Acids, bases and buffers

In a nutshell

This subtopic turns "acid" into numbers. You define acids and bases by proton transfer, measure acidity on the pH scale, and use K_a and K_w to calculate pH for strong acids, strong bases and weak acids.

It then explains buffers (solutions that resist pH change, including the one in your blood) and pH titration curves, which tell you which indicator to use for a titration.

Assumed knowledge: Acids; Chemical equilibrium; How far?.

Core content

Brønsted-Lowry acids and bases: proton donors and acceptors

A Brønsted-Lowry acid is a species that donates a proton.

A Brønsted-Lowry base is a species that accepts a proton.

The "proton" is an H⁺ ion: a hydrogen atom that has lost its only electron. So every acid-base reaction is an H⁺ transfer from the acid to the base.

Water can do either job:

• With HCl, water accepts a proton (it is the base): HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl^-(aq)
• With NH₃, water donates a proton (it is the acid): NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)

H₃O⁺ is the oxonium ion. In most equations it is simplified to H⁺(aq).

Brønsted-Lowry acids and bases: conjugate acid-base pairs

A conjugate acid-base pair is two species that differ by one proton (H⁺).

When an acid donates a proton, what is left is its conjugate base. When a base accepts a proton, it becomes its conjugate acid. So an acid-base equilibrium always contains two pairs:

How to label any equation:

1. Find the species on the left that loses an H⁺. That is acid 1; the same species minus H⁺ on the right is base 1.
2. The other species on the left gains that H⁺. That is base 2; the same species plus H⁺ on the right is acid 2.
3. Check each pair: the two formulae must differ by exactly one H and one positive charge.

The last row shows that one acid can act as a base when it meets a stronger acid: here H₂SO₄ donates a proton to HNO₃.

Brønsted-Lowry acids and bases: monobasic, dibasic and tribasic acids

The basicity of an acid is the number of protons (H⁺ ions) each molecule of the acid can donate.

Count only the acidic hydrogens. Ethanoic acid has four hydrogens, but only the one in the COOH group is donated, so CH₃COOH is monobasic.

Brønsted-Lowry acids and bases: the role of H⁺ in reactions of acids

In every reaction of an acid, the particle doing the work is H⁺(aq). The other ion from the acid is a spectator ion, so the ionic equations are the same whichever acid you use.

Insoluble solids such as CuO and CaCO₃ stay as whole formulae with (s) in the ionic equation; only dissolved ionic compounds are split into ions.

Brønsted-Lowry acids and bases: K_a and pK_a

A weak acid partially dissociates in water, so it sets up an equilibrium:

HA(aq) ⇌ H⁺(aq) + A^-(aq)

The equilibrium constant for this is the acid dissociation constant, K_a:

$K_a = \frac{[\text{H}^+(\text{aq})][\text{A}^-(\text{aq})]}{[\text{HA}(\text{aq})]}$

Units: (mol dm^-3 × mol dm^-3) ÷ mol dm^-3 = mol dm^-3.

K_a measures the extent of dissociation: the larger K_a, the stronger the acid.

K_a values are often tiny, so they are converted to pK_a:

$\text{p}K_a = -\log K_a \qquad K_a = 10^{-\text{p}K_a}$

A smaller pK_a means a larger K_a, so a stronger acid. pK_a has no units.

pH and [H⁺(aq)]: the pH expressions

pH is a convenient way to state [H⁺(aq)], which can range over many powers of ten.

$\text{pH} = -\log[\text{H}^+] \qquad [\text{H}^+] = 10^{-\text{pH}}$

Two consequences you must be able to use:

• Lower pH = higher [H⁺].
• A change of 1 pH unit is a 10-fold change in [H⁺]. pH 2 has ten times the [H⁺] of pH 3.

Give pH values to 2 decimal places unless the question says otherwise.

pH and [H⁺(aq)]: the ionic product of water, K_w

Water itself dissociates very slightly:

H₂O(l) ⇌ H⁺(aq) + OH^-(aq)

The concentration of water is so large that it is effectively constant, so it is folded into the constant. This gives the ionic product of water:

$K_w = [\text{H}^+(\text{aq})][\text{OH}^-(\text{aq})]$

K_w = 1.00 × 10^-14 mol² dm^-6 at 298 K (given on the data sheet).

