C16: Reactions of Acids
Acids react with metals, bases, and carbonates in predictable ways. Understanding these reactions allows you to predict products, write balanced equations, and prepare pure, dry samples of soluble and insoluble salts.
Acids, Bases and Alkalis
Acids are substances that produce hydrogen ions (H⁺) when dissolved in water. Bases are substances that can neutralise acids. Alkalis are bases that dissolve in water to produce hydroxide ions (OH⁻).
The key definitions to remember:
- Acid: A substance with a pH less than 7 that releases H⁺ ions in aqueous solution.
- Base: A substance that neutralises an acid (includes metal oxides, metal hydroxides, metal carbonates, and ammonia).
- Alkali: A base that dissolves in water to form a solution containing OH⁻ ions. Examples include sodium hydroxide, potassium hydroxide, and calcium hydroxide.
| Common Acids | Formula | Common Alkalis | Formula |
|---|---|---|---|
| Hydrochloric acid | HCl | Sodium hydroxide | NaOH |
| Sulfuric acid | H₂SO₄ | Potassium hydroxide | KOH |
| Nitric acid | HNO₃ | Calcium hydroxide | Ca(OH)₂ |
| Ethanoic acid | CH₃COOH | Ammonia solution | NH₃ (aq) |
All alkalis are bases, but not all bases are alkalis. Only bases that dissolve in water are alkalis. For example, copper oxide is a base but not an alkali because it does not dissolve in water.
The pH Scale
The pH scale runs from 0 to 14. A pH of 7 is neutral. Acids have a pH below 7. Alkalis have a pH above 7. The lower the pH, the more acidic the solution; the higher the pH, the more alkaline the solution.
The pH scale is a measure of the concentration of hydrogen ions in a solution:
- pH 1–6: Acidic (high concentration of H⁺ ions)
- pH 7: Neutral (equal concentrations of H⁺ and OH⁻ ions)
- pH 8–14: Alkaline (low concentration of H⁺ ions, high concentration of OH⁻ ions)
The pH scale is logarithmic, which means each decrease of 1 on the pH scale represents a tenfold increase in the concentration of H⁺ ions. A solution of pH 2 has ten times the concentration of H⁺ ions compared to a solution of pH 3.
| Indicator | Colour in Acid | Colour in Neutral | Colour in Alkali |
|---|---|---|---|
| Litmus paper | Red | Purple | Blue |
| Phenolphthalein | Colourless | Colourless | Pink |
| Methyl orange | Red | Orange | Yellow |
| Universal indicator | Red (pH 1) to Orange/Yellow (pH 3–6) | Green (pH 7) | Blue (pH 8–11) to Purple (pH 12–14) |
Universal indicator gives a range of colours across the pH scale, making it useful for estimating pH. Single indicators like litmus only tell you if a substance is acidic or alkaline, not how acidic or alkaline it is.
Neutralisation Reactions
Neutralisation is the reaction between an acid and a base to produce a salt and water. The general equation is: Acid + Base → Salt + Water. In terms of ions: H⁺(aq) + OH⁻(aq) → H₂O(l).
During neutralisation, the hydrogen ions from the acid react with the hydroxide ions from the alkali to form water. This is the essential ionic equation for all neutralisation reactions between acids and alkalis:
H⁺(aq) + OH⁻(aq) → H₂O(l)
This ionic equation is the same regardless of which acid and which alkali are used, because the spectator ions (the ions that do not change during the reaction) are not included.
Hydrochloric Acid + Sodium Hydroxide
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
The products are sodium chloride (a salt) and water. The H⁺ ions from the acid combine with the OH⁻ ions from the alkali to form water. The Na⁺ and Cl⁻ ions are spectator ions.
Sulfuric Acid + Potassium Hydroxide
H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)
Two moles of potassium hydroxide are needed to neutralise one mole of sulfuric acid because sulfuric acid is diprotic (it has two H⁺ ions that can be replaced).
Reactions of Acids with Metals
Acids react with metals above hydrogen in the reactivity series to produce a salt and hydrogen gas. The general equation is: Acid + Metal → Salt + Hydrogen.
