Chemistry · Simulation and virtual lab Free
Titration: how strong is the acid?
Titrate an unknown acid with 0.100 mol/dm³ sodium hydroxide, watch the indicator change colour and calculate the concentration.
- You will learn to
- Carry out a rough titration and accurate titrations until two titres agree.
- Calculate the concentration of an acid from the mean titre.
- Choose a suitable indicator using the pH curve.
- You will measure
- Titres (cm³) of 0.100 mol/dm³ NaOH; then the acid concentration.
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: burette, clamp stand, pipette, conical flask, white tile, indicator, NaOH solution.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To find the concentration of an acid by titration with 0.100 mol/dm³ sodium hydroxide, and to choose a suitable indicator.
What to do
- Keep Hydrochloric acid and Phenolphthalein. Read the burette start volume at eye level, at the bottom of the meniscus (see the magnifier), to 0.05 cm³.
- Rough titration: add 1.00 cm³ at a time until the flask turns pink. Tap record end point.
- Tap new trial. Add 1.00 cm³ portions until you are 1 cm³ before the rough titre, then add drop by drop until the first permanent pale pink. Record.
- Repeat until two titres agree within 0.10 cm³ (concordant).
- Calculate: moles of NaOH = 0.100 × titre ÷ 1000; moles of HCl are the same (1 : 1); concentration = moles ÷ 0.0250 dm³. Check your answer.
- Switch on the pH meter and look at the pH curve. Then try Vinegar with methyl orange and with phenolphthalein: which indicator gives the right end point?
Results
Questions
Write the equation for the reaction between hydrochloric acid and sodium hydroxide.
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). The ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).
A mean titre of 23.50 cm³ of 0.100 mol/dm³ NaOH neutralises 25.0 cm³ of HCl. Find the concentration of the acid.
n(NaOH) = 0.100 × 23.50/1000 = 2.35 × 10⁻³ mol = n(HCl). c = 2.35 × 10⁻³ ÷ 0.0250 = 0.094 mol/dm³.
Why is the rough titration not used in the average?
It is done quickly with large additions, so it overshoots the end point. It only tells you roughly where the end point is.
Why is methyl orange unsuitable for titrating vinegar (a weak acid) with sodium hydroxide?
For a weak acid and a strong base the steep part of the pH curve is from about pH 7 to 11. Methyl orange changes between pH 3.1 and 4.4, before the equivalence point, so it changes colour too early.
What does the pH curve show at the equivalence point?
A very steep rise in pH for a very small volume of alkali (one drop). For HCl it jumps from about pH 3 to pH 11; the equivalence point is at pH 7.
Why should you swirl the flask and add the alkali drop by drop near the end point?
To mix the solution so that the colour change is real, and so that you do not add more than one drop past the end point.
For teachers: expected results and how the model works
The hydrochloric acid is between 0.080 and 0.120 mol/dm³; the diluted vinegar (ethanoic acid, Ka = 1.8 × 10⁻⁵) between 0.068 and 0.092 mol/dm³. The burette starts at a random reading to practise subtraction. The pH is calculated from the charge balance, so the curves are realistic (weak acid: buffer region and half-neutralisation at pH ≈ 4.7).
Exam practice Exam set
National-exam style questions on this simulation (17 marks). Try each one on paper first, then open the marking scheme.
25.0 cm³ of hydrochloric acid was neutralised by 22.40 cm³ of 0.100 mol/dm³ sodium hydroxide. (a) Write the equation. (b) Calculate the moles of NaOH used. (c) Calculate the concentration of the acid. (5 marks)
Marking scheme: (a) HCl + NaOH → NaCl + H₂O (1). (b) n = 0.100 × 22.40 ÷ 1000 = 2.24 × 10⁻³ mol (1, 1). (c) Moles of HCl = 2.24 × 10⁻³ (1 : 1 ratio); c = 2.24 × 10⁻³ ÷ 0.0250 = 0.0896 mol/dm³ (1, 1 for working shown).