K_w applies to every aqueous solution, so if you know [OH^-] you can find [H⁺], and the other way round.

Neutral means [H⁺] = [OH^-], not "pH 7":

• In pure water, [H⁺]² = K_w, so [H⁺] = √K_w.
• At 298 K that gives [H⁺] = 1.00 × 10^-7 mol dm^-3 and pH = 7.00.
• The dissociation of water is endothermic, so at a higher temperature K_w is larger, [H⁺] is larger and the pH of pure water is below 7. It is still neutral, because [H⁺] still equals [OH^-].

pH and [H⁺(aq)]: strong monobasic acids and strong bases

A strong acid completely dissociates. For a strong monobasic acid, each HA gives one H⁺, so:

[H⁺] = [acid], then pH = -log[H⁺].

For 0.100 mol dm^-3 HCl: [H⁺] = 0.100 mol dm^-3, pH = 1.00.

A strong base also completely dissociates, giving OH^-, not H⁺. So you need K_w:

1. Find [OH^-] from the base concentration (NaOH, KOH: [OH^-] = [base]).
2. [H⁺] = K_w ÷ [OH^-]
3. pH = -log[H⁺]

For 0.100 mol dm^-3 NaOH: [H⁺] = 1.00 × 10^-14 ÷ 0.100 = 1.00 × 10^-13 mol dm^-3, pH = 13.00.

Watch for bases with two OH^- per formula: 0.0500 mol dm^-3 Ba(OH)₂ gives [OH^-] = 0.100 mol dm^-3.

pH and [H⁺(aq)]: weak monobasic acids and the two approximations

For a weak acid, [H⁺] is much less than [HA], so you cannot use [H⁺] = [acid]. Instead you use K_a with two approximations:

With both, the K_a expression simplifies to:

$K_a = \frac{[\text{H}^+]^2}{[\text{HA}]} \qquad \text{so} \qquad [\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$

This one equation, rearranged, finds pH, K_a or the acid concentration (Worked examples 4 and 5).

Still don't get it? · the weak acid approximations

Picture a stadium holding 10 000 fans. During half time, 13 of them leave to buy food.

How many fans are still in the stadium? Strictly 9 987, but if anyone asks, "about 10 000" is close enough. Taking 13 away from 10 000 barely changes it.

Now the chemistry, one step at a time:

1. The fans are HA molecules. The ones that leave are the molecules that dissociate.
2. Each molecule that leaves splits into exactly one H⁺ and one A^-. So the number of H⁺ ions equals the number of A^- ions. That is why [H⁺] = [A^-], and why the top of the K_a fraction becomes [H⁺]².
3. Water also makes a few H⁺ ions, but far fewer than the acid does, so we ignore them.
4. So few HA molecules leave that the concentration of HA at equilibrium is still about the starting concentration. That is why you can put the starting [HA] on the bottom of the fraction.
5. Now K_a = [H⁺]² ÷ [HA] has only one unknown. Rearrange and square root.

The exam wording: [H⁺] = [A^-] because the dissociation of water is negligible, and [HA]_eqm = [HA]_start because the dissociation of the weak acid is very small.

pH and [H⁺(aq)]: limitations of the approximations for "stronger" weak acids

The approximation [HA]_eqm = [HA]_{undissociated} only works when very little acid dissociates.

For a "stronger" weak acid (larger K_a), or a very dilute solution, a significant proportion of the HA dissociates. Then:

• [HA]_eqm is noticeably less than the starting concentration, so the approximation is no longer valid.
• Using the starting [HA] on the bottom of the K_a expression gives a [H⁺] that is too high, so the calculated pH is too low.

The numbers show it (0.0100 mol dm^-3 and 0.100 mol dm^-3 solutions; the "more exact" pH comes from a method you are not required to use):

For the stronger weak acid, the approximation predicts [H⁺] = 0.0100 mol dm^-3, which would mean 100% dissociation: the acid would not be weak at all.

The other approximation, [H⁺] = [A^-], also fails if the acid is so weak or so dilute that the H⁺ from water is no longer negligible (pH close to 7).