Acid + Metal → Salt + Hydrogen
The name of the salt depends on the acid used and the metal used. The salt name has two parts: the first part comes from the metal, and the second part comes from the acid.
| Acid | Salt Name Ending | Example |
|---|---|---|
| Hydrochloric acid (HCl) | -chloride | Sodium chloride, magnesium chloride |
| Sulfuric acid (H₂SO₄) | -sulfate | Sodium sulfate, magnesium sulfate |
| Nitric acid (HNO₃) | -nitrate | Sodium nitrate, magnesium nitrate |
Zinc + Sulfuric Acid
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
The products are zinc sulfate and hydrogen. Bubbles of hydrogen gas are produced, and the zinc gradually dissolves. The hydrogen can be confirmed using the squeaky pop test with a lit splint.
Magnesium + Hydrochloric Acid
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
The products are magnesium chloride and hydrogen. The reaction is vigorous with rapid fizzing. The magnesium ribbon dissolves quickly.
Reactions of Acids with Metal Oxides and Hydroxides
Acids react with metal oxides and metal hydroxides to produce a salt and water. These are neutralisation reactions. The general equations are: Acid + Metal oxide → Salt + Water; Acid + Metal hydroxide → Salt + Water.
Hydrochloric Acid + Copper Oxide
2HCl(aq) + CuO(s) → CuCl₂(aq) + H₂O(l)
The black copper oxide dissolves in the acid, producing a blue-green solution of copper(II) chloride and water. The solution turns blue-green because of the copper(II) ions.
Sulfuric Acid + Sodium Hydroxide
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
Two moles of sodium hydroxide are needed to neutralise one mole of sulfuric acid because sulfuric acid provides two H⁺ ions per molecule.
Nitric Acid + Calcium Hydroxide
2HNO₃(aq) + Ca(OH)₂(aq) → Ca(NO₃)₂(aq) + 2H₂O(l)
Calcium nitrate and water are produced. Two moles of nitric acid are needed because calcium hydroxide provides two OH⁻ ions per formula unit.
When balancing equations involving acids and bases, make sure the number of H⁺ ions from the acid equals the number of OH⁻ ions from the base. Diprotic acids like H₂SO₄ provide 2H⁺ per molecule, so they need 2 moles of a monobasic base like NaOH.
Reactions of Acids with Carbonates
Acids react with metal carbonates to produce a salt, water, and carbon dioxide gas. The general equation is: Acid + Metal carbonate → Salt + Water + Carbon dioxide.
Acid + Metal carbonate → Salt + Water + Carbon dioxide
The carbon dioxide produced can be detected by bubbling it through limewater (calcium hydroxide solution). The limewater turns cloudy because the carbon dioxide reacts with calcium hydroxide to form insoluble calcium carbonate.
Hydrochloric Acid + Sodium Carbonate
2HCl(aq) + Na₂CO₃(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)
The solution fizzes as carbon dioxide gas is produced. Sodium chloride and water are also formed. The CO₂ can be confirmed using the limewater test.
Sulfuric Acid + Copper Carbonate
H₂SO₄(aq) + CuCO₃(s) → CuSO₄(aq) + H₂O(l) + CO₂(g)
The green copper carbonate dissolves and effervesces, producing a blue solution of copper(II) sulfate, water, and carbon dioxide gas.
Hydrochloric Acid + Calcium Carbonate
2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g)
Fizzing is observed as the calcium carbonate (marble chips) react. The rate of this reaction can be measured by collecting the CO₂ gas in a gas syringe.
Preparing Soluble Salts
Soluble salts can be prepared by reacting an acid with a base, then crystallising the salt from the solution. The method depends on whether the base is soluble or insoluble.
Method 1: Using an insoluble base (metal oxide, metal hydroxide, or metal carbonate)
- Add the insoluble base to the acid and warm gently.
- The base will react and eventually dissolve. Continue adding the base until no more dissolves (some solid base is in excess).
- Filter off the excess unreacted solid base.
- Heat the filtrate gently to evaporate some of the water and concentrate the solution.
- Leave the solution to cool and crystallise. The salt crystals can then be filtered and dried.
