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Question 1 Rapport
(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.
(a) Storage of reagents
(b) Adding aqueous ammonia to
(c) Distinguishing \(Na_2CO_3\) from \(NaHCO_3\)
Détails de la réponse
(a) Storage of reagents
(b) Adding aqueous ammonia to
(c) Distinguishing \(Na_2CO_3\) from \(NaHCO_3\)
Question 2 Rapport
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.
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).
Détails de la réponse
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).
Question 3 Rapport
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)
(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.
Détails de la réponse
(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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