Buffers: what a buffer is and how to make one

A buffer solution is a system that minimises pH changes on addition of small amounts of an acid or a base.

A buffer does not keep the pH exactly constant, and it can be overwhelmed by large amounts of acid or base.

An acid buffer needs a weak acid, HA, and a large amount of its conjugate base, A^-. There are two ways to make one:

1. A weak acid + a salt of that weak acid, e.g. CH₃COOH + CH₃COONa. The salt completely dissociates, supplying lots of CH₃COO^-.
2. An excess of a weak acid + a strong alkali, e.g. excess CH₃COOH + NaOH. The NaOH reacts with some of the acid, converting it into CH₃COO^-, and the rest of the acid is left over: CH₃COOH(aq) + OH^-(aq) → CH₃COO^-(aq) + H₂O(l)

A weak acid on its own is not a buffer: it contains plenty of HA but only a tiny amount of A^-.

Adding 5 cm³ of alkali raises the pH of water by more than 4 units, but the buffer's pH rises by only 0.04.

Buffers: how the conjugate acid-base pair controls pH

Take the ethanoic acid / ethanoate buffer. It contains a large reservoir of both members of the conjugate pair, linked by this equilibrium:

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO^-(aq)

Adding a small amount of acid (H⁺):

1. The added H⁺ reacts with the conjugate base, CH₃COO^-: CH₃COO^-(aq) + H⁺(aq) → CH₃COOH(aq)
2. The equilibrium shifts to the left.
3. Most of the added H⁺ is removed, so the pH barely changes.

Adding a small amount of alkali (OH^-):

1. The added OH^- reacts with H⁺: H⁺(aq) + OH^-(aq) → H₂O(l)
2. CH₃COOH dissociates, so the equilibrium shifts to the right, replacing the H⁺ that was removed.
3. The pH barely changes.

(The OH^- can also be shown reacting with the weak acid directly: CH₃COOH(aq) + OH^-(aq) → CH₃COO^-(aq) + H₂O(l).)

The short version to remember: the conjugate base (A^-) removes added acid; the weak acid (HA) removes added alkali.

Still don't get it? · how a buffer works

Picture a fish tank with a clever water-level system. Next to the tank there are two big helpers:

• an empty overflow tank, ready to soak up any extra water poured in, and
• a full spare tank, ready to top the water up if some is scooped out.

Pour in a cup of water and the overflow tank takes it. Scoop out a cup and the spare tank refills it. The water level hardly moves either way. But if you tip in a whole bucket, the overflow tank fills up and the system fails.

Now the chemistry, one step at a time:

1. The water level is the pH, set by the amount of H⁺.
2. The "overflow tank" is the large amount of CH₃COO^-. It can grab extra H⁺ and turn it into CH₃COOH.
3. The "spare tank" is the large amount of CH₃COOH. It can release H⁺ to replace any that is removed.
4. Add acid: the extra H⁺ combines with CH₃COO^-. In equilibrium terms, the position moves to the left.
5. Add alkali: OH^- removes H⁺ to make water. More CH₃COOH dissociates to replace it. The position moves to the right.
6. A buffer works only for small additions, because each reservoir can run out. That is why the definition says "small amounts".

The exam wording: write the equilibrium, then say added H⁺ reacts with CH₃COO^- and the equilibrium shifts left, and added OH^- reacts with H⁺, CH₃COOH dissociates and the equilibrium shifts right.

Buffers: calculating the pH of a buffer

Use the K_a expression, but not the weak-acid shortcut. In a buffer [H⁺] is not equal to [A^-], because almost all the A^- comes from the salt (or from the partial neutralisation).

$[\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} \qquad \text{then} \qquad \text{pH} = -\log[\text{H}^+]$

• [HA] and [A^-] are the equilibrium concentrations of the conjugate pair, taken as the amounts you put in (or have left after a reaction).
• Both are in the same total volume, so you can use the ratio of their amounts in mol directly.
• When [HA] = [A^-], [H⁺] = K_a and pH = pK_a.
• Adding a little strong acid converts some A^- into HA; adding a little strong base converts some HA into A^-. Recalculate with the new amounts (Worked example 8).

