8.1.4 - Laboratory Activity : Constructing an ionic equation through the continuous variation method

Laboratory Activity 8.1.4: Constructing an ionic equation through the continuous variation method

Aim: To form a balanced ionic equation to represent the precipitation of barium chromate(VI) through the continuous variation method.
Problem statement: How can the ionic equation to represent the reaction between barium chloride and potassium chromate(VI) be determined?
Hypothesis: The height of the precipitate increases when the volume of BaCl2 solution is increased until it reaches a maximum height.
Variable:

	»	Fixed variable : Size of test tubes, concentration and volume of BaCrO4 solution, concentration of BaCl2 solution.
	»	Manipulated variable : Volume of BaCl2 solution
	»	Responding variable : Height of precipitate

Material: » Potassium chromate (VI) solution (0.5M) » Barium chloride solution (0.5M)

Apparatus: » 8 test tube » Test tube rack » Burette » Ruler » Retort stand

Procedure:

	►	The animation below shows the arrangement and the results of the experiment.
	►	1. 8 test tubes of the same size and height are prepared. 2. 5cm3 of potassium chromate(VI), K2CrO4 0.5M is poured into each test tube by using a burette. 3. Using another burette filled with barium chloride, BaCl2 solution. 1cm3 of BaCl2 solution, 0.5M is added into the first test tube, 2cm3 into the second test tube, and so on until 8cm3 into the eighth test tube. 4. Each test tube is shaken and left to stand on the rack until the precipitate settles. 5. The height of the precipitate in each test tube is measured with a ruler.

Observation:

	►	A yellow precipitate of barium chromate(VI) is formed when BaCl2 is added to K2CrO4 solution.
	►	The colour of the solution becomes lighter from test tubes 1 to 4 until it becomes colourless in test tube 5 to 8.

Results:

►

Test tube number 1 2 3 4 5 6 7 8 Volume of potassium chromate(VI) solution, K2CrO4 (cm3) 5 5 5 5 5 5 5 5 Volume of barium chloride solution, BaCl2 (cm3) 1 2 3 4 5 6 7 8 Height of precipitate (cm) 0.9 1.8 2.7 3.6 4.5 4.5 4.5 4.5

Discussion:

►

A graph of precipitation height is plotted against the volume of BaCl2 solution.

Calculation:

	►	The maximum height is obtained in test tube 5, showing that the reaction is complete in test tube 5. Thus 5cm3 of K2CrO4(0.5 M) reacts completely with 5cm3 of BaCl2(0.5M). The number of moles of K2CrO4 and BaCl2 is calculated as follows: Number of moles of K2CrO4 = $\frac{\mathrm{MV}}{1000}$ = $\frac{0.5\mathrm{\times }5}{1000}$ = 0.0025 moles Number of moles of BaCl2 = $\frac{\mathrm{MV}}{1000}$ = $\frac{0.5\mathrm{\times }5}{1000}$ = 0.0025 moles That is, if 0.0025 mole K2CrO4 reacts with 0.0025 moles BaCl2, then 1 mole K2CrO4 will react with 1 mole BaCl2.
	►	The chemical equation is written as: K2CrO4(aq) + BaCl2(aq) → BaCrO4(s) + 2KCl(aq) or the ionic equation involving Ba2+ and CrO2-4 ions only is: Ba2+(aq) + CrO2-4(aq) → BaCrO4(s)

Discussion:

	►	The yellow colour of the aqueous solution indicates excess CrO2-4 ions which have not reacted.
	►	The colourless solution(test tube 5 -8) shows that all the CrO2-4 ions have reacted with BaCl2 solution.

Conclusion:

	►	The height of the precipitate increases until all the CrO2-4 ions have reacted with the Ba2+ ions. ○ The ionic equation is: Ba2+(aq) + CrO2-4(aq) → BaCrO4(s)
	►	The hypothesis is accepted.

↶ 8.1 Salts

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