QADI (local)

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QUALITATIVE ANALYSIS DIAGNOSTIC INSTRUMENT*

Instructions

Choose the most suitable option and the reason for your choice in each question by filling the appropriate circles in the answer sheet. If you feel that all options given are inappropriate, indicate the question number and write down what you think the correct answer should be behind the answer sheet.

For Questions 1 to 4, refer to Experiment A:

Experiment A

Step Test Observations

a

b

c

To a sample of aqueous zinc chloride, add aqueous sodium hydroxide until a change is seen.

Add excess of aqueous sodium hydroxide to the mixture from (a).

Add dilute nitric(V) acid (HNO3) to the mixture from (b) until no further change is seen.

A white solid is obtained.

White solid disappears in excess reagent to give a colourless solution.

White solid reappears. When excess acid is added, the solid disappears giving a colourless solution.

1. What happens when aqueous sodium hydroxide is added to aqueous zinc chloride resulting in the white solid?

A Displacement B Precipitation C Redox

Reason/Justification

(1) The solution is too concentrated with sodium chloride so the sodium chloride comes out of the solution as a solid.

(2) Sodium hydroxide loses oxygen in forming sodium chloride and zinc chloride gains oxygen in forming zinc hydroxide.

(3) Sodium ion is more reactive than zinc ion. (4) Zinc ions combine with the hydroxide ions.

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2. In step (b), a colourless solution is obtained because the white solid _____ the excess sodium hydroxide.

A dissolves in B reacts with

(1) More solvent is added so there is more space for the white solid to dissolve.

(2) No further reaction is seen except for the disappearance of the white solid, and no new reagent is added.

(3) Sodium ion displaces the cation from the white solid.

(4) The white solid forms a new soluble compound with the excess sodium hydroxide.

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

A Yes B No

(1) Different reagents were used.

(2) The acid reacts with the excess sodium hydroxide to form sodium nitrate(V) which appears as the white solid.

(3) The acid reacts with the zinc compound to form zinc nitrate(V) which appears as the white solid.

(4) The acid reacts with the mixture to reverse the formation of the soluble compound in step (b).

(5) The acid removes the solvent which dissolves the white solid in step (b), so the white solid reappears.

4. The student also concludes that, in step (c), the white solid dissolves because more solvent (dilute nitric(V) acid) is added. Do you agree with the student’s conclusion?

A Yes B No

(1) Adding more acid dilutes the solution.

(2) Hydrogen is more reactive than the cation in the white solid, so it displaces the cation from the solid.

(3) No further reaction is seen except for the disappearance of the white solid, and no new reagent is added.

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For Questions 5 to 9, refer to Experiment B:

Experiment B

a

b

c(i)

c(ii)

Q is an aqueous solution of a compound which containing a cation and an anion which could possibly be chloride.

To a portion of Q, add an equal volume of aqueous silver nitrate(V) (AgNO3). Divide the mixture into 2 portions.

To the first portion, add dilute nitric(V) acid.

To the second portion, add aqueous ammonia.

To the mixture from c(i), add dilute nitric(V) acid until a change is seen.

A white solid is obtained.

No visible reaction. The white solid remains.

White solid disappears and a colourless liquid is formed.

White solid appears.

5. What is the white solid obtained in step (a)?

A AgCl

B ZnCl2 C Zn(NO3)2

(1) The silver ions are displaced by the more reactive cations from Q. (2) The silver ions combine with the chloride ions.

(3) The white solid disappears when aqueous ammonia is added in step c(i).

(4) The nitrate(V) ions are displaced by the more reactive chloride ions.

6. In step (b), the purpose of adding the dilute nitric(V) acid is to:

A acidify the mixture.

B determine if the sample contains carbonate ions after identifying the gas evolved (if any).

C dissolve the white solid.

(1) All nitrate(V) salts are soluble.

(2) Carbonate ions form a white solid with silver nitrate(V) which will react with nitric(V) acid.

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7. In step (b), can dilute hydrochloric acid be added instead of dilute nitric(V) acid?