Buffers: the carbonic acid-hydrogencarbonate buffer in blood

Blood plasma must stay between pH 7.35 and 7.45. The main buffer is the carbonic acid-hydrogencarbonate pair, H₂CO₃ / HCO₃^-:

H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃^-(aq)

Extra H₂CO₃ formed when acid is added is converted to H₂O and CO₂, and the CO₂ is breathed out through the lungs, so H₂CO₃ does not build up in the blood.

Neutralisation: pH titration curves

A pH titration curve plots pH against the volume of solution added. Its shape depends on whether the acid and base are strong or weak.

The equivalence point is where the acid and base have exactly reacted (amount of H⁺ = amount of OH^-). On a curve it is the middle of the vertical section.

What to read off (values for the 0.100 mol dm^-3 solutions plotted above):

To sketch a curve, get these right: the starting pH (set by the acid), the final pH (set by the base in excess), the volume at equivalence (from the amounts in mol) and the position and length of the vertical section.

For a weak acid with a strong base, the flatter region early on is a buffer region: the mixture contains both HA and the A^- being formed. At half the equivalence volume, half the acid has been converted to A^-, so [HA] = [A^-] and pH = pK_a.

Neutralisation: choosing an indicator and why it changes colour

An acid-base indicator is itself a weak acid, HIn, whose two forms have different colours:

HIn(aq) ⇌ H⁺(aq) + In^-(aq) (colour of HIn) ⇌ (colour of In^-)

• Add acid: extra H⁺ shifts the equilibrium to the left, so the solution shows the colour of HIn.
• Add alkali: OH^- removes H⁺, the equilibrium shifts to the right, so the solution shows the colour of In^-.
• The end point is where the indicator changes colour, when [HIn] = [In^-].

Choosing an indicator: the indicator's pH range must lie within the vertical section of the titration curve. Then one drop at the equivalence point changes the pH right through the indicator's range, and the end point matches the equivalence point.

(In the exam you will be given the pH range of any indicator you need. Use the range given.)

Reading the chart:

PAG11: pH measurement

PAG11 uses a pH meter (a pH probe connected to a meter or data logger), which gives a numerical pH reading, usually to 1 or 2 decimal places. A pH chart with universal indicator only gives a rough value.

Using a pH meter well:

1. Calibrate the meter first, using buffer solutions of known pH (for example pH 4.00 and pH 7.00). A probe drifts in storage, so an uncalibrated reading may be inaccurate.
2. Rinse the probe with distilled water before each new solution, so one solution does not contaminate the next.
3. Make sure the bulb is fully covered by the solution, and stir.
4. Wait for the reading to become steady before recording it.
5. Keep the temperature constant, because pH (and K_w) change with temperature.

Plotting a pH titration curve:

1. Pipette a known volume of acid (e.g. 25.0 cm³) into a beaker and record its starting pH.
2. Add the base from a burette in small portions (e.g. 1 cm³), stirring and recording the pH after each.
3. Add smaller portions (e.g. 0.1 to 0.2 cm³) near the equivalence point, where the pH changes fastest.
4. Carry on well past the equivalence point, until the pH levels off.

Identifying unknown solutions (the PAG11 activity): solutions of the same concentration have very different pH values, so measuring pH can tell them apart. At 0.100 mol dm^-3 and 298 K:

Worked examples

Every answer uses K_w = 1.00 × 10^-14 mol² dm^-6 (data sheet) where needed. pH is given to 2 decimal places; other answers to the significant figures of the least precise data.

1. pH of a strong monobasic acid, and [H⁺] from pH

(a) Calculate the pH of 0.0250 mol dm^-3 nitric acid.

$\text{HNO}_3 \text{ is a strong monobasic acid, so } [\text{H}^+] = [\text{HNO}_3] = 0.0250 \text{ mol dm}^{-3}$ $\text{pH} = -\log(0.0250) = 1.60$

(b) A solution of hydrochloric acid has pH 2.35. Calculate its concentration.

