Acids

In a nutshell

Acids release H⁺ ions in water, alkalis release OH^- ions, and when they meet they neutralise each other to make a salt and water.

This subtopic covers the common acids and alkalis, what "strong" and "weak" really mean, the neutralisation equations you must be able to write, and the titration skills (PAG2) that let you measure an unknown concentration accurately.

Assumed knowledge: Compounds, formulae and equations; Amount of substance.

Core content

Acids, bases, alkalis and neutralisation

An acid releases H⁺ ions in aqueous solution.

An alkali releases OH^- ions in aqueous solution. An alkali is a base that dissolves in water.

A base is any substance that neutralises an acid to form a salt. Bases include:

• metal oxides, e.g. CuO, MgO
• metal hydroxides, e.g. NaOH, KOH
• carbonates (and hydrogencarbonates), e.g. CaCO₃, Na₂CO₃, NaHCO₃
• ammonia, NH₃

So every alkali is a base, but not every base is an alkali (CuO and CaCO₃ are insoluble, so they are bases but not alkalis).

You must know these formulae:

How each one releases its ions in water:

• HCl(aq) → H⁺(aq) + Cl^-(aq)
• HNO₃(aq) → H⁺(aq) + NO₃^-(aq)
• NaOH(s) → Na⁺(aq) + OH^-(aq) (on dissolving)
• NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)

Ammonia has no OH in its formula. It makes OH^- ions by taking an H⁺ from a water molecule, which is why it still counts as an alkali.

Sulfuric acid has two hydrogens that can be released as H⁺, so 1 mol of H₂SO₄ neutralises 2 mol of NaOH. HCl and HNO₃ release only one H⁺ each.

Strong and weak acids

The difference is how much the acid dissociates (splits into ions) in water:

• A strong acid completely dissociates in aqueous solution. HCl, HNO₃ and H₂SO₄ are strong.
• A weak acid only partially dissociates in aqueous solution. CH₃COOH (ethanoic acid) is weak.

The equations show it:

• Strong: HCl(aq) → H⁺(aq) + Cl^-(aq) (one-way arrow: every molecule splits)
• Weak: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO^-(aq) (equilibrium arrow: most molecules stay whole)

So a solution of a strong acid contains a greater proportion of its acid as H⁺ ions than a weak acid of the same concentration.

"Strong" and "concentrated" are not the same thing:

You can have a dilute solution of a strong acid, or a concentrated solution of a weak acid.

Still don't get it? · strong vs concentrated

Picture a jar of wrapped sweets. Two separate questions:

• How many sweets are in the jar? That is about how full the jar is.
• Once they are in the jar, how many of them come out of their wrappers? That is about the type of sweet.

A jar could hold only a few sweets that all unwrap themselves, or be packed full of sweets that mostly stay wrapped.

Now swap the words back in:

1. The sweets are acid molecules and the jar is a dm³ of water.
2. "How many sweets per jar" is concentration (mol dm^-3). More acid per dm³ means more concentrated.
3. "Unwrapping" is dissociation: the acid splitting into H⁺ and its negative ion.
4. An acid where every molecule unwraps is strong (HCl). An acid where only a small proportion unwraps is weak (CH₃COOH).

The exam wording: a strong acid completely dissociates in aqueous solution; a weak acid partially dissociates. Never use "strong" to mean "concentrated".

Neutralisation

At its heart, every acid + alkali neutralisation is the same reaction:

H⁺(aq) + OH^-(aq) → H₂O(l)

For a strong acid with NaOH or KOH, the metal ion and the acid's negative ion are spectator ions: they cancel from the full equation and leave just this ionic equation.

