Worked examples: Acids, Bases and Salts

Work through each example by covering the solution, attempting it yourself, then checking every line, especially the units. This chapter is where careful working earns marks, so we set out each calculation the way the marking scheme expects. These are original questions in SPM style, not past-year papers. Relative atomic masses are given in each question, as they are in the real exam.

Example 1, Molarity from mass

Question. 4.0 g of sodium hydroxide, NaOH, is dissolved in water to make 500 cm³ of solution. Calculate the molarity. [Relative atomic mass: H = 1, O = 16, Na = 23]

Solution.

• Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹.
• Number of moles = mass ÷ molar mass = 4.0 g ÷ 40 g mol⁻¹ = 0.10 mol.
• Volume = 500 cm³ ÷ 1000 = 0.500 dm³.
• Molarity = moles ÷ volume = 0.10 mol ÷ 0.500 dm³ = 0.20 mol dm⁻³.

Common slip: leaving the volume in cm³, which makes the answer 1000 times too small.

Example 2, Converting g dm⁻³ to mol dm⁻³

Question. A solution of sulfuric acid has a concentration of 9.8 g dm⁻³. Express this in mol dm⁻³. [H = 1, O = 16, S = 32]

Solution.

• Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 98 g mol⁻¹.
• Molarity = concentration in g dm⁻³ ÷ molar mass = 9.8 g dm⁻³ ÷ 98 g mol⁻¹ = 0.10 mol dm⁻³.

Common slip: multiplying by the molar mass instead of dividing. To go from g dm⁻³ to mol dm⁻³ you divide; to go back you multiply.

Example 3, Moles in a measured volume

Question. Calculate the number of moles of hydrochloric acid in 25.0 cm³ of 0.10 mol dm⁻³ HCl.

Solution.

• Number of moles = molarity × volume in dm³ = 0.10 mol dm⁻³ × (25.0 ÷ 1000) dm³.
• = 0.10 × 0.0250 = 2.5 × 10⁻³ mol (0.0025 mol).

Common slip: forgetting to divide the volume by 1000, or writing the answer without a unit.

Example 4, Dilution

Question. What volume of water must be added to 25.0 cm³ of 2.0 mol dm⁻³ hydrochloric acid to dilute it to 0.25 mol dm⁻³?

Solution.

• Use M₁V₁ = M₂V₂, with M₁ = 2.0, V₁ = 25.0 cm³, M₂ = 0.25.
• V₂ = M₁V₁ ÷ M₂ = (2.0 × 25.0) ÷ 0.25 = 50.0 ÷ 0.25 = 200 cm³.
• This is the final volume, so volume of water added = 200 − 25.0 = 175 cm³.

Common slip: giving 200 cm³ as the answer when the question asks for the volume of water added, not the final volume.

Example 5, Titration of a monoprotic acid with an alkali

Question. 25.0 cm³ of sodium hydroxide solution is exactly neutralised by 20.0 cm³ of 0.10 mol dm⁻³ hydrochloric acid. Calculate the molarity of the sodium hydroxide. The equation is HCl + NaOH → NaCl + H₂O.

Solution.

• The mole ratio of acid to base is 1 : 1, so (Mₐ × Vₐ) / (M_b × V_b) = 1/1.
• Rearrange for the base: M_b = (Mₐ × Vₐ) ÷ V_b = (0.10 × 20.0) ÷ 25.0.
• = 2.0 ÷ 25.0 = 0.080 mol dm⁻³.

Common slip: using the wrong solution’s volume for M_b. Always match each molarity with its own volume.

Example 6, Titration with a 1 : 2 ratio

Question. 25.0 cm³ of 0.10 mol dm⁻³ sodium hydroxide is neutralised by 0.050 mol dm⁻³ sulfuric acid. Calculate the volume of acid needed. The equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

Solution.

• Ratio of acid : base = 1 : 2, so (Mₐ × Vₐ) / (M_b × V_b) = 1/2.
• Method by moles: moles of NaOH = 0.10 × 25.0 ÷ 1000 = 2.5 × 10⁻³ mol.
• From the equation, moles of H₂SO₄ = ½ × moles of NaOH = 1.25 × 10⁻³ mol.
• Volume of acid = moles ÷ molarity = 1.25 × 10⁻³ ÷ 0.050 = 0.0250 dm³ = 25.0 cm³.

Common slip: forgetting the 1 : 2 ratio and treating sulfuric acid as if it reacted 1 : 1 with the alkali.

Example 7, Preparing a standard solution

Question. Calculate the mass of anhydrous sodium carbonate, Na₂CO₃, needed to prepare 250 cm³ of a 0.10 mol dm⁻³ standard solution. [C = 12, O = 16, Na = 23]

Solution.

• Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g mol⁻¹.
• Moles required = molarity × volume in dm³ = 0.10 × (250 ÷ 1000) = 0.025 mol.
• Mass = moles × molar mass = 0.025 × 106 = 2.65 g.

Common slip: using 250 instead of 0.250 dm³ for the volume, or forgetting to multiply by the molar mass at the end.

Example 8, From molarity back to g dm⁻³

Question. Using the sodium hydroxide result from Example 5 (0.080 mol dm⁻³), express its concentration in g dm⁻³. [H = 1, O = 16, Na = 23]

Solution.

• Molar mass of NaOH = 40 g mol⁻¹.
• Concentration in g dm⁻³ = molarity × molar mass = 0.080 × 40 = 3.2 g dm⁻³.

Common slip: mixing up the two forms of concentration, remember that molarity is per mole and g dm⁻³ is per gram, linked by the molar mass.

Example 9, Choosing a salt-preparation method

Question. Suggest a method to prepare a pure, dry sample of copper(II) sulfate crystals, and explain why that method is chosen.

Solution. Copper(II) sulfate is a soluble salt of a metal that is not sodium, potassium or ammonium, so it is prepared by reacting an acid with an insoluble base. Add an excess of copper(II) oxide to warm dilute sulfuric acid and stir until no more dissolves; the excess ensures all the acid reacts. Filter to remove the unreacted copper(II) oxide, then heat the filtrate to the point of crystallisation and allow it to cool so crystals form. Finally, filter the crystals, wash with a little cold distilled water and dry between filter papers.

Common slip: recommending titration for a salt that is not sodium, potassium or ammonium, titration is reserved for those soluble bases because the excess cannot be filtered off.

From examples to marks

Notice that every calculation carries its units to the final line, converts cm³ to dm³ before dividing, and uses the mole ratio from a balanced equation. A one-to-one teacher can check that your titration layout and salt-preparation steps are complete. Rehearse these until the method is automatic, then test yourself with the practice questions for this chapter.