Making Soluble Salts, Neutralisation & Strong vs Weak Acids
Acids produce hydrogen ions (H⁺) in aqueous solution, whereas alkalis are soluble bases that produce hydroxide ions (OH⁻); in any neutralisation reaction, these ions combine according to the ionic equation H⁺(aq) + OH⁻(aq) -> H₂O(l). Pure, dry crystals of a soluble salt are prepared either by reacting an acid with an excess of an insoluble base (such as copper oxide) or by performing an accurate titration between an acid and a soluble alkali (such as sodium hydroxide). Examiners regularly test the 6-mark required practical method for making copper sulfate crystals, titration mole calculations, and the Higher Tier distinction between strong/weak acids and concentrated/dilute acids.
- Reactions of Acids to Make Salts
- 1. Acid + Metal -> Salt + Hydrogen (for metals above hydrogen in reactivity series). 2. Acid + Metal Oxide / Metal Hydroxide (Base/Alkali) -> Salt + Water. 3. Acid + Metal Carbonate -> Salt + Water + Carbon Dioxide. Hydrochloric acid (HCl) makes chlorides; Sulfuric acid (H₂SO₄) makes sulfates; Nitric acid (HNO₃) makes nitrates.
- Neutralisation & the pH Scale
- Acids have pH 0 to 6 (produce H⁺(aq) ions); neutral solutions (pure water) have pH 7; alkalis have pH 8 to 14 (soluble bases that produce OH⁻(aq) ions). Ionic equation for neutralisation: H⁺(aq) + OH⁻(aq) -> H₂O(l). Measured using a calibrated electronic pH probe (most accurate/precise) or Universal Indicator.
- Required Practical: Making a Soluble Salt from an Insoluble Base
- To make pure dry copper(II) sulfate crystals from copper(II) oxide and dilute sulfuric acid: 1. Gently warm dilute H₂SO₄ in a beaker (speeds up reaction). 2. Add insoluble CuO powder a spatula at a time with stirring until in excess (black solid remains at bottom) so all acid reacts. 3. Filter using filter paper and funnel to remove unreacted CuO. 4. Heat blue CuSO₄ filtrate gently over a water bath/electric heater to evaporate ~half the water. 5. Leave in an evaporating basin to cool and crystallise slowly. 6. Pat crystals dry between filter papers.
Moles, Mass & Solution Concentration Step-by-Step Calculator
Enter any mass, relative formula mass (Mr), and solution volume in cm³ to see the exact step-by-step calculation.
Step 1: Calculate Moles (n = m ÷ Mᵣ):
n = 11.7 g ÷ 58.5 g/mol = 0.2000 mol
Step 2, Convert Volume from cm³ to dm³ (÷ 1000):
V = 250 cm³ ÷ 1000 = 0.250 dm³
Step 3: Calculate Concentration (c = n ÷ V):
c = 0.2000 mol ÷ 0.250 dm³ = 0.800 mol/dm³ (or 46.80 g/dm³)
Exam Trap: Always divide cm³ by 1000 to convert into dm³ before using c = n / V!
Key Concepts & Exam Definitions (6)
- Reactions of Acids to Make Salts
- 1. Acid + Metal -> Salt + Hydrogen (for metals above hydrogen in reactivity series). 2. Acid + Metal Oxide / Metal Hydroxide (Base/Alkali) -> Salt + Water. 3. Acid + Metal Carbonate -> Salt + Water + Carbon Dioxide. Hydrochloric acid (HCl) makes chlorides; Sulfuric acid (H₂SO₄) makes sulfates; Nitric acid (HNO₃) makes nitrates.
- Neutralisation & the pH Scale
- Acids have pH 0 to 6 (produce H⁺(aq) ions); neutral solutions (pure water) have pH 7; alkalis have pH 8 to 14 (soluble bases that produce OH⁻(aq) ions). Ionic equation for neutralisation: H⁺(aq) + OH⁻(aq) -> H₂O(l). Measured using a calibrated electronic pH probe (most accurate/precise) or Universal Indicator.