Preparing Copper(II) Sulfate Crystals
1. Add copper oxide to warm sulfuric acid and stir. The copper oxide reacts and dissolves: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)
2. Keep adding copper oxide until no more dissolves (some excess solid remains on the bottom).
3. Filter the warm mixture to remove the excess copper oxide.
4. Heat the blue filtrate to evaporate some water.
5. Leave to cool. Blue copper(II) sulfate crystals form.
6. Filter the crystals, wash with cold distilled water, and dry between filter paper.
Method 2: Using an alkali (soluble base) — Titration Method
- Use a pipette to measure a known volume of alkali into a conical flask. Add a few drops of indicator.
- Slowly add the acid from a burette, swirling after each addition, until the indicator just changes colour (the end point).
- Note the volume of acid added. Repeat the experiment using exactly the same volumes of acid and alkali but without the indicator (so the salt is not contaminated with indicator dye).
- Evaporate some of the water from the neutral solution and leave to crystallise.
- Filter, wash, and dry the crystals.
When preparing a salt from a soluble base (alkali), you must use titration because you cannot simply add excess alkali — excess alkali would remain dissolved in the solution and contaminate the salt. The titration method ensures exactly the right amounts of acid and alkali are mixed.
Preparing Insoluble Salts
Insoluble salts can be prepared by precipitation. Two soluble salts are mixed, and the insoluble salt forms as a precipitate that can be filtered off, washed, and dried.
The general method for preparing an insoluble salt by precipitation:
- Choose two soluble salts that contain the required ions.
- Mix the two solutions together. The insoluble salt forms as a solid precipitate.
- Filter the mixture to collect the precipitate.
- Wash the precipitate with distilled water to remove any soluble impurities.
- Leave the precipitate to dry in a warm place or in a desiccator.
Preparing Lead(II) Iodide
Mix lead(II) nitrate solution with potassium iodide solution:
Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)
A bright yellow precipitate of lead(II) iodide forms immediately. Filter to collect the yellow solid, wash with distilled water, and dry.
Preparing Barium Sulfate
Mix barium chloride solution with sodium sulfate solution:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
A white precipitate of barium sulfate forms. Filter, wash, and dry.
Solubility rules: All sodium, potassium, and ammonium salts are soluble. All nitrates are soluble. Most chlorides are soluble (except silver chloride and lead chloride). Most sulfates are soluble (except barium sulfate, lead sulfate, and calcium sulfate). Most carbonates and hydroxides are insoluble (except sodium, potassium, and ammonium ones).
Strong vs Weak Acids
A strong acid is completely ionised in aqueous solution, so all the acid molecules split to release H⁺ ions. A weak acid is only partially ionised in aqueous solution, so only some of the acid molecules split to release H⁺ ions.
| Property | Strong Acids | Weak Acids |
|---|---|---|
| Ionisation in water | Complete (100%) | Partial (less than 100%) |
| pH of 0.1 mol/dm³ solution | pH 1 (HCl), pH 0.7 (H₂SO₄) | pH 2–4 (ethanoic acid ~pH 3) |
| Conductivity | Higher (more ions to carry charge) | Lower (fewer ions) |
| Rate of reaction with magnesium | Faster (more H⁺ ions available per unit time) | Slower |
| Examples | HCl, H₂SO₄, HNO₃ | Ethanoic acid (CH₃COOH), citric acid, carbonic acid |
For strong acids, the ionisation is shown with a single arrow (→) because it goes to completion:
HCl(aq) → H⁺(aq) + Cl⁻(aq)
For weak acids, the ionisation is shown with a reversible arrow (⇌) because it is an equilibrium:
CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)
For the same concentration, a strong acid has a lower pH than a weak acid because it has a higher concentration of H⁺ ions. However, equal volumes of a strong acid and a weak acid of the same concentration will produce the same amount of hydrogen gas when reacted with a metal, because the weak acid continues to ionise as H⁺ ions are used up.
The Effect of Concentration on pH
As the concentration of H⁺ ions increases by a factor of 10, the pH decreases by 1. A solution of pH 1 has 100 times the concentration of H⁺ ions compared to a solution of pH 3.
