Constructing ionic equations using the continuous variation method

Constructing ionic equations using the continuous variation method

1. The ionic equation for the formation of an insoluble salt can be constructed if we know the number of moles of cation and anion reacted together to form 1 mole of the insoluble salt.
2. For example:
   (a) 1 mole of silver chromate(VI) is formed from 2 moles of Ag⁺ ions and 1 mole of CrO₄^2- ions.
   Ionic equation:
   2Ag⁺(aq) + CrO₄^2- (aq) → Ag₂CrO₄ (s)
   (b) 1 mole of lead(II) bromide is formed from 1 mole of Pb²⁺ ions and 2 moles of Br ions. Ionic equation:
   Pb²⁺(aq) + 2Br^–(aq) → PbBr₂(s)
3. The number of moles of cation and anion which combine to form 1 mole of the insoluble salt can be determined from experiment by a continuous variation method.
4. The method involves the following steps.
   1. Carry out a reaction between a fixed volume of reactant A with varying volumes of a second reactant B.
   2. Determine the volume of reactant B required to react completely with the fixed volume of reactant A.
   3. Use the results of the experiment to calculate the number of moles of reactant A and number of moles of reactant B which reacted completely with each other.
   4. Determine the simplest mole ratio of reactant A to reactant B which combine to form one mole of the insoluble salt.
   5. Use the ratio to construct the ionic equation.

Constructing ionic equations example

1. 10 cm³ of 0.25 mol dm^-3 lead(II) nitrate solution reacts completely with 5 cm³ of 1.0 mol dm^-3 potassium iodide solution. A yellow precipitate of lead(II) iodide is formed. Construct the ionic equation for the formation of lead(II) iodide.
Solution:

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Constructing ionic equations experiment

Aim: To construct an ionic equation for the formation of lead(II) chromate(VI).
Problem statement: How to construct an ionic equation for the formation of lead(II) chromate(VI)?
Hypothesis: As the volume of potassium chromate(VI) solution increases, the height of the yellow precipitate increases until all the lead(II) nitrate has reacted.
Variables:
(a) Manipulated variable : Volume of potassium chromate(VI) solution
(b) Responding variable : Height of yellow lead(II) chromate(VI) precipitate
(c) Controlled variables : Volume and concentration of lead(II) nitrate solution, concentration of potassium chromate(VI) solution, size of test tubes
Materials: 0.5 mol dm^-3 lead(II) nitrate solution and 0.5 mol dm^-3 potassium chromate(VI) solution.
Apparatus: 7 test tubes, test tube rack, burettes, metre rule, stopper, dropper, retort stand and clamp.
Procedure:

1. Seven test tubes of the same size are labelled from 1 to 7 and are placed in a test tube rack.
2. A burette is filled with 0.5 mol dm^-3 lead(II) nitrate solution. 5.00 cm³ of the lead(II) nitrate solution is added to each of the seven test tubes.
3. A second burette is filled with 0.5 mol dm^-3 potassium chromate(VI) solution. The potassium chromate(VI) solution is added to each of the seven test tubes according to the volumes specified in Table.
4. Each test tube is stoppered and shaken well. The test tubes are left aside for about half an hour to allow the precipitate to settle.
5. The height of precipitate in each test tube is measured. The colour of the solution above the precipitate in each test tube is noted.
6. All readings and observations are recorded in Table.

Results:

Graph:
A graph of the height of precipitate against the volume of potassium chromate(VI) solution added is plotted.
Interpreting data:
From the graph:
Volume of potassium chromate(VI) solution required to completely react with 5.0 cm³ of lead(II) nitrate solution = 5.00 cm³

Number of moles of Pb²⁺ ions in 5.00 cm³ of 0.5 mol dm^-3 lead(II) nitrate solution in each of the test tubes

Discussion:

1. A yellow precipitate of lead(II) chromate(VI) is formed in each of the seven test tubes.
2. The height of the precipitate increases gradually from test tubes 1 to 5 because more and more lead(II) chromate(VI) is formed due to the increasing amount of potassium chromate(VI) added to the test tubes.
3. The first test tube to achieve maximum constant height of precipitate indicates a complete reaction has taken place. The volume of potassium chromate(VI) solution (5.00 cm³) added is just sufficient to react completely with the lead(II) nitrate in the test tube.
4. The colour of the solution above the precipitate in test tubes 1 to 4 are colourless due to the excess lead(II) nitrate.
5. In test tube 5, a complete reaction has taken place. No reactants are present in excess.
6. The colour of the solution above the precipitate in test tubes 6 and 7 are yellow due to the excess potassium chromate(VI).

Conclusion:
The ionic equation for the formation of lead(II) chromate(VI) can be obtained from a precipitation reaction. The hypothesis is accepted.