often work perfectly using the $K_a$ rearrangement.

Tip

Moles or concentrations?

If $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ are in the same final solution, their final volume is the same, so $[HA][A−]=n(HA)n(A−)\frac{[\mathrm{HA}]}{[\mathrm{A}^{-}]} = \frac{n(\mathrm{HA})}{n(\mathrm{A}^{-})}$ . This is why buffer calculations often use moles directly.

Example

Calculating the pH of an acidic buffer

A buffer contains 0.100 mol dm⁻³ ethanoic acid and 0.0800 mol dm⁻³ sodium ethanoate. For ethanoic acid, $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ . Calculate the pH.

Identify the weak acid and conjugate base: $HA\mathrm{HA}$ is $CH3COOH\mathrm{CH_3COOH}$ and $A−\mathrm{A}^{-}$ is $CH3COO−\mathrm{CH_3COO}^{-}$ .
Substitute into the buffer equation: $dm−3[\mathrm{H}^{+}] = 1.74 \times 10^{-5} \times \frac{0.100}{0.0800} = 2.18 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .
Convert to pH: $pH=−log⁡(2.18×10−5)=4.66\mathrm{pH} = -\log(2.18 \times 10^{-5}) = 4.66$ .

Common Mistake

Inverting the ratio

In $[H+]=Ka[HA][A−][\mathrm{H}^{+}] = K_a \frac{[\mathrm{HA}]}{[\mathrm{A}^{-}]}$ , the acid is on top. In $pH=pKa+log⁡[A−][HA]\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}$ , the base is on top.

Preparing buffers by partial neutralisation

A buffer can be made by adding some strong alkali to a weak acid. The alkali converts part of the weak acid into its conjugate base:

HA(aq)+OH−(aq)→A−(aq)+H2O(l)\mathrm{HA(aq)} + \mathrm{OH}^{-}\mathrm{(aq)} \to \mathrm{A}^{-}\mathrm{(aq)} + \mathrm{H_2O(l)}

After the reaction, if both $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ remain, you have a buffer.

Example

Preparing a buffer by partial neutralisation

25.0 cm³ of 0.200 mol dm⁻³ ethanoic acid is mixed with 10.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide. Calculate the pH, using $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

Calculate initial moles using $n = c V$ , with volume in dm³: $moln(\mathrm{CH_3COOH}) = 0.200 \times 0.0250 = 0.00500\ \mathrm{mol}$ and $moln(\mathrm{OH}^{-}) = 0.100 \times 0.0100 = 0.00100\ \mathrm{mol}$ .
Use the neutralisation stoichiometry: $OH−\mathrm{OH}^{-}$ is the limiting reagent, so it uses up 0.00100 mol of acid and forms 0.00100 mol of ethanoate. Final moles are $moln(\mathrm{HA}) = 0.00400\ \mathrm{mol}$ and $moln(\mathrm{A}^{-}) = 0.00100\ \mathrm{mol}$ .
Use the mole ratio in the buffer equation: $dm−3[\mathrm{H}^{+}] = 1.74 \times 10^{-5} \times \frac{0.00400}{0.00100} = 6.96 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ , so $pH=4.16\mathrm{pH} = 4.16$ .

Adding acid or alkali to a buffer

When strong acid or strong alkali is added to a buffer, deal with the stoichiometric reaction first, then use the buffer equation.

Added acid decreases $A−\mathrm{A}^{-}$ and increases $HA\mathrm{HA}$ .
Added alkali decreases $HA\mathrm{HA}$ and increases $A−\mathrm{A}^{-}$ .

Example

Adding strong acid to a buffer

1.000 dm³ of a buffer contains 0.100 mol of $HA\mathrm{HA}$ and 0.100 mol of $A−\mathrm{A}^{-}$ . Then 0.0100 mol of HCl is added. Use $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

HCl is a strong acid, so it supplies 0.0100 mol of $H+\mathrm{H}^{+}$ , which reacts with $A−\mathrm{A}^{-}$ : $A−+H+→HA\mathrm{A}^{-} + \mathrm{H}^{+} \to \mathrm{HA}$ .
Update the moles: $moln(\mathrm{A}^{-}) = 0.100 - 0.0100 = 0.0900\ \mathrm{mol}$ and $moln(\mathrm{HA}) = 0.100 + 0.0100 = 0.110\ \mathrm{mol}$ .
Calculate the new hydrogen ion concentration: $dm−3[\mathrm{H}^{+}] = 1.74 \times 10^{-5} \times \frac{0.110}{0.0900} = 2.13 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .
Convert to pH: $pH=−log⁡(2.13×10−5)=4.67\mathrm{pH} = -\log(2.13 \times 10^{-5}) = 4.67$ . The pH falls only slightly; the original pH was about 4.76.