$[\text{H}^+] = 10^{-2.35} = 4.47 \times 10^{-3} \text{ mol dm}^{-3}$ $\text{HCl is strong and monobasic, so } [\text{HCl}] = 4.5 \times 10^{-3} \text{ mol dm}^{-3} \text{ (2 s.f.)}$

A pH given to 2 decimal places fixes [H⁺] to only 2 significant figures, so the final concentration is given to 2 s.f.

2. pH of a strong base, using K_w

Calculate the pH of 0.0400 mol dm^-3 sodium hydroxide at 298 K.

$\text{NaOH completely dissociates, so } [\text{OH}^-] = 0.0400 \text{ mol dm}^{-3}$ $[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.0400} = 2.50 \times 10^{-13} \text{ mol dm}^{-3}$ $\text{pH} = -\log(2.50 \times 10^{-13}) = 12.60$

The common slip is to take -log of [OH^-] and write pH = 1.40. A strong base must have a pH above 7, so check your answer against that.

3. pH of pure water at a different temperature

At 323 K, K_w = 5.48 × 10^-14 mol² dm^-6. Calculate the pH of pure water at 323 K.

$\text{In pure water } [\text{H}^+] = [\text{OH}^-], \text{ so } K_w = [\text{H}^+]^2$ $[\text{H}^+] = \sqrt{5.48 \times 10^{-14}} = 2.34 \times 10^{-7} \text{ mol dm}^{-3}$ $\text{pH} = -\log(2.34 \times 10^{-7}) = 6.63$

The water is still neutral, because [H⁺] = [OH^-]. Saying it is "slightly acidic" loses the mark.

4. pH of a weak acid from K_a

Calculate the pH of 0.150 mol dm^-3 propanoic acid, CH₃CH₂COOH. K_a = 1.35 × 10^-5 mol dm^-3.

$\text{CH}_3\text{CH}_2\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)}$ $K_a = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{COOH}]} \quad \text{(using both approximations)}$ $[\text{H}^+]^2 = 1.35 \times 10^{-5} \times 0.150 = 2.025 \times 10^{-6} \text{ mol}^2 \text{ dm}^{-6}$ $[\text{H}^+] = \sqrt{2.025 \times 10^{-6}} = 1.42 \times 10^{-3} \text{ mol dm}^{-3}$ $\text{pH} = -\log(1.42 \times 10^{-3}) = 2.85$

Forgetting the square root gives [H⁺] = 2.025 × 10^-6 and pH 5.69. Treating the acid as strong gives pH 0.82. Both lose the answer mark.

5. K_a and pK_a of a weak acid from its pH

A 0.200 mol dm^-3 solution of methanoic acid, HCOOH, has pH 2.22. Calculate K_a and pK_a.

$[\text{H}^+] = 10^{-2.22} = 6.03 \times 10^{-3} \text{ mol dm}^{-3}$ $K_a = \frac{[\text{H}^+]^2}{[\text{HCOOH}]} = \frac{(6.03 \times 10^{-3})^2}{0.200} = 1.8 \times 10^{-4} \text{ mol dm}^{-3}$ $\text{p}K_a = -\log(1.8 \times 10^{-4}) = 3.74$

K_a is given to 2 significant figures because the pH (2.22) carries only 2 significant figures of [H⁺]. Check the approximation: [H⁺] ÷ [HA] = 6.03 × 10^-3 ÷ 0.200 = 3%, so very little acid has dissociated and the approximation holds.

6. pH of a buffer made from a weak acid and its salt

3.28 g of sodium ethanoate, CH₃COONa, is dissolved in 250 cm³ of 0.100 mol dm^-3 ethanoic acid. K_a = 1.75 × 10^-5 mol dm^-3. Calculate the pH of the buffer.

$M(\text{CH}_3\text{COONa}) = (2 \times 12.0) + (3 \times 1.0) + (2 \times 16.0) + 23.0 = 82.0 \text{ g mol}^{-1}$ $n(\text{CH}_3\text{COO}^-) = \frac{3.28}{82.0} = 0.0400 \text{ mol}$ $n(\text{CH}_3\text{COOH}) = 0.100 \times \frac{250}{1000} = 0.0250 \text{ mol}$ $[\text{H}^+] = K_a \times \frac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]} = 1.75 \times 10^{-5} \times \frac{0.0250}{0.0400} = 1.09 \times 10^{-5} \text{ mol dm}^{-3}$ $\text{pH} = -\log(1.09 \times 10^{-5}) = 4.96$

Both species are in the same 250 cm³, so the ratio of amounts in mol equals the ratio of concentrations. Putting the ratio upside down (A^- over HA) gives pH 4.55 and loses the mark.

