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Acids, alkalis and titrations

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Question 12

A student performed a titration to find the concentration of a solution of sulfuric acid, H2SO4\text{H}_2\text{SO}_4H2​SO4​, using a standard solution of sodium hydroxide, NaOH\text{NaOH}NaOH.

a.

The student repeated the titration of sulfuric acid solution with the sodium hydroxide solution and recorded these results:

Run1234
Burette reading after adding alkali / cm3\text{cm}^3cm323.9022.2023.3022.65
Burette reading before adding alkali / cm3\text{cm}^3cm31.250.901.450.70
Volume of alkali added / cm3\text{cm}^3cm322.6521.3021.8521.95

The average (mean) volume of sodium hydroxide added should be calculated using only concordant results. Concordant results are those volumes that differ from each other by 0.10 cm30.10\text{ cm}^30.10 cm3 or less.

Identify the concordant results by stating which runs should be selected.

[1]
b.

Calculate the average (mean) volume of sodium hydroxide added using your selected results.

[1]
c.

The student used the same method to find the concentration of another sulfuric acid solution. The equation for the reaction is:

H2SO4+2NaOH→Na2SO4+2H2O \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} H2​SO4​+2NaOH→Na2​SO4​+2H2​O

These are the student's experimental results:

  • Volume of sulfuric acid solution: 20.0 cm320.0\text{ cm}^320.0 cm3
  • Volume of sodium hydroxide solution added: 18.60 cm318.60\text{ cm}^318.60 cm3
  • Concentration of sodium hydroxide solution: 0.150 mol/dm30.150\text{ mol/dm}^30.150 mol/dm3

The student used these results to calculate the concentration of the sulfuric acid solution:

  • Step 1: amount of NaOH=0.150×18.60100=0.0279 mol\text{amount of NaOH} = \frac{0.150 \times 18.60}{100} = 0.0279\text{ mol}amount of NaOH=1000.150×18.60​=0.0279 mol
  • Step 2: amount of H2SO4=0.0279×2=0.0558 mol\text{amount of H}_2\text{SO}_4 = 0.0279 \times 2 = 0.0558\text{ mol}amount of H2​SO4​=0.0279×2=0.0558 mol
  • Step 3: concentration of H2SO4=0.055818.60×1000=3.00 mol/dm3\text{concentration of H}_2\text{SO}_4 = \frac{0.0558}{18.60} \times 1000 = 3.00\text{ mol/dm}^3concentration of H2​SO4​=18.600.0558​×1000=3.00 mol/dm3

There is one mistake in each step of the calculation. What correction should the student make in Step 1?

[1]
d.

What correction should the student make in Step 2?

[1]
e.

What correction should the student make in Step 3?

[1]

Acids, alkalis and titrations Questions

  1. IGCSE
  2. /Chemistry
  3. /Acids, alkalis and titrations