A Yes B No

(1) Both nitrate(V) and chloride ions are already present in the mixture. (2) Chloride ions from dilute hydrochloric acid will interfere with the

test.

(3) Dilute hydrochloric acid can be used as it has similar properties and reactions as dilute nitric(V) acid.

(4) Dilute hydrochloric acid will not react with the white solid. (5) Dilute hydrochloric acid will react with the white solid.

8. In step (c (i)), it can be concluded that the white solid disappeared because it reacted with the aqueous ammonia to form ammonium chloride.

A True B False

(1) Ammonium chloride is a soluble salt.

(2) Chloride ions still present in solution react with the aqueous ammonia.

(3) The ammonium ion is more reactive than the silver ion, so it displaces the silver ion.

(4) The aqueous ammonia reacts with the white solid to produce a soluble compound which is not ammonium chloride.

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

A True B False

(1) The acid decomposes the ammonium chloride to liberate chloride ions which react with the silver ions present in solution.

(2) The acid reacts with the aqueous ammonia to produce a new insoluble salt.

(3) The acid reacts with the aqueous ammonia and the soluble compound in step (c (i)) to produce the same white solid as in step (a).

(4) The acid removes the ammonium ions so that silver ions can react with hydroxide ions.

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10. Will you test for oxygen gas when you heat an unknown solid?

A Yes B No

(1) All gases must be tested when a substance is heated.

(2) Oxygen gas is produced only when acids react with compounds containing oxygen.

(3) Oxygen gas will not be liberated during heating, instead it would be used up.

(4) Some compounds decompose on heating to give oxygen gas.

11. A student is given a solution which may contain carbonate ions. She decides to add aqueous barium nitrate(V) (Ba(NO3)2) to the solution first, followed by dilute nitric(V) acid. Is it possible for her to determine whether carbonate ions are present when she carries out the above procedures followed by the identification of the gas involved (if any)?

A Yes B No

(1) An insoluble carbonate would be formed leaving no free carbonate ions in solution to react with the acid.

(2) Dilute nitric(V) acid will react with both a soluble and an insoluble carbonate.

(3) The above procedure is strictly to test for the presence of sulphate(VI) (SO42-) only.

(4) The acid must be added directly to the solution to test for the presence of a carbonate.

(5) The unknown compound will only react with the dilute nitric(V) acid and not with the barium nitrate(V).

12. Instead of dilute nitric(V) acid, can the student add dilute sulphuric(VI) acid (H2SO4) after the addition of barium nitrate(V) in Question 11?

A Yes B No

(1) Any acid can be used as they have similar properties and reactions. (2) Sulphate(VI) ions from dilute sulphuric(VI) acid will interfere with

the test.

(3) Dilute sulphuric(VI) acid will react with the nitrate(V) ions.

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For Questions 13 to 16, refer to Experiment C:

Experiment C

a

b

c

To a sample of aqueous copper(II)

sulphate(VI), add aqueous ammonia until a change is seen.

Add excess of aqueous ammonia to the mixture from (a).

Add dilute sulphuric(VI) acid to the mixture from (b) until no further change is seen.

A light blue solid is obtained.

Light blue solid disappears in excess aqueous ammonia to give a deep blue solution.

A light blue solid appears. When excess acid is added, the solid disappears giving a light blue solution.

13. What happens when aqueous ammonia is added to aqueous copper(II) sulphate(VI) in step (a)?

A Displacement B Precipitation C Redox

(1) Aqueous ammonia gains oxygen in forming ammonium sulphate(VI) but copper(II) sulphate(VI) loses oxygen in forming copper(II) hydroxide.

(2) Copper(II) ions combine with the hydroxide ions.

(3) Copper(II) ion is more reactive than the ammonium ion. (4) Copper(II) ion is less reactive than the ammonium ion.

14. In step (b), why does the light blue solid disappear?

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

(1) Ammonium ion displaces the cation from the light blue solid.

(2) More solvent is added so there is more volume for the light blue solid to dissolve in.

(3) No further reaction is seen except for the disappearance of the light blue solid, and no new reagent is added.

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

A True B False

(1) Different reagents were used.