A salt forms when the H⁺ of an acid is replaced by a metal ion or an ammonium ion. The acid decides the type of salt:

The four patterns, with full equations (you must be able to write these, balanced, with state symbols):

Acid + alkali → salt + water

• HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
• H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)
• CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

Acid + metal oxide → salt + water

• 2HNO₃(aq) + CuO(s) → Cu(NO₃)₂(aq) + H₂O(l) (copper(II) nitrate)
• H₂SO₄(aq) + MgO(s) → MgSO₄(aq) + H₂O(l)

Acid + carbonate → salt + water + carbon dioxide

• 2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g)
• H₂SO₄(aq) + Na₂CO₃(aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)
• HCl(aq) + NaHCO₃(aq) → NaCl(aq) + H₂O(l) + CO₂(g)

Observations: effervescence (fizzing, from the CO₂) and a solid carbonate dissolves. For a dissolved carbonate the ionic equation is 2H⁺(aq) + CO₃^2-(aq) → H₂O(l) + CO₂(g).

Acid + ammonia → ammonium salt (no water formed)

• HNO₃(aq) + NH₃(aq) → NH₄NO₃(aq)
• HCl(aq) + NH₃(aq) → NH₄Cl(aq)

Acids also react with reactive metals to make a salt and hydrogen, but that is a redox reaction, covered in Redox (2.1.5).

Acid-base titrations

A titration measures exactly what volume of one solution reacts with a known volume of another. If you know the concentration of one, you can calculate the other.

The solution of known concentration is a standard solution. You make one by dissolving an accurately weighed solid and making it up to an exact volume in a volumetric flask.

Making a standard solution (e.g. 250.0 cm³ of Na₂CO₃(aq)):

1. Calculate the mass of solid needed (Worked example 2).
2. Weigh the solid in a weighing boat on a balance. Reweigh the empty boat after tipping the solid out, and use the difference as the mass transferred.
3. Dissolve the solid in a small volume of distilled water in a beaker, stirring with a glass rod.
4. Transfer the solution to a 250.0 cm³ volumetric flask through a funnel.
5. Rinse the beaker, glass rod and funnel with distilled water and add the washings to the flask, so no solute is left behind.
6. Make up to the graduation mark with distilled water, adding the last part dropwise until the bottom of the meniscus sits on the line, read at eye level.
7. Stopper and invert the flask several times so the solution is evenly mixed.

PAG2: Acid-base titration

Preparing the apparatus:

• Rinse the burette with the solution it will hold, and rinse the pipette with the solution it will deliver. Water left inside would dilute them.
• Rinse the conical flask with distilled water only. Water does not change the amount in mol of the solution you pipette in.
• Fill the burette using a funnel, remove the funnel, then run a little solution out so the jet below the tap is full with no air bubble.

Carrying out the titration:

1. Use a pipette filler and a volumetric pipette to transfer 25.0 cm³ of one solution into the conical flask.
2. Add a few drops of indicator (only a few: indicators are themselves weak acids).
3. Stand the flask on a white tile so the colour change is easy to see.
4. Read the burette to the bottom of the meniscus at eye level, and record the initial reading.
5. Do a rough titration first to find the approximate end point.
6. Repeat accurately: add the solution quickly at first while swirling, then dropwise near the end point, until the indicator just changes colour (the end point).
7. Record the final reading. Titre = final reading - initial reading.
8. Repeat until you have concordant titres, i.e. titres within 0.10 cm³ of each other.

Recording and processing results:

• Record both burette readings for every titration, in a table.
• Record every reading to 2 decimal places, ending in 0 or 5 (a burette is read to the nearest 0.05 cm³), e.g. 24.20, not 24.2.
• Calculate the mean titre from the concordant titres only. Leave out the rough titre and any titre that is not concordant.

Here titrations 1 and 3 (24.20 and 24.20 cm³) are concordant. The rough titre (24.80) and titration 2 (24.55) are left out of the mean. These are the results used in Worked example 1.

Choosing an indicator:

The reasons behind these choices (pH curves) are covered at A level in Acids, bases and buffers (5.1.3).

Titration calculations

Every titration calculation follows the same chain:

1. Amount of the solution you know both values for: $n = c \times V$, with V in dm³ (cm³ ÷ 1000).
2. Use the balanced equation's mole ratio to get the amount of the other substance.
3. Answer the question asked: concentration ($c = \frac{n}{V}$), mass ($m = n \times M$), molar mass ($M = \frac{m}{n}$) or a percentage.