Common Mistake

When the buffer stops working

The buffer formula is not valid if the added strong acid uses up nearly all the $A−\mathrm{A}^{-}$ , or the added strong alkali uses up nearly all the $HA\mathrm{HA}$ . At that point, the solution is no longer a buffer.

Buffer capacity

Buffer capacity is how much acid or alkali a buffer can absorb before its pH changes significantly.

Definition

Buffer capacity

Buffer capacity increases when the total concentrations of $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ are larger. The pH depends mainly on their ratio; the capacity depends on how much of both is present.

A buffer works best when the concentrations of acid and conjugate base are similar. If $[HA]=[A−][\mathrm{HA}] = [\mathrm{A}^{-}]$ , then:

[H+]=Ka[\mathrm{H}^{+}] = K_a

So:

pH=pKa\mathrm{pH} = \mathrm{p}K_a

A useful rule of thumb is that a buffer is most effective within about one pH unit of its $pKa\mathrm{p}K_a$ .

Buffers in titration curves

During a weak acid–strong base titration, a buffer region forms before the equivalence point. For example, when sodium hydroxide is added to ethanoic acid, some ethanoic acid is converted into ethanoate ions. The mixture then contains both $CH3COOH\mathrm{CH_3COOH}$ and $CH3COO−\mathrm{CH_3COO}^{-}$ .

Titration curve for ethanoic acid with sodium hydroxide showing the buffer region, half-neutralisation point and equivalence point

At the half-neutralisation point, exactly half the original weak acid has been converted into conjugate base. Therefore $[HA]=[A−][\mathrm{HA}] = [\mathrm{A}^{-}]$ , so $pH=pKa\mathrm{pH} = \mathrm{p}K_a$ .

Example

Finding pKa from a titration curve

A weak acid is titrated with sodium hydroxide. The equivalence point is at 24.0 cm³ of NaOH added. The pH at 12.0 cm³ is 4.80. Find $K_a$ .

Recognise that 12.0 cm³ is half of 24.0 cm³, so this is the half-neutralisation point.
At half-neutralisation, the moles of $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ are equal, so $pH=pKa\mathrm{pH} = \mathrm{p}K_a$ .
Therefore $pKa=4.80\mathrm{p}K_a = 4.80$ and $dm−3K_a = 10^{-4.80} = 1.6 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

Alkaline buffers

An alkaline buffer contains a weak base and its conjugate acid. A common example is ammonia and ammonium chloride:

weak base: $NH3\mathrm{NH_3}$
conjugate acid: $NH4+\mathrm{NH_4}^{+}$

Added acid is removed by ammonia:

NH3(aq)+H+(aq)→NH4+(aq)\mathrm{NH_3(aq)} + \mathrm{H}^{+}\mathrm{(aq)} \to \mathrm{NH_4}^{+}\mathrm{(aq)}

Added alkali is removed by ammonium ions:

NH4+(aq)+OH−(aq)→NH3(aq)+H2O(l)\mathrm{NH_4}^{+}\mathrm{(aq)} + \mathrm{OH}^{-}\mathrm{(aq)} \to \mathrm{NH_3(aq)} + \mathrm{H_2O(l)}

If you are given the $K_a$ of the conjugate acid, $BH+\mathrm{BH}^{+}$ , you can treat it like an acid buffer:

pH=pKa(BH+)+log⁡([B][BH+])\mathrm{pH} = \mathrm{p}K_a(\mathrm{BH}^{+}) + \log\left(\frac{[\mathrm{B}]}{[\mathrm{BH}^{+}]}\right)

Exam technique

In the exam

Identify the weak acid/conjugate base pair, then write the relevant equilibrium and $K_a$ expression.
If any strong acid or alkali reacts, calculate the new moles first; do not put initial amounts straight into the buffer equation.
Sense-check your answer: adding acid should lower pH slightly, adding alkali should raise pH slightly, and the pH should usually be close to $pKa\mathrm{p}K_a$ .