The relationship between H⁺ ion concentration and pH is logarithmic:
Each decrease of 1 in pH = 10× increase in H⁺ concentration
Comparing H⁺ Concentrations
A solution of pH 2 has 10 times the H⁺ concentration of a solution of pH 3, and 100 times the H⁺ concentration of a solution of pH 4.
A solution of pH 1 has 10,000 (10⁴) times the H⁺ concentration of a solution of pH 5.
Factor of change = 10^(pH difference)
Diluting a Strong Acid
If 10 cm³ of 1.0 mol/dm³ hydrochloric acid (pH 1) is diluted with 90 cm³ of water to make 100 cm³ of 0.1 mol/dm³ solution, the concentration of H⁺ ions decreases by a factor of 10 and the pH increases from 1 to 2.
Practice Questions
1. Write a balanced symbol equation for the reaction between hydrochloric acid and calcium carbonate. State the names of all products.
2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g)
The products are calcium chloride, water, and carbon dioxide.
2. Describe how you would prepare a pure, dry sample of copper(II) sulfate crystals starting from copper oxide and dilute sulfuric acid.
Add copper oxide to warm dilute sulfuric acid and stir. Continue adding copper oxide until no more dissolves (some excess solid remains). Filter the mixture to remove excess copper oxide. Heat the filtrate to evaporate some water and concentrate the solution. Leave to cool so that crystals form. Filter off the crystals, wash with cold distilled water, and dry between filter paper.
3. Explain the difference between a strong acid and a weak acid, using hydrochloric acid and ethanoic acid as examples.
Hydrochloric acid is a strong acid because it is completely ionised in water: all HCl molecules split to form H⁺ and Cl⁻ ions. Ethanoic acid is a weak acid because it is only partially ionised in water: only some CH₃COOH molecules split to form H⁺ and CH₃COO⁻ ions, while most remain as intact molecules. For equal concentrations, hydrochloric acid has a lower pH, higher conductivity, and reacts faster with magnesium because it has a higher concentration of H⁺ ions.
4. Write the ionic equation for the neutralisation reaction between any acid and any alkali.
H⁺(aq) + OH⁻(aq) → H₂O(l)
5. A solution of pH 3 is diluted by a factor of 100. What is the new pH?
Diluting by a factor of 100 means the H⁺ concentration decreases by a factor of 100, which is 10². Since each factor of 10 increases the pH by 1, the pH increases by 2. The new pH is 3 + 2 = 5.
6. Describe how you would prepare a pure, dry sample of the insoluble salt lead(II) sulfate by precipitation.
Mix lead(II) nitrate solution with sodium sulfate solution. A white precipitate of lead(II) sulfate forms: Pb(NO₃)₂(aq) + Na₂SO₄(aq) → PbSO₄(s) + 2NaNO₃(aq). Filter the mixture to collect the precipitate. Wash the precipitate with distilled water to remove soluble sodium nitrate and any unreacted starting solutions. Dry the precipitate in a warm oven or desiccator.
Required Practical: Preparing a Pure Dry Sample of a Soluble Salt
To prepare copper(II) sulfate crystals: add copper oxide to warm dilute sulfuric acid and stir. Continue adding until no more dissolves (some excess solid remains). Filter to remove excess copper oxide. Heat the filtrate gently to evaporate some water. Leave to cool and crystallise. Filter off the crystals, wash with cold distilled water, and dry between filter paper.
Key techniques: adding the base in excess ensures all the acid has reacted, filtration removes the unreacted excess base, evaporation concentrates the solution, and crystallisation produces pure crystals. Washing with cold water removes impurities without dissolving the crystals.
Safety: wear eye protection because acids are corrosive. Heat the acid gently, not to boiling. If using a Bunsen burner, follow standard safety procedures.
This method works for insoluble bases (metal oxides, hydroxides, carbonates). For soluble bases (alkalis), you must use titration to get exactly the right amounts of acid and alkali.
Maths Skills
Writing Balanced Equations for Neutralisation
Writing balanced equations for acid reactions requires careful counting of atoms and charges. For neutralisation: ensure the number of H+ ions from the acid equals the number of OH− ions from the base. Diprotic acids like H2SO4 need 2 mol of a monobasic base like NaOH.