7. pH of a buffer made from excess weak acid and a strong alkali

50.0 cm³ of 0.200 mol dm^-3 ethanoic acid is mixed with 20.0 cm³ of 0.150 mol dm^-3 sodium hydroxide. K_a = 1.75 × 10^-5 mol dm^-3. Calculate the pH of the buffer formed.

$\text{CH}_3\text{COOH(aq)} + \text{OH}^-\text{(aq)} \longrightarrow \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_2\text{O(l)}$ $n(\text{CH}_3\text{COOH}) \text{ at start} = 0.200 \times \frac{50.0}{1000} = 0.0100 \text{ mol}$ $n(\text{OH}^-) = 0.150 \times \frac{20.0}{1000} = 3.00 \times 10^{-3} \text{ mol}$ $n(\text{CH}_3\text{COO}^-) \text{ formed} = 3.00 \times 10^{-3} \text{ mol}$ $n(\text{CH}_3\text{COOH}) \text{ left} = 0.0100 - 3.00 \times 10^{-3} = 7.00 \times 10^{-3} \text{ mol}$ $[\text{H}^+] = 1.75 \times 10^{-5} \times \frac{7.00 \times 10^{-3}}{3.00 \times 10^{-3}} = 4.08 \times 10^{-5} \text{ mol dm}^{-3}$ $\text{pH} = -\log(4.08 \times 10^{-5}) = 4.39$

The mark is in the fifth line: use the acid left over (7.00 × 10^-3 mol), not the starting amount (0.0100 mol, which gives pH 4.23). If you work in concentrations instead, divide both amounts by the total volume, 70.0 cm³ ([HA] = 0.100, [A^-] = 0.0429 mol dm^-3); the ratio, and the pH, are the same.

8. pH of a buffer after adding a small amount of strong acid

5.00 cm³ of 1.00 mol dm^-3 hydrochloric acid is added to the buffer in Worked example 6 (0.0250 mol CH₃COOH and 0.0400 mol CH₃COO^-). Calculate the new pH.

$n(\text{H}^+) \text{ added} = 1.00 \times \frac{5.00}{1000} = 5.00 \times 10^{-3} \text{ mol}$ $\text{CH}_3\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} \longrightarrow \text{CH}_3\text{COOH(aq)}$ $n(\text{CH}_3\text{COO}^-) = 0.0400 - 0.00500 = 0.0350 \text{ mol}$ $n(\text{CH}_3\text{COOH}) = 0.0250 + 0.00500 = 0.0300 \text{ mol}$ $[\text{H}^+] = 1.75 \times 10^{-5} \times \frac{0.0300}{0.0350} = 1.50 \times 10^{-5} \text{ mol dm}^{-3}$ $\text{pH} = -\log(1.50 \times 10^{-5}) = 4.82$

The pH falls only from 4.96 to 4.82. The same acid added to 250 cm³ of water would give [H⁺] = 5.00 × 10^-3 ÷ 0.255 = 0.0196 mol dm^-3, pH 1.71. Remember to change both amounts: A^- goes down and HA goes up.

9. Mass of salt needed for a buffer of a chosen pH

What mass of sodium ethanoate must be added to 200 cm³ of 0.100 mol dm^-3 ethanoic acid to make a buffer of pH 5.00? K_a = 1.75 × 10^-5 mol dm^-3. Assume the volume does not change.