Two extra steps turn up in harder (non-structured) questions:

• Scaling up: if you titrated a 25.0 cm³ portion taken from 250.0 cm³ of solution, the full solution contains 10 times the amount in your portion.
• Back titration: add a known excess of acid to a solid base, then titrate the acid left over. The acid that reacted = acid added - acid left over (Worked example 5).

Worked examples

A_r values are from the Periodic Table on the data sheet. Every step is on its own line with units carried through. Final answers are given to 3 significant figures to match the least precise data (e.g. 25.0 cm³, 0.100 mol dm^-3).

1. Mean titre and concentration (1 : 1)

25.0 cm³ of hydrochloric acid was titrated with 0.100 mol dm^-3 sodium hydroxide. Calculate the concentration of the acid.

$\text{HCl(aq)} + \text{NaOH(aq)} \longrightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$ $\text{mean titre (concordant titres 1 and 3)} = \frac{24.20 + 24.20}{2} = 24.20 \text{ cm}^3$ $n(\text{NaOH}) = c \times V = 0.100 \times \frac{24.20}{1000} = 2.420 \times 10^{-3} \text{ mol}$ $1\,\text{NaOH} : 1\,\text{HCl} \implies n(\text{HCl}) = 2.420 \times 10^{-3} \text{ mol}$ $c(\text{HCl}) = \frac{n}{V} = \frac{2.420 \times 10^{-3}}{25.0 / 1000} = 0.0968 \text{ mol dm}^{-3}$

Averaging all four titres gives 24.44 cm³ and an answer of 0.0978 mol dm^-3, which loses the mean-titre mark.

2. Mass of solid for a standard solution

Calculate the mass of anhydrous sodium carbonate, Na₂CO₃, needed to make 250.0 cm³ of a 0.0800 mol dm^-3 standard solution.

$n(\text{Na}_2\text{CO}_3) = c \times V = 0.0800 \times \frac{250.0}{1000} = 0.0200 \text{ mol}$ $M(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1}$ $m = n \times M = 0.0200 \times 106.0 = 2.12 \text{ g}$

In practice: weighing boat + solid = 14.37 g, empty boat afterwards = 12.25 g, so the mass transferred = 14.37 - 12.25 = 2.12 g.

3. A diprotic acid, in mol dm^-3 and g dm^-3

25.0 cm³ of dilute sulfuric acid needed a mean titre of 21.60 cm³ of 0.150 mol dm^-3 sodium hydroxide. Calculate the concentration of the acid in mol dm^-3 and in g dm^-3.

$\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH(aq)} \longrightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$ $n(\text{NaOH}) = 0.150 \times \frac{21.60}{1000} = 3.240 \times 10^{-3} \text{ mol}$ $1\,\text{H}_2\text{SO}_4 : 2\,\text{NaOH} \implies n(\text{H}_2\text{SO}_4) = \frac{3.240 \times 10^{-3}}{2} = 1.620 \times 10^{-3} \text{ mol}$ $c(\text{H}_2\text{SO}_4) = \frac{1.620 \times 10^{-3}}{25.0 / 1000} = 0.0648 \text{ mol dm}^{-3}$ $M(\text{H}_2\text{SO}_4) = (2 \times 1.0) + 32.1 + (4 \times 16.0) = 98.1 \text{ g mol}^{-1}$ $\text{mass concentration} = 0.0648 \times 98.1 = 6.36 \text{ g dm}^{-3}$

The mark is in the third line: halve the amount of NaOH. Using 1 : 1 doubles the answer to 0.130 mol dm^-3.

4. Scaling up to identify an unknown carbonate

2.32 g of a Group 1 carbonate, M₂CO₃, was dissolved in water and made up to 250.0 cm³. A 25.0 cm³ portion needed 22.40 cm³ of 0.150 mol dm^-3 hydrochloric acid. Identify M.

