1. 1
   A student wants to calculate the volume of sodium hydroxide solution, of concentration 0.100 mol/dm3, needed to exactly neutralise 25.0 cm3 of hydrochloric acid of concentration 0.150 mol/dm3, using the reaction HCl + NaOH -> NaCl + H2O.

   (a)

   Which equation should be used first to find the number of moles of hydrochloric acid present?

   [1 mark]

   • ANumber of moles = concentration x volume (dm3)
   • BNumber of moles = volume (dm3) / concentration
   • CNumber of moles = concentration / volume (dm3)
   • DNumber of moles = concentration + volume (dm3)

   (b)

   What is the mole ratio of hydrochloric acid to sodium hydroxide in this reaction?

   [1 mark]

   • A1 : 1
   • B1 : 2
   • C2 : 1
   • D1 : 3

   (c)

   Calculate the number of moles of hydrochloric acid present in the 25.0 cm3 sample, showing your working.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A different student wants to calculate the volume of sulfuric acid solution, of concentration 0.200 mol/dm3, needed to exactly neutralise 30.0 cm3 of sodium hydroxide solution of concentration 0.150 mol/dm3, using the reaction H2SO4 + 2NaOH -> Na2SO4 + 2H2O.

   (a)

   What is the mole ratio of sulfuric acid to sodium hydroxide in this reaction?

   [1 mark]

   • A1 : 2
   • B1 : 1
   • C2 : 1
   • D2 : 2

   (b)

   Which equation should be used to calculate the volume of sulfuric acid needed, once the number of moles of acid required has been found?

   [1 mark]

   • AVolume (dm3) = number of moles / concentration
   • BVolume (dm3) = number of moles x concentration
   • CVolume (dm3) = concentration / number of moles
   • DVolume (dm3) = number of moles + concentration

   (c)

   Calculate the number of moles of sodium hydroxide present in the 30.0 cm3 sample, showing your working.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A student neutralises 50.0 cm3 of hydrochloric acid of concentration 0.200 mol/dm3 with excess sodium hydroxide, then evaporates and crystallises the resulting solution to obtain solid sodium chloride, NaCl (Mr = 58.5). When the student actually weighs the dry crystals obtained, the mass is less than the calculated (theoretical) value.

   (a)

   Calculate the theoretical (calculated) mass of sodium chloride this reaction should produce, showing your working.

   [3 marks]

   (b)

   Explain why the actual mass of sodium chloride crystals the student obtains in the laboratory is likely to be less than the theoretical mass calculated in part (a).

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A pharmaceutical manufacturer is producing sodium sulfate, Na2SO4 (Mr = 142), on a large scale by reacting 2.00 dm3 of sulfuric acid solution (concentration 0.100 mol/dm3) completely with sodium hydroxide solution (concentration 0.500 mol/dm3), using the reaction H2SO4 + 2NaOH -> Na2SO4 + 2H2O.

   (a)

   A pharmaceutical manufacturer needs to react 2.00 dm3 of sulfuric acid solution, of concentration 0.100 mol/dm3, completely with sodium hydroxide solution of concentration 0.500 mol/dm3, using the reaction H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Calculate the volume, in cm3, of sodium hydroxide solution required, showing full working.

   [6 marks]

   (b)

   Calculate the theoretical (calculated) mass of sodium sulfate, Na2SO4 (Mr = 142), this large-scale reaction should produce, showing full working. Then explain why the manufacturer is unlikely to obtain exactly this mass of product in practice.

   [6 marks]

   Total for question 4: 12 marks

End of questions