Self review

Check yourself

Why does an ethanoic acid/ethanoate buffer remove added $H+\mathrm{H}^{+}$ ions?
If a buffer has equal moles of $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ , what is the relationship between pH and $pKa\mathrm{p}K_a$ ?
After adding a small amount of sodium hydroxide to an acidic buffer, which component increases and which decreases?

What you'll learn

What a buffer solution is, and why it resists pH change.
How weak acid/conjugate base pairs such as ethanoic acid/ethanoate work.
How to calculate buffer pH using $K_a$ and mole ratios.
How buffers appear in weak acid–strong base titration curves.

Prerequisites: pH and weak acid equilibria

Before buffers, you need the idea of a weak acid. A weak acid only partially dissociates in water, so an equilibrium is set up:

HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA(aq)} \rightleftharpoons \mathrm{H}^{+}\mathrm{(aq)} + \mathrm{A}^{-}\mathrm{(aq)}

For this equilibrium:

Ka=[H+][A−][HA]K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}

The pH is linked to hydrogen ion concentration by:

pH=−log⁡[H+]\mathrm{pH} = -\log[\mathrm{H}^{+}]

Definition

Conjugate acid-base pair

A conjugate acid-base pair differs by one proton, $H+\mathrm{H}^{+}$ . In $HA/A−\mathrm{HA}/\mathrm{A}^{-}$ , $HA\mathrm{HA}$ is the acid and $A−\mathrm{A}^{-}$ is its conjugate base.

Definition

pKa

$pKa=−log⁡Ka\mathrm{p}K_a = -\log K_a$ . A lower $pKa\mathrm{p}K_a$ means a stronger acid. Buffers work best when the desired pH is close to the acid’s $pKa\mathrm{p}K_a$ .

What is a buffer?

A buffer solution resists changes in pH when small amounts of acid or alkali are added.

Definition

Buffer solution

A buffer solution contains a mixture that can remove added $H+\mathrm{H}^{+}$ and added $OH−\mathrm{OH}^{-}$ , so the hydrogen ion concentration changes only slightly.

The most common A-Level buffer is an acidic buffer, made from:

a weak acid, $HA\mathrm{HA}$
a salt containing its conjugate base, $A−\mathrm{A}^{-}$

For example, an ethanoic acid/ethanoate buffer contains:

ethanoic acid, $CH3COOH\mathrm{CH_3COOH}$
sodium ethanoate, $CH3COONa\mathrm{CH_3COONa}$ , which supplies $CH3COO−\mathrm{CH_3COO}^{-}$ ions

The sodium ions are spectator ions; they do not control the pH.

Common Mistake

A weak acid alone is not a good buffer

How an acidic buffer works

In an acidic buffer, the weak acid and conjugate base act as two “reservoirs”.

If acid is added, the conjugate base removes the added $H+\mathrm{H}^{+}$ :

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

If alkali is added, the weak acid removes the added $OH−\mathrm{OH}^{-}$ :

HA(aq)+OH−(aq)→A−(aq)+H2O(l)\mathrm{HA(aq)} + \mathrm{OH}^{-}\mathrm{(aq)} \to \mathrm{A}^{-}\mathrm{(aq)} + \mathrm{H_2O(l)}

The key point is that the ratio of $HA\mathrm{HA}$ to $A−\mathrm{A}^{-}$ changes only slightly, so $[H+][\mathrm{H}^{+}]$ and pH change only slightly.

Diagram showing how an acidic buffer removes added hydrogen ions and hydroxide ions

Key Idea

The ratio controls the pH

For an acidic buffer, the pH mainly depends on the ratio of weak acid to conjugate base, not just the total amount present.

The buffer pH equation

Start with the weak acid equilibrium:

Ka=[H+][A−][HA]K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}

Rearrange for hydrogen ion concentration:

[H+]=Ka[HA][A−][\mathrm{H}^{+}] = K_a \frac{[\mathrm{HA}]}{[\mathrm{A}^{-}]}

Then calculate pH using:

pH=−log⁡[H+]\mathrm{pH} = -\log[\mathrm{H}^{+}]

You may also see the same idea written as:

pH=pKa+log⁡([A−][HA])\mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)

This is the Henderson–Hasselbalch form. Edexcel questions often work perfectly using the $K_a$ rearrangement.

Tip

Moles or concentrations?