Calculating the mass of salt from given reactant masses: use the three-step method. (1) Calculate moles of the limiting reactant, (2) use the balanced equation to find moles of salt, (3) calculate mass of salt = moles × Mr of the salt.
Calculating Concentration After Dilution
When a solution is diluted, the number of moles of solute stays the same but the volume increases, so the concentration decreases. Use: C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume after dilution.
For example, if 25 cm³ of 2.0 mol/dm³ HCl is diluted to 100 cm³: C2 = (C1 × V1) / V2 = (2.0 × 25) / 100 = 0.50 mol/dm³. The concentration has decreased by a factor of 4 because the volume increased by a factor of 4.
Common Misconceptions
Strength vs Concentration
Wrong: Strong acids are more concentrated than weak acids Correct: Strength and concentration are different properties. Strength refers to how completely an acid ionises in water (strong = fully ionised, weak = partially ionised). Concentration refers to how much acid is dissolved per unit volume. A dilute strong acid (e.g. 0.01 mol/dm³ HCl) is completely ionised but has fewer H+ ions than a concentrated weak acid
Acid Danger
Wrong: All acids are equally dangerous Correct: The danger of an acid depends on both its concentration and its strength. A concentrated strong acid like 5 mol/dm³ HCl is highly corrosive and dangerous. However, a dilute weak acid like the ethanoic acid in vinegar (about 0.8 mol/dm³) is safe to consume. Even a dilute strong acid (0.01 mol/dm³ HCl) is harmless. Always consider both strength and concentration when assessing risk
6-Mark Extended Question
Strong vs Weak Acids
Explain the difference between a strong acid and a weak acid. Discuss why this matters when choosing acids for reactions.
A strong acid is completely ionised in aqueous solution, meaning all acid molecules split to release H+ ions. Examples include HCl, H2SO4 and HNO3. A weak acid is only partially ionised, meaning only some molecules release H+ ions while most remain as intact molecules. Examples include ethanoic acid (CH3COOH) and citric acid. [2 marks]
The difference matters because at the same concentration, a strong acid has a lower pH, higher electrical conductivity, and reacts faster with metals and carbonates than a weak acid. This is because a strong acid produces more H+ ions per unit volume. For example, 0.1 mol/dm³ HCl has a pH of 1, while 0.1 mol/dm³ CH3COOH has a pH of about 3. [2 marks]
When choosing acids for reactions, strength affects the rate and the pH of the product. A strong acid gives a faster reaction, which is useful in industrial processes. However, a weak acid provides a milder, more controllable reaction, which is preferable in food preparation (vinegar for pickling) and when safety is a concern. Crucially, at the same concentration, equal volumes of strong and weak acids will produce the same total amount of product because the weak acid continues to ionise as H+ ions are used up. [2 marks]
AO3: Analyse and Evaluate
Identifying Strong vs Weak Acids from pH Data
A student measures the pH of four different acid solutions, all at the same concentration of 0.1 mol/dm³. The results are: Acid P = pH 1, Acid Q = pH 2.9, Acid R = pH 0.7, Acid S = pH 3.2. (a) Which acids are strong and which are weak? Explain your answer. (b) The student then dilutes Acid P by a factor of 10. Predict the new pH.
(a) Strong acids are completely ionised, so they produce more H+ ions per unit volume and have a lower pH. Acids P (pH 1) and R (pH 0.7) are strong acids — their pH values are very low for 0.1 mol/dm³. R is an even stronger acid or is diprotic (H2SO4 provides 2H+ per molecule). Acids Q (pH 2.9) and S (pH 3.2) are weak acids — their pH is much higher than a strong acid at the same concentration because only a small fraction of molecules are ionised. [Explanation of ionisation and H+ concentration difference]
(b) Diluting by a factor of 10 reduces the H+ concentration by a factor of 10, so the pH increases by 1. The new pH of Acid P is 1 + 1 = 2. This is because pH is logarithmic: each decrease of 1 on the pH scale represents a 10-fold increase in H+ concentration, so diluting by 10 increases pH by 1.
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