$[\text{H}^+] = 10^{-5.00} = 1.00 \times 10^{-5} \text{ mol dm}^{-3}$ $[\text{CH}_3\text{COO}^-] = \frac{K_a \times [\text{CH}_3\text{COOH}]}{[\text{H}^+]} = \frac{1.75 \times 10^{-5} \times 0.100}{1.00 \times 10^{-5}} = 0.175 \text{ mol dm}^{-3}$ $n(\text{CH}_3\text{COONa}) = 0.175 \times \frac{200}{1000} = 0.0350 \text{ mol}$ $m = n \times M = 0.0350 \times 82.0 = 2.87 \text{ g}$

Sense check: the pH (5.00) is above pK_a (4.76), so the buffer needs more A^- than HA. Here 0.0350 mol of A^- against 0.0200 mol of HA fits. A common slip is to stop at 0.175 and multiply the concentration by M; you must convert to an amount in mol first.

Common exam mistakes

• Defining a Brønsted-Lowry acid as a "hydrogen donor" or "H donor". It is a proton donor (H⁺ donor). A hydrogen atom is not a proton.
• Pairing species on the same side of the equation as a conjugate pair. A conjugate pair is one species on the left and one on the right that differ by one H⁺ (e.g. HCl and Cl^-, not HCl and H₂O).
• Calling ethanoic acid tetrabasic because it has four H atoms. Only the COOH hydrogen is donated, so it is monobasic.
• Writing the K_a expression wrongly: including [H₂O], putting [HA] on top, or missing the square brackets. K_a = [H⁺][A^-] ÷ [HA].
• Wrong or missing units. K_a is in mol dm^-3, K_w in mol² dm^-6. pH and pK_a have no units.
• Dropping the minus sign in pH = -log[H⁺], or using ln instead of log. A pH that comes out negative for a dilute acid is the giveaway.
• Giving pH to 1 decimal place, or [H⁺] to too many figures. Give pH to 2 decimal places unless told otherwise.
• Using [H⁺] = [acid] for a weak acid. That only works for a strong monobasic acid. A weak acid needs K_a and a square root.
• Taking -log[OH^-] as the pH of a strong base. Convert first with [H⁺] = K_w ÷ [OH^-]. A strong base cannot have a pH below 7.
• Forgetting that Ba(OH)₂ (or Ca(OH)₂) releases two OH^- per formula, so [OH^-] is double the base concentration.
• Saying water above 298 K is "acidic" because its pH is below 7. It is still neutral, because [H⁺] = [OH^-].
• Stating the weak-acid approximations without the reasons. The marks need "dissociation of water is negligible" for [H⁺] = [A^-], and "the weak acid dissociates only slightly" for [HA]_eqm = [HA]_start.
• Explaining the limitation vaguely ("the acid is too strong"). Say that a significant proportion of the acid dissociates, so [HA]_eqm is no longer approximately equal to [HA]_start.
• Defining a buffer as a solution that "keeps the pH constant" or "neutralises acids and bases". A buffer minimises pH changes on addition of small amounts of acid or base.
• Explaining buffer action without naming the species. Name what reacts: added H⁺ reacts with CH₃COO^-; added OH^- reacts with H⁺ (or with CH₃COOH). Then give the direction of the equilibrium shift for each.
• Thinking a strong acid and its salt (HCl + NaCl) make a buffer. A buffer needs a weak acid and its conjugate base.
• Using the starting amount of weak acid in an "excess weak acid + NaOH" buffer. Subtract the amount of OH^- to get the acid left over, and use the amount of OH^- as the amount of A^- formed.
• Changing only one amount when acid or alkali is added to a buffer. One member of the pair goes down and the other goes up by the same amount.
• Using [H⁺] = √(K_a × [HA]) for a buffer. In a buffer [H⁺] ≠ [A^-]; use [H⁺] = K_a × [HA] ÷ [A^-].
• Sketching every titration curve with equivalence at pH 7. Weak acid + strong base: above 7. Strong acid + weak base: below 7.
• Saying an indicator is suitable because it "changes colour at pH 7". The marks need the indicator's pH range to lie within the vertical (steep) section of the curve.
• Choosing an indicator for a weak acid + weak base titration. There is no vertical section, so no indicator is suitable.
• Explaining an indicator's colour change without the equilibrium. Write HIn ⇌ H⁺ + In^-, then say acid shifts it left (colour of HIn) and alkali shifts it right (colour of In^-).
• Writing "clear" for phenolphthalein in acid. It is colourless.
• Using a pH meter without calibrating it, or not rinsing the probe with distilled water between solutions.