Example

Calculating the pH of an acidic buffer

A buffer contains 0.100 mol dm⁻³ ethanoic acid and 0.0800 mol dm⁻³ sodium ethanoate. For ethanoic acid, $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ . Calculate the pH.

Identify the weak acid and conjugate base: $HA\mathrm{HA}$ is $CH3COOH\mathrm{CH_3COOH}$ and $A−\mathrm{A}^{-}$ is $CH3COO−\mathrm{CH_3COO}^{-}$ .
Substitute into the buffer equation: $dm−3[\mathrm{H}^{+}] = 1.74 \times 10^{-5} \times \frac{0.100}{0.0800} = 2.18 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .
Convert to pH: $pH=−log⁡(2.18×10−5)=4.66\mathrm{pH} = -\log(2.18 \times 10^{-5}) = 4.66$ .

Common Mistake

Inverting the ratio

Preparing buffers by partial neutralisation

A buffer can be made by adding some strong alkali to a weak acid. The alkali converts part of the weak acid into its conjugate base:

HA(aq)+OH−(aq)→A−(aq)+H2O(l)\mathrm{HA(aq)} + \mathrm{OH}^{-}\mathrm{(aq)} \to \mathrm{A}^{-}\mathrm{(aq)} + \mathrm{H_2O(l)}

After the reaction, if both $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ remain, you have a buffer.

Example

Preparing a buffer by partial neutralisation

25.0 cm³ of 0.200 mol dm⁻³ ethanoic acid is mixed with 10.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide. Calculate the pH, using $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

Calculate initial moles using $n = c V$ , with volume in dm³: $moln(\mathrm{CH_3COOH}) = 0.200 \times 0.0250 = 0.00500\ \mathrm{mol}$ and $moln(\mathrm{OH}^{-}) = 0.100 \times 0.0100 = 0.00100\ \mathrm{mol}$ .
Use the neutralisation stoichiometry: $OH−\mathrm{OH}^{-}$ is the limiting reagent, so it uses up 0.00100 mol of acid and forms 0.00100 mol of ethanoate. Final moles are $moln(\mathrm{HA}) = 0.00400\ \mathrm{mol}$ and $moln(\mathrm{A}^{-}) = 0.00100\ \mathrm{mol}$ .
Use the mole ratio in the buffer equation: $dm−3[\mathrm{H}^{+}] = 1.74 \times 10^{-5} \times \frac{0.00400}{0.00100} = 6.96 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ , so $pH=4.16\mathrm{pH} = 4.16$ .

Adding acid or alkali to a buffer

When strong acid or strong alkali is added to a buffer, deal with the stoichiometric reaction first, then use the buffer equation.

Added acid decreases $A−\mathrm{A}^{-}$ and increases $HA\mathrm{HA}$ .
Added alkali decreases $HA\mathrm{HA}$ and increases $A−\mathrm{A}^{-}$ .

Example

Adding strong acid to a buffer

1.000 dm³ of a buffer contains 0.100 mol of $HA\mathrm{HA}$ and 0.100 mol of $A−\mathrm{A}^{-}$ . Then 0.0100 mol of HCl is added. Use $dm−3K_a = 1.74 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

HCl is a strong acid, so it supplies 0.0100 mol of $H+\mathrm{H}^{+}$ , which reacts with $A−\mathrm{A}^{-}$ : $A−+H+→HA\mathrm{A}^{-} + \mathrm{H}^{+} \to \mathrm{HA}$ .
Update the moles: $moln(\mathrm{A}^{-}) = 0.100 - 0.0100 = 0.0900\ \mathrm{mol}$ and $moln(\mathrm{HA}) = 0.100 + 0.0100 = 0.110\ \mathrm{mol}$ .
Calculate the new hydrogen ion concentration: $dm−3[\mathrm{H}^{+}] = 1.74 \times 10^{-5} \times \frac{0.110}{0.0900} = 2.13 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .
Convert to pH: $pH=−log⁡(2.13×10−5)=4.67\mathrm{pH} = -\log(2.13 \times 10^{-5}) = 4.67$ . The pH falls only slightly; the original pH was about 4.76.

Common Mistake

When the buffer stops working

Buffer capacity

Buffer capacity is how much acid or alkali a buffer can absorb before its pH changes significantly.