$\text{M}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \longrightarrow 2\text{MCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$ $n(\text{HCl}) = 0.150 \times \frac{22.40}{1000} = 3.360 \times 10^{-3} \text{ mol}$ $n(\text{M}_2\text{CO}_3) \text{ in } 25.0 \text{ cm}^3 = \frac{3.360 \times 10^{-3}}{2} = 1.680 \times 10^{-3} \text{ mol}$ $n(\text{M}_2\text{CO}_3) \text{ in } 250.0 \text{ cm}^3 = 1.680 \times 10^{-3} \times 10 = 0.01680 \text{ mol}$ $M(\text{M}_2\text{CO}_3) = \frac{m}{n} = \frac{2.32}{0.01680} = 138 \text{ g mol}^{-1}$ $\text{mass of CO}_3 = 12.0 + (3 \times 16.0) = 60.0, \quad A_r(\text{M}) = \frac{138.1 - 60.0}{2} = 39.0 \text{ (using the unrounded } 138.1\text{)}$ $\text{M is potassium (}A_r = 39.1\text{); the carbonate is K}_2\text{CO}_3$

The fourth line is the one most often missed: the 2.32 g was dissolved in the whole 250.0 cm³, so multiply by 10 before dividing the mass by the amount.

5. Back titration: purity of a carbonate

A 1.25 g sample of impure calcium carbonate was added to 50.0 cm³ of 1.00 mol dm^-3 hydrochloric acid (an excess). When the reaction stopped, the mixture was made up to 250.0 cm³ with distilled water. 25.0 cm³ portions of this solution needed a mean titre of 26.40 cm³ of 0.100 mol dm^-3 sodium hydroxide. Calculate the percentage by mass of CaCO₃ in the sample. (Assume the impurities do not react with acid.)

$\text{HCl(aq)} + \text{NaOH(aq)} \longrightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$ $n(\text{NaOH}) = 0.100 \times \frac{26.40}{1000} = 2.640 \times 10^{-3} \text{ mol} = n(\text{HCl left}) \text{ in } 25.0 \text{ cm}^3$ $n(\text{HCl left}) \text{ in } 250.0 \text{ cm}^3 = 2.640 \times 10^{-3} \times 10 = 0.02640 \text{ mol}$ $n(\text{HCl added}) = 1.00 \times \frac{50.0}{1000} = 0.0500 \text{ mol}$ $n(\text{HCl reacted with CaCO}_3) = 0.0500 - 0.02640 = 0.02360 \text{ mol}$ $\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \longrightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$ $n(\text{CaCO}_3) = \frac{0.02360}{2} = 0.01180 \text{ mol}$ $M(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1 \text{ g mol}^{-1}$ $m(\text{CaCO}_3) = 0.01180 \times 100.1 = 1.181 \text{ g}$ $\text{percentage by mass} = \frac{1.181}{1.25} \times 100 = 94.5\%$

The titration measures the acid left over, not the acid that reacted. Using 0.02640 mol as the acid that reacted gives a CaCO₃ mass of 1.32 g, more than the whole sample.

Still don't get it? · back titrations

Imagine you give a friend £20 to buy some snacks. You can't see what they bought, but they hand you £5.60 in change.

You don't need the receipt: they spent £20 - £5.60 = £14.40.

A back titration works the same way:

1. The £20 is the acid you added. You know exactly how much, from its concentration and volume: n = c × V.
2. The snacks are the solid base. It reacts with some of the acid, but you can't titrate a solid in a conical flask directly.
3. The change is the acid left over. This is what your titration measures, because the NaOH in the burette only reacts with the acid still in the flask.
4. Acid used by the solid = acid added - acid left over.
5. Only then use the equation for the solid (e.g. CaCO₃ : HCl = 1 : 2) to get the amount of the solid.

In the exam, label each amount clearly ("HCl added", "HCl left", "HCl reacted"). The marks are for each of those steps, and the most common error is treating "acid left over" as "acid reacted".

