Virtual CrezLab - Qualitative Analysis

Experiment A
	Test	Observations
a	To a sample of aqueous zinc chloride, add aqueous sodium hydroxide until a change is observed.	A white ppt. is formed.
b	Add excess of aqueous sodium hydroxide to the mixture from (a).	White ppt. dissolves to form a colourless solution.
c	Add dilute nitric acid to the mixture from (b) until no further change is observed.	White ppt. reappears. When excess nitric acid is added, white ppt. dissolves to form a colourless solution.

For Questions 1 to 8, refer to Experiment A.
What happens when aqueous sodium hydroxide is added to aqueous zinc chloride resulting in the white precipitate?

A.	Displacement
B.	Precipitation
C.	Redox

For Question 1, choose the reason/justification.

A.	The solution is too concentrated with sodium chloride so that sodium chloride comes out of the solution as a solid.
B.	Sodium chloride loses oxygen in forming sodium chloride and zinc chloride.
C.	Sodium ion is more reactive than zinc ion.
D.	Zinc ions combine with the hydroxide ions.

In step(b), a colourless solution is obtained because the white precipitate ____________.

A.	dissolves in sodium hydroxide
B.	reacts with sodium hydroxide

For Question 3, choose the reason/justification.

A.	More solvent is added so there is more space for the white precipitate to dissolve.
B.	No further reaction is seen except for the disappearance of the white precipitate, and no new reagent is added.
C.	The white precipitate forms a new soluble compound with the excess sodium hydroxide.
D.	Hydroxide ion displaces the cation from the white precipitate.

A student concludes that the white precipitate obtained in step(c) is the same as the white precipitate obtained in step (a). Do you agree with the student’s conclusion?

A.	Yes
B.	No

For Question 5, choose the reason/justification.

A.	Different reagents were used.
B.	The acid reacts with the zinc compound to form zinc nitrate which appears as the white precipitate.
C.	The acid reacts with the mixture to reverse the formation of the soluble compound in step(b).
D.	The acid removes the solvent which dissolves the white precipitate in step (b), so the white solid reappears.

The student also concludes that, in step(c), the white precipitate dissolves because more solvent is added. Do you agree with the student's conclusion?

A.	Yes
B.	No

For Question 7, choose the reason/justification.

A.	Hydrogen is more reactive than the cation in the white precipitate, so it displaces the cation from the solid.
B.	No further reaction is seen except for the disappearance of the white precipitate, and no new reagent is added.
C.	The acid contains hydrogen ions which reduce the white precipitate.
D.	The acid reacts with the white precipitate to form a new soluble compound.

Experiment B
	Test	Observations
	Q is an aqueous solution of a compound containing a cation and an anion which could possibly be chloride.
a	To a sample of Q, add an equal volume of aqueous silver nitrate. Divide the mixture into 2 portions.	A white ppt. is formed
b	To the first portion, add dilute nitric acid.	No visible reaction. The white ppt. remains.
c (i)	To the second portion, add aqueous ammonia.	White ppt. dissolves to form a colourless solution.
c (ii)	To the mixture from c(i), add dilute nitric acid until a change is observed.	White ppt. reappears.

For Questions 9 to 18, refer to Experiment B.
What is the white precipitate obtained in step(a)?

A.	AgCl
B.	ZnCl
C.	Zn(NO)

10.

For Question 9, choose the reason/justification.

A.	The silver ions are displaced by the more reactive cations from Q.
B.	The silver ions combine with the chloride ions.
C.	The white precipitate disappears when aqueous ammonia is added in step c(i)
D.	The nitrate ions are displaced by the more reactive chloride ions.

11.

In step (b), the purpose of adding the dilute nitric acid is to

A.	acidify the mixture
B.	determine if the sample contains carbonate ions
C.	dissolve the white solid

12.

For Question 11, choose the reason/justification.

A.	All nitrate salts are soluble.
B.	Carbonate ions form a white precipitate with silver nitrate which will react with nitric acid.
C.	Dilute nitric acid is a strong oxidising agent.
D.	Dilute nitric acid is a good solvent.

13.

