Free Concentration & titration questions for IGCSE Chemistry (0620), part of Stoichiometry. Each question has the answer and the marking points an examiner would look for.
Generate a full Concentration & Titration worksheet →All Stoichiometry questions25.0 cm³ of 0.100 mol/dm³ sodium hydroxide is exactly neutralised by 27.5 cm³ of hydrochloric acid, HCl.
What is the concentration of the acid?
Answer: B
Which piece of apparatus is used to add acid gradually in a titration?
Answer: C
25.0 cm³ of 0.100 mol/dm³ aqueous sodium hydroxide is titrated with dilute sulfuric acid. The mean titre is 23.91 cm³.
2NaOH + H2SO4 → Na2SO4 + 2H2O
Calculate the concentration of the sulfuric acid in mol/dm³.
Name a suitable indicator and give the colour change at the end-point.
A sample of hydrated sodium carbonate, Na2CO3•xH2O, of mass 2.50 g is heated until all the water is removed. The mass of the anhydrous salt is 0.927 g.
Calculate the value of x.
Describe how the student can make sure that all of the water has been removed.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.0523 mol/dm³ moles NaOH = 0.0025 M1 moles H2SO4 = 0.00125 M1 0.0523 A1 | 3 |
| (b) | methyl orange: yellow to orange/red (or phenolphthalein: pink to colourless) suitable indicator B1 correct colour change for that indicator, in the right direction B1 | 2 |
| (c) | 10 mass of water = 1.57 g B1 moles water = 0.0874; moles salt = 0.00874 M1 ratio water : salt M1 10 A1 | 4 |
| (d) | Heat, cool and weigh repeatedly until the mass is constant heat to constant mass B1 weigh after cooling, repeat until two readings agree B1 | 2 |
25.0 cm³ of sodium hydroxide solution is titrated with 0.100 mol/dm³ hydrochloric acid. The average titre is 20.00 cm³. NaOH + HCl → NaCl + H₂O
Calculate the number of moles of hydrochloric acid used.
Calculate the concentration of the sodium hydroxide in mol/dm³.
Name a suitable indicator and state the colour change at the end-point.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.002 mol 0.002 mol B1 | 1 |
| (b) | 0.08 mol/dm³ moles NaOH = 0.002 C1 ÷ 0.0250 = 0.08 mol/dm³ A1 | 2 |
| (c) | methyl orange yellow → orange/red (or phenolphthalein pink → colourless) Any 2 of: methyl orange; yellow to orange (or phenolphthalein; pink to colourless) B2 | 2 |
This question is about the reaction between aqueous sodium hydroxide and hydrochloric acid.
Write the ionic equation for the neutralisation of an acid by an alkali.
Write the symbol equation for the reaction between sodium hydroxide and hydrochloric acid.
A student titrates 25.0 cm³ of aqueous sodium hydroxide with 0.100 mol/dm³ hydrochloric acid, using methyl orange indicator. Fig. 5.1 shows the burette readings at the start and end of the first (rough) titration.
Use Fig. 5.1 to find the volume of acid added in the rough titration.
Name the piece of apparatus used to measure exactly 25.0 cm³ of sodium hydroxide.
Give the colour change at the end-point.
The mean volume of acid used is 25.00 cm³. Calculate the concentration of the sodium hydroxide in mol/dm³.
Convert your answer to (c)(iv) into a concentration in g/dm³. [Mᵣ: NaOH, 40]
Explain why the student repeats the titration until two concordant results are obtained.
The student wants to make pure sodium sulfate crystals. Explain why the indicator method must be changed before evaporating the solution.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | H⁺ + OH⁻ → H₂O H⁺ + OH⁻ → H₂O B1 | 1 |
| (b) | NaOH + HCl → NaCl + H₂O correct formulae B1 balanced B1 | 2 |
| (c)(i) | 25.40 cm³ readings 0.80 and 26.20 cm³ B1 25.40 cm³ B1 | 2 |
| (ii) | volumetric pipette (volumetric) pipette B1 | 1 |
| (iii) | yellow to orange / red yellow to orange (or red) B1 | 1 |
| (iv) | 0.100 mol/dm³ moles of acid = 0.100 × 25.00/1000 = 2.50 × 10−3 M1 moles NaOH = moles acid = 2.50 × 10−3 M1 ÷ 0.0250 = 0.100 mol/dm³ A1 | 3 |
| (v) | 4.00 g/dm³ 0.100 × 40 = 4.00 g/dm³ B1 | 1 |
| (vi) | to check reliability / identify anomalies to make sure the results are repeatable / to identify anomalous results B1 | 1 |
| (d) | repeat without indicator using the same volumes, indicator would contaminate repeat the titration with the same volumes but without the indicator B1 otherwise the crystals would be contaminated with indicator B1 | 2 |
A solution is made by dissolving sodium hydroxide in water. (Mr NaOH = 40)
Calculate the mass of sodium hydroxide needed to make 500 cm³ of a 0.50 mol/dm³ solution.
Convert this concentration into g/dm³.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 10 g moles = 0.5 × 0.5 C1 × 40 = 10 g A1 | 2 |
| (b) | 20 g/dm³ 20 g/dm³ B1 | 1 |
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