- Required Practical: Making a Soluble Salt from an Insoluble Base
- To make pure dry copper(II) sulfate crystals from copper(II) oxide and dilute sulfuric acid: 1. Gently warm dilute H₂SO₄ in a beaker (speeds up reaction). 2. Add insoluble CuO powder a spatula at a time with stirring until in excess (black solid remains at bottom) so all acid reacts. 3. Filter using filter paper and funnel to remove unreacted CuO. 4. Heat blue CuSO₄ filtrate gently over a water bath/electric heater to evaporate ~half the water. 5. Leave in an evaporating basin to cool and crystallise slowly. 6. Pat crystals dry between filter papers.
- Strong vs Weak Acids & pH (Higher Tier)
- Strong acids (hydrochloric, nitric, sulfuric) completely ionise in water: HCl(aq) -> H⁺(aq) + Cl⁻(aq). Weak acids (ethanoic, citric, carbonic) only partially ionise in water in a reversible equilibrium: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq). At the same concentration (mol/dm³), a strong acid has a higher H⁺ concentration and therefore a lower pH and faster reaction rate than a weak acid.
- Required Practical: Acid-Alkali Titration (Triple Science)
- 1. Use a volumetric pipette and safety filler to measure an exact volume (e.g., 25.0 cm³) of alkali into a conical flask on a white tile. 2. Add a few drops of a single-colour-change indicator (phenolphthalein or methyl orange, not universal indicator). 3. Fill a burette with acid and record initial meniscus reading to 2 d.p. 4. Run acid in while swirling, adding dropwise near the end-point until a permanent colour change occurs. 5. Repeat to obtain concordant titres (within 0.10 cm³ of each other).
- Titration Concentration Calculations (Triple / Higher)
- Core equations: Moles (mol) = Concentration (mol/dm³) x Volume (dm³), where Volume in dm³ = Volume in cm³ / 1000. Use the stoichiometric mole ratio from the balanced chemical equation (e.g., 2NaOH + H₂SO₄ -> Na₂SO₄ + 2H₂O is a 2:1 ratio) to find unknown moles, then divide by volume in dm³. To convert mol/dm³ to g/dm³, multiply by Relative Formula Mass (Mr).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Write the ionic equation for any neutralisation reaction between an aqueous acid and an aqueous alkali.
Name the salt and other products formed when calcium carbonate reacts with dilute nitric acid.
Why can you not use the 'excess insoluble base + filtration' method to make sodium chloride from sodium hydroxide and hydrochloric acid?
Why do you use phenolphthalein or methyl orange in a titration rather than Universal Indicator?
If the pH of a solution changes from pH 5 to pH 2, how does the concentration of H⁺(aq) ions change?
Explain why 0.1 mol/dm³ hydrochloric acid reacts faster with magnesium ribbon than 0.1 mol/dm³ ethanoic acid.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Describe a complete laboratory method to prepare pure, dry crystals of magnesium sulfate starting from insoluble magnesium oxide powder and dilute sulfuric acid. [6 marks]
Click “Check Mark Scheme” to compare your answer, 1. Measure a fixed volume of dilute sulfuric acid into a beaker and warm it gently over a water bath (to increase the rate of reaction) [1 mark]. 2. Add magnesium oxide powder a spatula at a time while stirring with a glass rod until the magnesium oxide is in excess (unreacted white powder remains at the bottom of the beaker), ensuring all the acid has reacted [1 mark]. 3. Filter the mixture using a filter funnel and filter paper to remove the excess insoluble magnesium oxide [1 mark]. 4. Pour the clear magnesium sulfate filtrate into an evaporating basin and heat gently over a boiling water bath until crystals start to appear around the edge (crystallisation point) [1 mark]. 5. Remove from the heat and leave the saturated solution in a warm place to cool slowly so magnesium sulfate crystals form [1 mark]. 6. Filter off the crystals and pat them dry between sheets of filter paper (or leave in a warm oven) [1 mark].
