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Spot the mistake.

In a titration, 25.00cm3of sodium hydroxide solution were pipetted into a conical flask. A 0.100 mol dm-3 solution of sulphuric acid was run into the flask from a burette. An indicator in the flask changed colour when 22.00cm3 of the acid had been added. H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(aq) Calculate the concentration of the sodium hydroxide solution.

Moles of H2SO4 added to the conical flask: Amount = concentration x volume Amount = 0.100 x 22.00 x 10-3 Amount = 2.200 x 10-3 mol

Moles of NaOH in the conical flask: 1 mol of H2SO4 reacts with 1 mol of NaOH -3 -3 2.200 x 10 mol of H2SO4 reacts with 2.200 x 10 mol of NaOH Amount of NaOH = 2.200 x 10-3 mol

Concentration of NaOH: Concentration = amount volume Concentration = 2.200 x 10-3 25.00 x 10-3 Concentration = 0.0880 mol dm-3

1 Spot the mistake.

In a titration, 25.00cm3of sodium hydroxide solution were pipetted into a conical flask. A 0.100 mol dm-3 solution of sulphuric acid was run into the flask from a burette. An indicator in the flask changed colour when 22.00cm3 of the acid had been added. H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(aq) Calculate the concentration of the sodium hydroxide solution.

Moles of H2SO4 added to the conical flask: Amount = concentration x volume Amount = 0.100 x 22.00 Amount = 2.200 mol

Moles of NaOH in the conical flask: 1 mol of H2SO4 reacts with 2 mol of NaOH 2.200 mol of H2SO4 reacts with 2 x 2.200 mol of NaOH Amount of NaOH = 4.400 mol

Concentration of NaOH: Concentration = amount volume Concentration = 4.400 25.00 Concentration = 0.176 mol dm-3

2 Spot the mistake.

In a titration, 25.00cm3of sodium hydroxide solution were pipetted into a conical flask. A 0.100 mol dm-3 solution of sulphuric acid was run into the flask from a burette. An indicator in the flask changed colour when 22.00cm3 of the acid had been added. H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(aq) Calculate the concentration of the sodium hydroxide solution.

Moles of H2SO4 added to the conical flask: Amount = concentration x volume Amount = 0.100 x 25.00 x 10-3 Amount = 2.500 x 10-3 mol

Moles of NaOH in the conical flask: 1 mol of H2SO4 reacts with 2 mol of NaOH -3 2.500 x 10 mol of H2SO4 reacts with 2 x 2.500 x 10-3 mol of NaOH Amount of NaOH = 5.000 x 10-3 mol

Concentration of NaOH: Concentration = amount volume Concentration = 5.000 x 10-3 22.00 x 10-3 Concentration = 0.227 mol dm-3

3 Spot the mistake.

In a titration, 25.00cm3of sodium hydroxide solution were pipetted into a conical flask. A 0.100 mol dm-3 solution of sulphuric acid was run into the flask from a burette. An indicator in the flask changed colour when 22.00cm3 of the acid had been added. H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(aq) Calculate the concentration of the sodium hydroxide solution.

Moles of H2SO4 added to the conical flask: Amount = concentration x volume Amount = 0.100 x 22.00 x 10-3 Amount = 2.200 x 10-3 mol

Moles of NaOH in the conical flask: 1 mol of H2SO4 reacts with 2 mol of NaOH -3 2.200 x 10 mol of H2SO4 reacts with 2 x 2.200 x 10-3 mol of NaOH Amount of NaOH = 4.400 x 10-3 mol

Concentration of NaOH: Concentration = amount volume Concentration = 4.400 x 10-3 25.00 Concentration = 1.76 x 10-4 mol dm-3

4 Spot the mistake.

In a titration, 25.00cm3of sodium hydroxide solution were pipetted into a conical flask. A 0.100 mol dm-3 solution of sulphuric acid was run into the flask from a burette. An indicator in the flask changed colour when 22.00cm3 of the acid had been added. H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(aq) Calculate the concentration of the sodium hydroxide solution.

Moles of H2SO4 added to the conical flask: Amount = concentration x volume Amount = 0.100 x 22.00 Amount = 2.200 mol

Moles of NaOH in the conical flask: 1 mol of H2SO4 reacts with 2 mol of NaOH 2.200 mol of H2SO4 reacts with 2 x 2.200 mol of NaOH Amount of NaOH = 4.400 mol

Concentration of NaOH: Concentration = amount volume Concentration = 4.400 25.00 x 10-3 Concentration = 176 mol dm-3

5 Spot the mistake.

In a titration, 25.00cm3of sodium hydroxide solution were pipetted into a conical flask. A 0.100 mol dm-3 solution of sulphuric acid was run into the flask from a burette. An indicator in the flask changed colour when 22.00cm3 of the acid had been added. H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(aq) Calculate the concentration of the sodium hydroxide solution.

Moles of H2SO4 added to the conical flask: Amount = concentration x volume Amount = 0.100 x 22.00 x 10-3 Amount = 2.200 x 10-3 mol

Moles of NaOH in the conical flask: 1 mol of H2SO4 reacts with 2 mol of NaOH -3 - 2.200 x 10 mol of H2SO4 reacts with 2 x 2.200 x10 3 mol of NaOH Amount of NaOH = 4.400 x 10-3 mol

Concentration of NaOH: Concentration = amount volume Concentration = 4.400 x 10-3 25.00 x 10-3 Concentration = 0.2 mol dm-3

6 Spot the mistake.

In a titration, 25.00cm3of sodium hydroxide solution were pipetted into a conical flask. A 0.100 mol dm-3 solution of sulphuric acid was run into the flask from a burette. An indicator in the flask changed colour when 22.00cm3 of the acid had been added. H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(aq) Calculate the concentration of the sodium hydroxide solution.

Moles of H2SO4 added to the conical flask: Amount = concentration x volume Amount = 0.100 x 22.00 x 10-3 Amount = 2.200 x 10-3 mol

Moles of NaOH in the conical flask: 1 mol of H2SO4 reacts with 2 mol of NaOH -3 2.200 x 10 mol of H2SO4 reacts with 2 x 2.200 x 10-3 mol of NaOH Amount of NaOH = 4.400 x 10-3 mol

Concentration of NaOH: Concentration = amount x volume Concentration = 4.400 x 10-3 x 25.00 x 10-3 Concentration = 1.10 x 10-4 mol dm-3

7 What were the problems? 1. Stoichiometric equation ignored. 2. Forgot to convert to dm3 for both solutions. 3. 22 and 25 mixed up. 4. 25 instead of 25 x 10-3 5. 22 instead of 22 x 10-3 6. Only 1 sf instead of 3. 7. Relationship not rearranged in the last step.

Correct answer is 0.176 mol dm-3

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