Acids Bases and Salts Questions and Answers (O level Chemistry)

Question 1

Acids are neutralised by alkalis.

  1. Write the ionic equation for the reaction between an acid and an alkali.
  2. Sodium sulphate is a soluble salt that can be prepared using a titration method.(i) Name a sodium compound and an acid that can be used to make sodium sulphate by titration.

    (ii) Describe how titration is used to prepare a colourless solution of sodium sulphate.

    (iii) Describe how a sample of pure sodium sulphate crystals can be made from aqueous sodium sulphate.

  3. Aqueous sodium sulphate can be used to prepare barium sulphate as shown by the ionic equation below: Ba2+(aq) + SO42–(aq) → BaSO4(s)

    In an experiment, 20.0cm3 of 0.550mol/dm3 of barium nitrate was added to excess aqueous sodium sulphate.

    (i) Calculate the maximum mass of barium sulphate that could be made. [The relative formula mass of BaSO4 is 233.]

    (ii) A mass of 1.92g of dry barium sulphate was obtained. Calculate the percentage yield of barium sulphate.

Solution 1

Acids are neutralised by alkalis.

  1. The ionic equation for the reaction between an acid and an alkali is: H+(aq) + OH(aq) → H2O(l)

    Here is how it is derived:

    • First write the chemical equation for the reaction between any acid and alkali, for example:
    • HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
    • Slip the aqueous substances into ions:
    • H+(aq) Cl(aq) + Na+(aq) OH(aq) → Na+(aq) Cl(aq) + H2O(l)
    • Remove the spectator ions:
    • H+(aq) Cl(aq) + Na+(aq) OH(aq) → Na+(aq) Cl(aq) + H2O(l)
    • To produce:
    • H+(aq) + OH(aq) → H2O(l)
  2. (i) Sulphuric acid AND sodium hydroxide(ii)
    • Place a measured volume of the alkali (sodium hydroxide) in conical flask and the acid (sulphuric acid) in the burette.
    • Slowly add the acid to the alkali until the indicator shows that the alkali has been neutralised.
    • Record the volume of the acid required to neutralise the alkali.
    • Repeat the titration using same volumes but with no indicator.
    • The resulting solution is sodium hydroxide.

    (iii)

    • Evaporate the sodium hydroxide solution by heating it.
    • Continue heating the solution until the first signs of crystallisation appear.
    • Leave the saturated solution to crystallise.
    • Dry the crystals.
  3. Aqueous sodium sulphate can be used to prepare barium sulphate as shown by the ionic equation below: Ba2+(aq) + SO42–(aq) → BaSO4(s)

    In an experiment, 20.0cm3 of 0.550mol/dm3 of barium nitrate was added to excess aqueous sodium sulphate.

    (i)

    • Number of moles = concentration × volume
    • Number of moles of barium nitrate = 0.020dm3 × 0.550
    • = 0.011
    • Mass = number of moles × molar mass
    • Mass = 0.011 × 233
    • Mass = 2.563 g

    (ii) Percentage yield = (actual yield ÷ expected yield) × 100

    = (1.92g ÷ 2.563g) × 100

    = 74.91%


Sydney Chako

Mathematics, Chemistry and Physics teacher at Sytech Learning Academy. From Junior Secondary School to Tertiary Level Engineering Mathematics and Engineering Science.

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