Key definitions

• Brønsted-Lowry acid: a species that donates a proton (H⁺).
• Brønsted-Lowry base: a species that accepts a proton (H⁺).
• Conjugate acid-base pair: two species that differ by one proton (H⁺), e.g. CH₃COOH and CH₃COO^-.
• Monobasic / dibasic / tribasic acid: an acid that can donate one / two / three protons per molecule (e.g. HCl / H₂SO₄ / H₃PO₄).
• Strong acid: an acid that completely dissociates in aqueous solution.
• Weak acid: an acid that partially dissociates in aqueous solution.
• Acid dissociation constant, K_a: the equilibrium constant for the dissociation of a weak acid, HA(aq) ⇌ H⁺(aq) + A^-(aq); K_a = [H⁺(aq)][A^-(aq)] ÷ [HA(aq)].
• pK_a: pK_a = -log K_a; K_a = 10^-pK_a.
• pH: pH = -log[H⁺]; [H⁺] = 10^-pH.
• Ionic product of water, K_w: K_w = [H⁺(aq)][OH^-(aq)] = 1.00 × 10^-14 mol² dm^-6 at 298 K.
• Neutral solution: a solution in which [H⁺] = [OH^-].
• Buffer solution: a system that minimises pH changes on addition of small amounts of an acid or a base.
• Equivalence point: the point at which the acid and base have exactly reacted (amount of H⁺ = amount of OH^-); the middle of the vertical section of a pH curve.
• End point: the point at which the indicator changes colour, when [HIn] = [In^-].

Specification

• I can define a Brønsted-Lowry acid as a proton donor and a Brønsted-Lowry base as a proton acceptor. (5.1.3 a i)
• I can identify the two conjugate acid-base pairs in an acid-base equilibrium. (5.1.3 a ii)
• I can classify acids as monobasic, dibasic or tribasic. (5.1.3 a iii)
• I can write ionic equations showing the role of H⁺ when acids react with metals, carbonates, metal oxides and alkalis. (5.1.3 b)
• I can write the K_a expression for a weak acid, use it as a measure of the extent of dissociation, and convert between K_a and pK_a. (5.1.3 c)
• I can use pH = -log[H⁺] and [H⁺] = 10^-pH. (5.1.3 d)
• I can use the expression K_w = [H⁺][OH^-]. (5.1.3 e)
• I can calculate pH, or related quantities, for strong monobasic acids and for strong bases using K_w. (5.1.3 f)
• I can calculate pH, K_a or concentration for a weak monobasic acid, stating the two approximations used. (5.1.3 g)
• I can explain why the approximations break down for "stronger" weak acids. (5.1.3 h)
• I can define a buffer solution. (5.1.3 i)
• I can describe how to make a buffer from a weak acid and its salt, or from excess weak acid and a strong alkali. (5.1.3 j)
• I can explain how the conjugate acid-base pair in an acid buffer controls pH when small amounts of acid or alkali are added. (5.1.3 k)
• I can calculate the pH of a buffer from K_a and the concentrations of the conjugate pair, and related quantities such as the pH after adding acid or alkali, or the mass of salt needed. (5.1.3 l)
• I can explain how the carbonic acid-hydrogencarbonate buffer controls blood pH (7.35 to 7.45). (5.1.3 m)
• I can sketch and interpret pH titration curves for strong and weak acids with strong and weak bases. (5.1.3 n i)
• I can explain the choice of indicator from its pH range and the curve, and explain its colour change in terms of the HIn ⇌ H⁺ + In^- equilibrium. (5.1.3 n ii, iii)
• I can describe how to measure pH with a pH meter, including calibration. (5.1.3 o, PAG11)

Related notes

• Acids: strong and weak acids, neutralisation equations and titrations (the AS foundation for this note)
• Amount of substance: n = c × V and n = m ÷ M, used in every buffer calculation
• Redox: acids reacting with metals as a redox reaction
• Chemical equilibrium: le Chatelier's principle, behind buffer action and indicator colours
• How far?: K_c expressions and units, which K_a and K_w build on
• Carboxylic acids and esters: the weak acids used throughout this note