Acids Bases and Salts Questions and Answers (O level Chemistry)
Question 1
Acids are neutralised by alkalis.
- Write the ionic equation for the reaction between an acid and an alkali.
- 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.
- 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.
- 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)
- (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.
- 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%
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