A learner obtained titres of 23.60 (rough), 22.65, 22.30 and 22.40 cm³. Calculate the mean titre to use. Explain your choice. (3 marks)
Marking scheme: Ignore the rough titre and 22.65 cm³ (1); use the concordant titres, which agree within 0.10 cm³: 22.30 and 22.40 (1); mean = (22.30 + 22.40) ÷ 2 = 22.35 cm³ (1).
Give three precautions when using a burette in a titration. (3 marks)
Marking scheme: Rinse it with the solution it will contain (1); remove the air bubble from the tip before reading (1); read the bottom of the meniscus at eye level (1) (also: add dropwise near the end point, swirl the flask, use a white tile).
Why is phenolphthalein suitable for titrating a strong acid with a strong base, and what colour change is seen when alkali is added from the burette? (3 marks)
Marking scheme: The pH changes very sharply (about pH 3 to 11) at the end point (1), so the indicator changes colour within this range (1); colourless to pink (permanent pale pink) (1).
25.0 cm³ of 0.050 mol/dm³ sulfuric acid is neutralised by sodium hydroxide of concentration 0.100 mol/dm³. Calculate the volume of NaOH needed. (3 marks)
Marking scheme: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O; moles of acid = 0.050 × 0.0250 = 1.25 × 10⁻³ mol (1); moles of NaOH = 2.50 × 10⁻³ mol (1); volume = 2.50 × 10⁻³ ÷ 0.100 = 0.0250 dm³ = 25.0 cm³ (1).
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Titration: how strong is the acid?
Aim: To find the concentration of an acid by titration with 0.100 mol/dm³ sodium hydroxide, and to choose a suitable indicator.
Apparatus (real laboratory): burette, clamp stand, pipette, conical flask, white tile, indicator, NaOH solution.
What to do (with the simulation on www.donatstem.co.rw)
- Keep Hydrochloric acid and Phenolphthalein. Read the burette start volume at eye level, at the bottom of the meniscus (see the magnifier), to 0.05 cm³.
- Rough titration: add 1.00 cm³ at a time until the flask turns pink. Tap record end point.
- Tap new trial. Add 1.00 cm³ portions until you are 1 cm³ before the rough titre, then add drop by drop until the first permanent pale pink. Record.
- Repeat until two titres agree within 0.10 cm³ (concordant).
- Calculate: moles of NaOH = 0.100 × titre ÷ 1000; moles of HCl are the same (1 : 1); concentration = moles ÷ 0.0250 dm³. Check your answer.
- Switch on the pH meter and look at the pH curve. Then try Vinegar with methyl orange and with phenolphthalein: which indicator gives the right end point?
Results
| # | Trial | Initial reading (cm³) | Final reading (cm³) | Titre (cm³) |
|---|
| 1 | | | | |
| 2 | | | | |
| 3 | | | | |
| 4 | | | | |
| 5 | | | | |
| 6 | | | | |
Graph
Optional: plot pH (vertical axis) against volume of NaOH added (cm³) and mark the end point.
Questions
- Write the equation for the reaction between hydrochloric acid and sodium hydroxide. [2]
- A mean titre of 23.50 cm³ of 0.100 mol/dm³ NaOH neutralises 25.0 cm³ of HCl. Find the concentration of the acid. [2]
- Why is the rough titration not used in the average? [2]
- Why is methyl orange unsuitable for titrating vinegar (a weak acid) with sodium hydroxide? [2]
- What does the pH curve show at the equivalence point? [2]
- Why should you swirl the flask and add the alkali drop by drop near the end point? [2]
- 25.0 cm³ of hydrochloric acid was neutralised by 22.40 cm³ of 0.100 mol/dm³ sodium hydroxide. (a) Write the equation. (b) Calculate the moles of NaOH used. (c) Calculate the concentration of the acid. [5]
- A learner obtained titres of 23.60 (rough), 22.65, 22.30 and 22.40 cm³. Calculate the mean titre to use. Explain your choice. [3]
- Give three precautions when using a burette in a titration. [3]
- Why is phenolphthalein suitable for titrating a strong acid with a strong base, and what colour change is seen when alkali is added from the burette? [3]
- 25.0 cm³ of 0.050 mol/dm³ sulfuric acid is neutralised by sodium hydroxide of concentration 0.100 mol/dm³. Calculate the volume of NaOH needed. [3]
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Titration: how strong is the acid?
