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Question 68 of 68 in Chemical Tests

Chemical Tests

6 marks · Normal

A student tests an unknown powder. To show if the powder contains water and carbonate ions, the following experiments are carried out.

(a)

Describe the chemical test used to show if the powder contains water.

[2 marks]
(b)

A sample of the powder is added to dilute acid and gas is released. The gas is bubbled through limewater, which becomes cloudy.

(i)

Identify the ion present responsible for the release of this gas.

[1 mark]
(ii)

Write the ionic equation for the reaction between this ion and dilute acid.

[2 marks]
(iii)

Suggest a physical test to confirm if the water in the sample is pure.

[1 mark]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) Add the powder to white anhydrous copper(II) sulfate. If water is present, it will turn the powder blue.

(b(i)) Carbonate ion, CO3^2-

(b(ii)) CO3^2-(s) + 2H^+(aq) → CO2(g) + H2O(l)

(b(iii)) Measure its boiling point or freezing point and compare to those of pure water (100 °C/0 °C).

Common mistakes: Giving a flame test for water; Incorrect ionic formula

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Describes adding the powder to anhydrous copper(II) sulfate

A1 (a) · 1 mark · States that the powder turns blue if water is present

B1 (b(i)) · 1 mark · Identifies the carbonate ion (CO3^2-)

B2 (b(ii)) · 2 marks · Writes correct ionic equation for carbonate with acid

B3 (b(iii)) · 1 mark · Suggests accurate physical test for purity (boiling point or freezing point)

Worked solution

a. To test for water chemically, add the powder to white anhydrous copper(II) sulfate. If water (or a hydrated substance) is present, the white solid turns blue.

b(i). The ion responsible is the carbonate ion, CO3^2-.

b(iii). To check if the water is pure, measure its boiling point (should be 100 °C) or freezing point (should be 0 °C) at 1 atm; if different, the water is impure.

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