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QUALITATIVE ANALYSIS DIAGNOSTIC INSTRUMENT*

Instructions

Choose the most suitable option and the reason for your choice in each question byfilling 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 givinga 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 sodiumchloride comes out of the solution as a solid.

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

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

* Developed by Daniel Tan, National Institute of Education, Singapore.

Email: [email protected]

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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 todissolve.

(2) No further reaction is seen except for the disappearance of the whitesolid, 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 asthe white solid obtained in step (a). Do you agree with the students

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 thesoluble 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 students

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 whitesolid, and no new reagent is added.

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

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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.

(3) Dilute nitric(V) acid is a strong oxidising agent.(4) Dilute nitric(V) acid is a good solvent.(5)

So that Q can react properly with silver nitrate(V).

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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 andreactions 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 itreacted 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 itdisplaces the silver ion.(4) The aqueous ammonia reacts with the white solid to produce asoluble 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 chlorideions which react with the silver ions present in solution.

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

(3) The acid reacts with the aqueous ammonia and the soluble compoundin 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 reactwith hydroxide ions.

(5) The white solid is the element, silver.

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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 beused 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 theidentification of the gas involved (if any)?

A Yes B No


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

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

(3) The above procedure is strictly to test for the presence ofsulphate(VI) (SO4

2-) only.

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

(5) The unknown compound will only react with the dilute nitric(V) acidand 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.(4) The addition of barium nitrate(V) has already invalidated (spoilt) the

test for carbonate or sulphate(IV) 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 givinga 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 lightblue solid, and no new reagent is added.

(4) There is a chemical reaction between the light blue solid and excessammonia forming product(s) which is/are soluble.

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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 thesoluble compound in step (b).

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

16. Excess dilute sulphuric(VI) acid acts as a solvent for the light blue solid instep (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 lightblue 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.

Yellow solid is obtained.

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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 onheating.(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 givewater and/or carbon dioxide.

(5) Those ionic compounds containing carbonate, hydrogen or hydroxideions will liberate carbon dioxide or hydrogen on heating.

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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 addaqueous 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 givinga colourless solution

white ppt., soluble in excessgiving 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

sulphur dioxide (SO2) turns aqueous potassium dichromate(VI) green

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