Common exam mistakes

• Using "strong" to mean "concentrated". Strong means the acid completely dissociates; concentrated means lots of acid per dm³. A dilute solution of HCl is still a strong acid.
• Saying a weak acid "does not dissociate". It partially dissociates. Its equation needs the equilibrium arrow ⇌, while a strong acid uses →.
• Writing "H⁺ ions" without "in aqueous solution" (or "in water") when defining an acid or alkali. The ions are only released when the substance is dissolved.
• Using a 1 : 1 ratio for sulfuric acid. H₂SO₄ reacts with 2 NaOH; the amount of H₂SO₄ is half the amount of NaOH.
• Leaving products out of equations. Acid + carbonate gives salt + water + CO₂; acid + ammonia gives an ammonium salt and no water.
• Mixing up ammonia and ammonium. Ammonia is NH₃; the ammonium ion is NH₄⁺; ammonium chloride is NH₄Cl.
• Averaging every titre. The mean uses concordant titres only (within 0.10 cm³). The rough titre and any outlier are left out.
• Recording burette readings to 1 decimal place. Write 24.20, not 24.2. Readings go to 2 decimal places, ending in 0 or 5.
• Writing "clear" for phenolphthalein in acid. The colour is colourless. "Clear" means see-through, which a pink solution also is.
• Rinsing the wrong thing with the wrong liquid. Burette and pipette are rinsed with their own solutions; the conical flask with distilled water. Rinsing the flask with the solution adds extra moles; rinsing the burette with water dilutes it.
• Vague standard-solution methods. "Dissolve it and put it in a flask" scores little. The steps that earn credit: weigh by difference, rinse washings into the volumetric flask, make up to the mark (bottom of the meniscus), invert to mix.
• Forgetting the scale-up factor. If a 25.0 cm³ portion came from 250.0 cm³, multiply the amount by 10 before working out the original mass or concentration.
• Not converting cm³ to dm³ in n = c × V. Divide by 1000 first.
• Wrong number of significant figures. Give the final answer to the same number of significant figures as the least precise data (usually 3).

Key definitions

• Acid: a substance that releases H⁺ ions in aqueous solution.
• Alkali: a substance that releases OH^- ions in aqueous solution (a soluble base).
• Base: a substance that neutralises an acid to form a salt (e.g. metal oxides, metal hydroxides, carbonates, ammonia).
• Strong acid: an acid that completely dissociates in aqueous solution.
• Weak acid: an acid that partially dissociates in aqueous solution.
• Neutralisation: the reaction of H⁺ and OH^- to form H₂O; H⁺(aq) + OH^-(aq) → H₂O(l).
• Salt: a compound formed when the H⁺ ion of an acid is replaced by a metal ion or an ammonium ion.
• Standard solution: a solution of known concentration.
• Titre: the volume of solution added from the burette to reach the end point.
• End point: the point at which the indicator just changes colour.
• Concordant titres: titres that agree within 0.10 cm³ of each other.

Specification

• I can give the formulae of HCl, H₂SO₄, HNO₃, CH₃COOH, NaOH, KOH and NH₃, and explain that acids release H⁺ ions and alkalis release OH^- ions in aqueous solution. (2.1.4 a)
• I can explain the difference between strong and weak acids in terms of how much they dissociate. (2.1.4 b)
• I can write the ionic equation for neutralisation, H⁺ + OH^- → H₂O. (2.1.4 c i)
• I can write full balanced equations for acids reacting with alkalis, metal oxides, carbonates and ammonia to form salts. (2.1.4 c ii)
• I can describe how to prepare a standard solution of a required concentration using a volumetric flask. (2.1.4 d, PAG2)
• I can describe how to carry out an acid-base titration, including rinsing, reading the burette, using an indicator and getting concordant titres. (2.1.4 d, PAG2)
• I can do structured and non-structured titration calculations, including mean titres, mole ratios, scaling up and back titrations, for familiar and unfamiliar acids and bases. (2.1.4 e)

Related notes

• Compounds, formulae and equations: formulae of ions and writing ionic equations
• Amount of substance: n = c × V, molar mass and mole ratios
• Redox: acids reacting with metals
• Qualitative analysis: the test for carbonate ions with acid
• Acids, bases and buffers (A level): pH, K_a and choosing indicators from pH curves