(2) The acid reacts with the copper(II) compound to form copper(II) sulphate (VI) which appears as the light blue solid.

(3) The acid reacts with the mixture to reverse the formation of the soluble compound in step (b).

(4) The acid removes the solvent which dissolves the light blue solid in step (b), so the light blue solid reappears.

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

A True B False

(1) Adding more acid dilutes the mixture.

(2) Hydrogen is more reactive than the cation in the light blue solid, so it displaces the cation from the solid.

(3) No further reaction is seen except for the disappearance of the light blue solid, and no new reagent is added.

(4) The acid contains hydrogen ions which reduce the light blue solid. (5) The acid reacts with the light blue solid to form a soluble compound.

For Questions 17 to 18, refer to Experiment D:

Experiment D

a

b

R is a colourless aqueous solution of a compound containing a cation and an anion which could possibly be iodide.

To a portion of R, add an equal volume of dilute nitric(V) acid.

To the mixture from (a), add a small volume of aqueous lead(II) nitrate(V) (Pb(NO3)2).

No visible reaction, a colourless solution remains.

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17. The purpose of adding the dilute nitric(V) acid in step (a) is to

A acidify the mixture.

B determine if R reacts with the acid.

(1) All nitrate(V) salts are soluble.

(2) Dilute nitric(V) acid is a strong oxidising agent. (3) Dilute nitric(V) acid is a good solvent.

(4) So that R can react properly with lead(II) nitrate(V).

(5) The acid will react with any carbonate ions if they are present in R.

18. How is the yellow solid formed?

A Displacement B Precipitation

(1) The lead(II) ion is more reactive than the cation from R. (2) The lead(II) ion is less reactive than the cation from R. (3) The lead(II) ions combine with iodide ions.

(4) The iodide ions are more reactive than the nitrate(V) ions. (5) The iodide ions are less reactive than the nitrate(V) ions.

19. A student believes that she has to test for hydrogen and carbon dioxide when she heats an unknown ionic compound (in powder form). Do you agree?

A Yes B No

(1) Ionic compounds have strong bonds and do not decompose on heating.

(2) Gases are only evolved when covalent compounds are heated. (3) Some carbonates decompose on heating to give carbon dioxide but

hydrogen is never liberated when any ionic compound is heated. (4) The ionic compound would react with oxygen on heating to give

water and/or carbon dioxide.

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APPENDIX A

NOTES FOR USE IN QUALITATIVE ANALYSIS

Test for anions

anion test test results

carbonate (CO32-) add dilute acid effervescence, carbon

dioxide produced chloride (Cl-)

[in solution]

acidify with dilute nitric acid, then add aqueous silver nitrate

white ppt.

iodide (I-) [in solution]

acidify with dilute nitric acid, then add aqueous lead(II) nitrate

yellow ppt.

nitrate(NO3-) [in solution]

add aqueous sodium hydroxide then aluminium foil; warm carefully

ammonia produced

sulphate(SO42-) [in solution]

acidify with dilute nitric acid, then add aqueous barium nitrate

white ppt.

Test for aqueous cations

cation effect of aqueous sodium hydroxide effect of aqueous ammonia

aluminium (Al3+) white ppt., soluble in excess giving a colourless solution

white ppt., insoluble in excess

ammonium (NH4+) ammonia produced on warming -

calcium (Ca2+) white ppt., insoluble in excess no ppt. or very slight white ppt

copper(II) (Cu2+) light blue ppt., insoluble in excess light blue ppt., soluble in excess giving a dark blue solution

iron(II) (Fe2+) green ppt., insoluble in excess green ppt., insoluble in excess

iron(III) (Fe3+) red-brown ppt., insoluble in excess red-brown ppt., insoluble in excess

zinc (Zn2+) white ppt., soluble in excess giving a colourless solution

white ppt., soluble in excess giving a colourless solution

Test for gases

gas test and test result

ammonia (NH3) turns damp red litmus paper blue

carbon dioxide (CO2) turns limewater milky

chlorine (Cl2) bleaches damp litmus paper

hydrogen (H2) “pops” with a lighted splint

oxygen (O2) relights a glowing splint