Can dilute hydrochloric acid be added in step (b) instead of dilute nitric acid?

A.	Yes
B.	No

14.

For Question 13, choose the reason/justification.

A.	Both nitrate and chloride ions are already present in the mixture.
B.	Chloride ions from dilute hydrochloric acid will interfere with the test.
C.	Dilute hydrochloric acid can be used as it has similar properties and reactions as dilute nitric acid.
D.	Dilute hydrochloric acid will react with the white precipitate formed.

15.

In step c (i), it can be concluded that the white precipitate dissolved to form a colourless solution because it reacted with the aqueous ammonia to form ammonium chloride.

A. True

B. False

16.

For Question 15, choose the reason/justification.

A.	Ammonium chloride is a soluble salt.
B.	Chloride ions still present in solution react with the aqueous ammonia.
C.	The ammonium ion is more reactive than the silver ion, so it displaces the silver ion.
D.	The aqueous ammonia reacts with the white precipitate to produce a soluble compound which is not ammonium chloride.

17.

In step c(ii), the hydrogen ions in the acid reduce the silver ions present to form the white precipitate.

A. True

B. False

18.

For Question 17, choose the reason/justification.

A.	The acid reacts with the aqueous ammonia and the soluble compound in step c(i) produces the same white precipitate as in step (a).
B.	The acid decomposes ammonium chloride to liberate chloride ions which react with the silver ions present in solution.
C.	The acid removes the ammonium ions so that silver ions can react with hydroxide ions.
D.	The acid reacts with aqueous ammonia to form a new insoluble salt.

19.

Experiment C
	Test	Observations
a	To a sample of aqueous copper(II) sulphate, add aqueous ammonia until a change is observed.	A blue ppt. is formed.
b	Add excess aqueous ammonia to the mixture from (a).	Blue ppt. dissolves to form a deep blue solution.
c	Add sulphuric acid to the mixture from (b) until no further change is observed.	A blue ppt. reappears. Blue ppt. dissolves in excess sulphuric acid to form a light blue solution.

For Questions 19 to 26, refer to Experiment C.
What happens when aqueous ammonia is added to aqueous copper(II) sulphate in step(a)?

A.	Displacement
B.	Precipitation
C.	Redox

20.

For Question 19, choose the reason/justification.

A.	Aqueous ammonia gains oxygen in forming ammonium sulphate but copper(II) sulphate loses oxygen in forming copper(II) hydroxide.
B.	Copper(II) ion is more reactive than ammonium ion.
C.	Copper(II) ion combine with hydroxide ions.
D.	Copper(II) ion is less reactive than ammonium ion.

21.

In step (b), why does the light blue precipitate disappear?

A.	It dissolves in aqueous ammonia
B.	It reacts with aqueous ammonia

22.

For Question 21, choose the reason/justification.

A.	Ammonium ion displaces the cation from the light blue precipitate.
B.	More solvent is added so there is more volume for the light blue precipitate to dissolve in it.
C.	There is a chemical reaction between the light blue precipitate and excess ammonia forming product(s) which is/are soluble.
D.	No further reaction is seen except for the disappearance of the light blue precipitate, and no new reagent is added.

23.

The light blue precipitate obtained in step(a) and in step(c) is the same.

A. True

B. False

24.

For Question 23, choose the reason/justification.

A.	Different reagents are used
B.	The acid reacts with the copper(II) compound to form copper(II) sulphate which appears as the light blue solid.
C.	The acid reacts with the mixture to reverse the formation of the soluble compound in step (b).
D.	The acid removes the solvent which dissolves the light blue precipitate in step(b) so the light blue precipitate reappears.

25.

Excess dilute sulphuric acid acts as a solvent for the light blue precipitate in step(c).

A. True

B. False

26.

For Question 25, choose the reason/justification.

A.	The acid reacts with the light blue precipitate to form a new soluble compound.
B.	Hydrogen ion is more reactive than the cation in the light blue precipitate, so it displaces the cation from the solid.
C.	The acid contains hydrogen ions which reduce the light blue solid.
D.	Adding more acid dilutes the mixture.

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