Home / iGCSE Chemistry (0620) Theory (Core):12.2 Acid–base titrations: Exam Style Questions Paper 3
The hydroxides of the Group I metals are soluble in water. Most other metal hydroxides are insoluble in water.
(a) (i) Crystals of lithium chloride can be prepared from lithium hydroxide by titration.
25.0 cm3 of aqueous lithium hydroxide is pipetted into the conical flask.
A few drops of an indicator are added. Dilute hydrochloric acid is added slowly to the alkali until the indicator just changes colour. The volume of acid needed to neutralize the lithium hydroxide is noted.
A neutral solution of lithium chloride, which still contains the indicator, is left. Describe how you could obtain a neutral solution of lithium chloride which does not contain an indicator.
(ii) You cannot prepare a neutral solution of magnesium chloride by the same method. Describe how you could prepare a neutral solution of magnesium chloride.
(b) The concentration of the hydrochloric acid was 2.20 mol/dm3. The volume of acid needed to neutralize the 25.0 cm3 of lithium hydroxide was 20.0 cm3. Calculate the concentration of the aqueous lithium hydroxide.
LiOH + HCl → LiCl + H2O
(c) Lithium chloride forms three hydrates. They are LiCl.H2O, LiCl.2H2O and LiCl.3H2O.
Which one of these three hydrates contains 45.9% of water?
Show how you arrived at your answer.
(a) (i) To remove the indicator from the neutral solution of lithium chloride, you can either:
1. Add activated carbon (charcoal) to adsorb the indicator and then filter the solution.
2. Repeat the titration without the indicator, using the same volume of acid (20.0 cm³) to ensure neutrality.
(ii) To prepare a neutral solution of magnesium chloride:
1. React magnesium metal/carbonate/oxide/hydroxide with hydrochloric acid until no more reaction occurs.
2. Filter the solution to remove any unreacted solid, leaving a neutral MgCl₂ solution.
(b) Using the equation \(\text{LiOH} + \text{HCl} \rightarrow \text{LiCl} + \text{H}_2\text{O}\):
Moles of HCl = \(2.20 \times 0.020 = 0.044\) mol.
Moles of LiOH = 0.044 mol (1:1 ratio).
Concentration of LiOH = \(\frac{0.044}{0.025} = 1.76 \text{ mol/dm}^3\).
(c) For LiCl·2H₂O:
Molar mass = \(42.5 (\text{LiCl}) + 36 (\text{2H}_2\text{O}) = 78.5 \text{ g/mol}\).
% water = \(\frac{36}{78.5} \times 100 = 45.9\%\).
Thus, LiCl·2H₂O is the correct hydrate.