- Q2
Explain the difference between a strong acid and a weak acid, and explain why 1.0 mol/dm³ hydrochloric acid has a pH of 0 whereas 1.0 mol/dm³ ethanoic acid has a pH of 2.4. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. A strong acid (such as hydrochloric acid) completely ionises (dissociates) into ions when dissolved in water: HCl -> H⁺ + Cl⁻ [1 mark]. 2. A weak acid (such as ethanoic acid) only partially ionises in aqueous solution in a reversible reaction: CH₃COOH ⇌ H⁺ + CH₃COO⁻, so most molecules remain un-ionised [1 mark]. 3. Therefore, for the same 1.0 mol/dm³ concentration of acid, hydrochloric acid has a much higher concentration of H⁺(aq) ions than ethanoic acid [1 mark]. 4. Because every 10-fold increase in H⁺(aq) concentration lowers the pH by 1 unit, the higher H⁺ concentration in hydrochloric acid gives it a much lower pH (0 vs 2.4) [1 mark].
- Q3
In a titration, 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide (NaOH) solution was exactly neutralised by 20.0 cm³ of dilute sulfuric acid (H₂SO₄). The equation is: 2NaOH + H₂SO₄ -> Na₂SO₄ + 2H₂O. Calculate the concentration of the sulfuric acid in mol/dm³ and in g/dm³. (Relative atomic masses: H = 1, O = 16, S = 32). (Triple Science) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Convert volume of NaOH to dm³: 25.0 / 1000 = 0.0250 dm³; Moles of NaOH = Concentration x Volume = 0.200 x 0.0250 = 0.00500 mol [1 mark]. 2. From the balanced equation, the mole ratio of NaOH to H₂SO₄ is 2 : 1, so moles of H₂SO₄ = 0.00500 / 2 = 0.00250 mol [1 mark]. 3. Convert volume of H₂SO₄ to dm³: 20.0 / 1000 = 0.0200 dm³; Concentration of H₂SO₄ = 0.00250 / 0.0200 = 0.125 mol/dm³ [1 mark]. 4. Calculate relative formula mass (Mr) of H₂SO₄ = (2 x 1) + 32 + (4 x 16) = 98 [1 mark]. 5. Concentration in g/dm³ = 0.125 mol/dm³ x 98 g/mol = 12.25 g/dm³ [1 mark].
- Q4
Describe three techniques a student should use during an acid-alkali titration to ensure the final titre volume is as accurate and reliable as possible. (Triple Science) [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Place the conical flask on a white tile and swirl continuously while adding acid dropwise near the end-point so the exact drop that causes the permanent indicator colour change is seen clearly [1 mark]. 2. Read the bottom of the meniscus in the burette and pipette at eye level to avoid parallax error (and remove the filter funnel from the top of the burette before taking readings) [1 mark]. 3. Repeat the titration after a rough trial until at least two concordant titres (within 0.10 cm³ of each other) are obtained, and calculate the mean using only the concordant titres [1 mark].
Common Exam Questions (4)
Why do you add excess insoluble metal oxide when making a soluble salt like copper(II) sulfate?
You add the insoluble black copper(II) oxide powder in excess (until no more dissolves and black powder remains at the bottom) to ensure that 100% of the sulfuric acid has reacted and been neutralised. The unreacted insoluble copper(II) oxide is then easily removed by filtration, leaving a pure aqueous solution of copper(II) sulfate.
Why do you gently heat the salt solution and then leave it to cool rather than boiling it completely dry?
Heating the filtrate gently over a water bath evaporates some of the water until the solution reaches its crystallisation point (saturated solution). Leaving it to cool slowly allows large, regular hydrated copper(II) sulfate crystals to form; boiling it completely to dryness would drive off the water of crystallisation and leave anhydrous white powder instead of crystals.
What is the difference between a strong acid and a concentrated acid? (Higher Tier)
Strength refers to degree of ionisation in water: a strong acid (HCl, H₂SO₄, HNO₃) completely ionises (dissociates) in aqueous solution, whereas a weak acid (ethanoic acid, citric acid, carbonic acid) only partially ionises in water (reversible reaction). Concentration refers to the amount of acid (in moles or grams) dissolved per dm³ of solution. You can have a dilute strong acid or a concentrated weak acid.
How does H⁺ ion concentration relate to changes on the pH scale? (Higher Tier)
The pH scale is logarithmic (base 10). As the pH decreases by 1 unit, the hydrogen ion (H⁺) concentration of the solution increases by a factor of 10 (10¹). Therefore, an acid of pH 1 has a H⁺ concentration 100 times (10²) greater than an acid of pH 3, and 1,000 times (10³) greater than an acid of pH 4.
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