WAEC SSCE - Chemistry - 2002 (Practical)

Question 1 Report

All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account or expeimental procedure is required. All calculations must be done in your answer book.

A is \(0.125\ \text{mol dm}^{-3}\) H\(_2\)SO\(_4\). B is a solution containing X \(\text{g dm}^{-3}\) of NaOH.

(a) Put A into the burette and titrate it against \(20.0\ \text{cm}^3\) or \(25.0\ \text{cm}^3\) portions of B using methyl orange as indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is H\(_2\)SO\(_4\) + 2NaOH\(_{(aq)}\) \(\to\) Na\(_2\)SO\(_4\) + 2H\(_2\)O\(_{(l)}\)

(b) From your results and the information provided above, calculate the;

(i) amount of H\(_2\)SO\(_4\) in the average volume of A used

(ii) Concentration of B in \(\text{mol dm}^{-3}\)

(iii) value of X.

[H = 1: O = 16; Na = 23]

(c) Describe briefly a suitable laboratory procedure for obtaining pure water from the titration mixture. (No diagram is required)

Answer

Answer Details

(a) Titration results (pipette volume = 25.0 cm³ of B; indicator methyl orange, endpoint yellow to orange)

	Rough	1st	2nd	3rd
Final reading (cm³)	24.00	23.25	23.15	23.20
Initial reading (cm³)	0.00	0.00	0.00	0.00
Volume of A used (cm³)	24.00	23.25	23.15	23.20

Average volume of A = (23.25 + 23.15 + 23.20) / 3 = 69.60 / 3 = 23.20 cm³.

(b)(i) Amount of H₂SO₄ in the average volume of A

n(H₂SO₄) = 0.125 × 23.20 / 1000 = 2.90 × 10^-3 mol.

(b)(ii) Concentration of B in mol dm^-3

From the equation H₂SO₄ + 2NaOH gives Na₂SO₄ + 2H₂O, the mole ratio of acid to base is 1 : 2.

n(NaOH) = 2 × 2.90 × 10^-3 = 5.80 × 10^-3 mol in the 25.0 cm³ pipetted.

[B] = 5.80 × 10^-3 / (25.0 / 1000) = 0.232 mol dm^-3.

(b)(iii) Value of X

Molar mass of NaOH = 23 + 16 + 1 = 40 g mol^-1.

X = concentration × molar mass = 0.232 × 40 = 9.28 g dm^-3.

(c) Obtaining pure water from the titration mixture

The neutralised mixture is a solution of sodium tetraoxosulphate(VI) in water. Set it up for simple distillation: heat the mixture so that the water boils off, passes into the condenser and is collected as pure distillate, while the non-volatile sodium sulphate remains behind in the flask.

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Question 2 Report

Credit will be given for strict adherence to the instructions, for observations precisely recorded and for accurate inferences. All tests, observations, and inferences must be clearly entered in your answer book, in ink, at the i.itne time they are made.

C is a sample of copper (II) tetraoxosulphate (VI) crystals. Carry out the following exercises on C. Record your observations and identify any gases evolved. State the conclusion you draw from the result of each test.

(a) Put half of C in a test tube and heat strongly

(b) Make solution of about 10 cm\(^{-3}\) with the sécond half of C and divide it into three portions

(i) To the first portion, add Sodium hydroxide solution in drops and then in excess. Heat the mixture

(ii) To the second portion, add aqueous ammonia in drops and then in excess followed by few drops of moderately concentrated HCI.

(iii) To the third portion, add all the zinc dust provided and stir thoroughly until there is a visible change.

Answer

C is hydrated copper(II) tetraoxosulphate(VI), \(CuSO_4\cdot5H_2O\). The expected observations and inferences are:

Test	Observation	Inference / Conclusion
(a) Heat half of C strongly	The blue crystals turn white and colourless droplets condense on the cooler part of the tube; on stronger heating the residue turns black and a colourless gas of choking smell (\(SO_2/SO_3\)) is given off	Water of crystallisation is present (blue \(\to\) white anhydrous salt, water evolved); on strong heating it decomposes to black \(CuO\)
(b)(i) Solution + \(NaOH\) dropwise then excess, heat	A pale blue gelatinous precipitate forms, insoluble in excess; on heating it turns black	\(Cu^{2+}\) present (\(Cu(OH)_2\), which becomes black \(CuO\) on heating)
(b)(ii) Solution + \(NH_3\) dropwise then excess, then HCl	With a few drops of ammonia a pale blue precipitate forms; in excess ammonia it dissolves to a deep (royal) blue solution; adding \(HCl\) discharges the deep blue colour giving a pale blue/green solution	\(Cu^{2+}\) present, forming the soluble tetraamminecopper(II) complex \([Cu(NH_3)_4]^{2+}\), which is decomposed by acid
(b)(iii) Third portion + zinc dust, stir	The blue colour fades to colourless and a reddish-brown solid (copper) is deposited	Zinc displaces copper (Zn is above Cu in the activity series): \(Zn + CuSO_4 \to ZnSO_4 + Cu\)

General conclusion: C contains \(Cu^{2+}\) and \(SO_4^{2-}\) ions and water of crystallisation, confirming hydrated copper(II) tetraoxosulphate(VI).