Definition

Buffer capacity

A buffer works best when the concentrations of acid and conjugate base are similar. If $[HA]=[A−][\mathrm{HA}] = [\mathrm{A}^{-}]$ , then:

[H+]=Ka[\mathrm{H}^{+}] = K_a

So:

pH=pKa\mathrm{pH} = \mathrm{p}K_a

A useful rule of thumb is that a buffer is most effective within about one pH unit of its $pKa\mathrm{p}K_a$ .

Buffers in titration curves

Titration curve for ethanoic acid with sodium hydroxide showing the buffer region, half-neutralisation point and equivalence point

Example

Finding pKa from a titration curve

A weak acid is titrated with sodium hydroxide. The equivalence point is at 24.0 cm³ of NaOH added. The pH at 12.0 cm³ is 4.80. Find $K_a$ .

Recognise that 12.0 cm³ is half of 24.0 cm³, so this is the half-neutralisation point.
At half-neutralisation, the moles of $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ are equal, so $pH=pKa\mathrm{pH} = \mathrm{p}K_a$ .
Therefore $pKa=4.80\mathrm{p}K_a = 4.80$ and $dm−3K_a = 10^{-4.80} = 1.6 \times 10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

Alkaline buffers

An alkaline buffer contains a weak base and its conjugate acid. A common example is ammonia and ammonium chloride:

weak base: $NH3\mathrm{NH_3}$
conjugate acid: $NH4+\mathrm{NH_4}^{+}$

Added acid is removed by ammonia:

NH3(aq)+H+(aq)→NH4+(aq)\mathrm{NH_3(aq)} + \mathrm{H}^{+}\mathrm{(aq)} \to \mathrm{NH_4}^{+}\mathrm{(aq)}

Added alkali is removed by ammonium ions:

NH4+(aq)+OH−(aq)→NH3(aq)+H2O(l)\mathrm{NH_4}^{+}\mathrm{(aq)} + \mathrm{OH}^{-}\mathrm{(aq)} \to \mathrm{NH_3(aq)} + \mathrm{H_2O(l)}

If you are given the $K_a$ of the conjugate acid, $BH+\mathrm{BH}^{+}$ , you can treat it like an acid buffer:

pH=pKa(BH+)+log⁡([B][BH+])\mathrm{pH} = \mathrm{p}K_a(\mathrm{BH}^{+}) + \log\left(\frac{[\mathrm{B}]}{[\mathrm{BH}^{+}]}\right)

Exam technique

In the exam

Identify the weak acid/conjugate base pair, then write the relevant equilibrium and $K_a$ expression.
If any strong acid or alkali reacts, calculate the new moles first; do not put initial amounts straight into the buffer equation.
Sense-check your answer: adding acid should lower pH slightly, adding alkali should raise pH slightly, and the pH should usually be close to $pKa\mathrm{p}K_a$ .

Self review

Check yourself

Why does an ethanoic acid/ethanoate buffer remove added $H+\mathrm{H}^{+}$ ions?
If a buffer has equal moles of $HA\mathrm{HA}$ and $A−\mathrm{A}^{-}$ , what is the relationship between pH and $pKa\mathrm{p}K_a$ ?
After adding a small amount of sodium hydroxide to an acidic buffer, which component increases and which decreases?

What you'll learn

Prerequisites: pH and weak acid equilibria

What is a buffer?

How an acidic buffer works

The buffer pH equation

Preparing buffers by partial neutralisation

Adding acid or alkali to a buffer

Buffer capacity

Buffers in titration curves

Alkaline buffers

Review Buffers by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Buffers

What you'll learn

Prerequisites: pH and weak acid equilibria

What is a buffer?

How an acidic buffer works

The buffer pH equation

Preparing buffers by partial neutralisation

Adding acid or alkali to a buffer

Buffer capacity

Buffers in titration curves

Alkaline buffers

Review Buffers by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Buffers

What you'll learn

Prerequisites: pH and weak acid equilibria

What is a buffer?

How an acidic buffer works

The buffer pH equation

Preparing buffers by partial neutralisation

Adding acid or alkali to a buffer

Buffer capacity

Buffers in titration curves

Alkaline buffers

Review Buffers by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Buffers Revision Guide

Buffers

What you'll learn

Prerequisites: pH and weak acid equilibria

What is a buffer?

How an acidic buffer works

The buffer pH equation

Preparing buffers by partial neutralisation

Adding acid or alkali to a buffer

Buffer capacity

Buffers in titration curves

Alkaline buffers

Review Buffers by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Buffers Revision Guide