Expected results
The hydrochloric acid is between 0.080 and 0.120 mol/dm³; the diluted vinegar (ethanoic acid, Ka = 1.8 × 10⁻⁵) between 0.068 and 0.092 mol/dm³. The burette starts at a random reading to practise subtraction. The pH is calculated from the charge balance, so the curves are realistic (weak acid: buffer region and half-neutralisation at pH ≈ 4.7).
Answers
- Write the equation for the reaction between hydrochloric acid and sodium hydroxide. [2]
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). The ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).
- A mean titre of 23.50 cm³ of 0.100 mol/dm³ NaOH neutralises 25.0 cm³ of HCl. Find the concentration of the acid. [2]
n(NaOH) = 0.100 × 23.50/1000 = 2.35 × 10⁻³ mol = n(HCl). c = 2.35 × 10⁻³ ÷ 0.0250 = 0.094 mol/dm³.
- Why is the rough titration not used in the average? [2]
It is done quickly with large additions, so it overshoots the end point. It only tells you roughly where the end point is.
- Why is methyl orange unsuitable for titrating vinegar (a weak acid) with sodium hydroxide? [2]
For a weak acid and a strong base the steep part of the pH curve is from about pH 7 to 11. Methyl orange changes between pH 3.1 and 4.4, before the equivalence point, so it changes colour too early.
- What does the pH curve show at the equivalence point? [2]
A very steep rise in pH for a very small volume of alkali (one drop). For HCl it jumps from about pH 3 to pH 11; the equivalence point is at pH 7.
- Why should you swirl the flask and add the alkali drop by drop near the end point? [2]
To mix the solution so that the colour change is real, and so that you do not add more than one drop past the end point.
- 25.0 cm³ of hydrochloric acid was neutralised by 22.40 cm³ of 0.100 mol/dm³ sodium hydroxide. (a) Write the equation. (b) Calculate the moles of NaOH used. (c) Calculate the concentration of the acid. [5]
(a) HCl + NaOH → NaCl + H₂O (1). (b) n = 0.100 × 22.40 ÷ 1000 = 2.24 × 10⁻³ mol (1, 1). (c) Moles of HCl = 2.24 × 10⁻³ (1 : 1 ratio); c = 2.24 × 10⁻³ ÷ 0.0250 = 0.0896 mol/dm³ (1, 1 for working shown).
- A learner obtained titres of 23.60 (rough), 22.65, 22.30 and 22.40 cm³. Calculate the mean titre to use. Explain your choice. [3]
Ignore the rough titre and 22.65 cm³ (1); use the concordant titres, which agree within 0.10 cm³: 22.30 and 22.40 (1); mean = (22.30 + 22.40) ÷ 2 = 22.35 cm³ (1).
- Give three precautions when using a burette in a titration. [3]
Rinse it with the solution it will contain (1); remove the air bubble from the tip before reading (1); read the bottom of the meniscus at eye level (1) (also: add dropwise near the end point, swirl the flask, use a white tile).
- Why is phenolphthalein suitable for titrating a strong acid with a strong base, and what colour change is seen when alkali is added from the burette? [3]
The pH changes very sharply (about pH 3 to 11) at the end point (1), so the indicator changes colour within this range (1); colourless to pink (permanent pale pink) (1).
- 25.0 cm³ of 0.050 mol/dm³ sulfuric acid is neutralised by sodium hydroxide of concentration 0.100 mol/dm³. Calculate the volume of NaOH needed. [3]
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O; moles of acid = 0.050 × 0.0250 = 1.25 × 10⁻³ mol (1); moles of NaOH = 2.50 × 10⁻³ mol (1); volume = 2.50 × 10⁻³ ÷ 0.100 = 0.0250 dm³ = 25.0 cm³ (1).