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Answer Details

C is hydrated copper(II) tetraoxosulphate(VI), \(CuSO_4\cdot5H_2O\). The expected observations and inferences are:

Test	Observation	Inference / Conclusion
(a) Heat half of C strongly	The blue crystals turn white and colourless droplets condense on the cooler part of the tube; on stronger heating the residue turns black and a colourless gas of choking smell (\(SO_2/SO_3\)) is given off	Water of crystallisation is present (blue \(\to\) white anhydrous salt, water evolved); on strong heating it decomposes to black \(CuO\)
(b)(i) Solution + \(NaOH\) dropwise then excess, heat	A pale blue gelatinous precipitate forms, insoluble in excess; on heating it turns black	\(Cu^{2+}\) present (\(Cu(OH)_2\), which becomes black \(CuO\) on heating)
(b)(ii) Solution + \(NH_3\) dropwise then excess, then HCl	With a few drops of ammonia a pale blue precipitate forms; in excess ammonia it dissolves to a deep (royal) blue solution; adding \(HCl\) discharges the deep blue colour giving a pale blue/green solution	\(Cu^{2+}\) present, forming the soluble tetraamminecopper(II) complex \([Cu(NH_3)_4]^{2+}\), which is decomposed by acid
(b)(iii) Third portion + zinc dust, stir	The blue colour fades to colourless and a reddish-brown solid (copper) is deposited	Zinc displaces copper (Zn is above Cu in the activity series): \(Zn + CuSO_4 \to ZnSO_4 + Cu\)

General conclusion: C contains \(Cu^{2+}\) and \(SO_4^{2-}\) ions and water of crystallisation, confirming hydrated copper(II) tetraoxosulphate(VI).

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Rates Of Chemical Reaction

Acids, Bases And Salts

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Question 3 Report

(a) Suggest how the following liquid reagents can be suitably stored in the laboratory

(i) X which fumes in moist air;

(ii) Y which is slowly decomposed by sunlight in ordinary reagent bottles.

(b) State what is observed when aqueous ammonia is added to:

(i) litmus paper;

(ii) \(\mathrm{Pb(NO_3)_2}\) solution in drops until in excess

(iii) freshly precipitated AgCI in excess.

(c) A salt sample was suspected to be either \(\mathrm{Na_2CO_2}\) or \(\mathrm{NaHCO_3}\). A student who was required to identify it tested a portion for solubility in water and then for effect on litmus paper.

(i) What was observed in each case?

(ii) Give the reason why the student's procedure was unsuitable

(iii) Describe briefly how you would have identified the salt.

Answer

Answer Details

(a) Storage of reagents

(i) X, which fumes in moist air (e.g. concentrated hydrochloric or trioxonitrate(V) acid): store in a tightly stoppered (airtight) bottle so that it does not react with atmospheric moisture.
(ii) Y, which is slowly decomposed by sunlight (e.g. silver trioxonitrate(V), hydrogen peroxide): store in a dark (amber/brown) coloured bottle kept away from light.

(b) Adding aqueous ammonia to

(i) Litmus paper: red litmus turns blue (ammonia solution is alkaline).
(ii) \(Pb(NO_3)_2\) solution, drops then excess: a white precipitate of lead(II) hydroxide forms, which is insoluble in excess ammonia.
(iii) Freshly precipitated \(AgCl\) in excess: the white \(AgCl\) dissolves (it forms the soluble diamminesilver(I) complex \([Ag(NH_3)_2]^+\)).

(c) Distinguishing \(Na_2CO_3\) from \(NaHCO_3\)

(i) In each case the salt dissolves in water and the solution turns red litmus blue (both are soluble and alkaline).
(ii) The procedure is unsuitable because both salts give identical results (both are soluble and both are alkaline to litmus), so the tests cannot tell them apart.
(iii) A better method is to heat a dry sample: sodium hydrogentrioxocarbonate(IV) decomposes readily, giving off \(CO_2\) that turns limewater milky \((2NaHCO_3 \to Na_2CO_3 + H_2O + CO_2)\), whereas sodium trioxocarbonate(IV) is stable to heat and gives no gas. (Alternatively, warm each with dilute acid; \(NaHCO_3\) effervesces more readily.)

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Acids, Bases And Salts

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