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Frage 1 Bericht
All your burette readings (initials and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
State what would be observed if the following reactions are carried out in the laboratory:
(i) methyl orange is dropped into a solution of lime juice:
(ii) hydrogen sulphide gas is bubbled through Iron (III) chloride solution:
(iii) sulphur (IV) oxide gas is bubbled into acidified solution of KMnO\(_4\):
(iv) ethanoic acid is added to a solution of Ka\(_2\)CO\(_3\)
(i) Lime juice is acidic, so methyl orange turns red (pink).
(ii) The yellow/brown iron(III) chloride solution turns pale green (Fe3+ is reduced to Fe2+) and a pale yellow precipitate of sulphur forms, making the solution turbid:
\[2FeCl_3 + H_2S \rightarrow 2FeCl_2 + 2HCl + S\]
(iii) The purple colour of the KMnO4 is decolourized. Sulphur(IV) oxide is a reducing agent and reduces MnO4- (purple) to Mn2+ (colourless).
(iv) There is effervescence and a colourless gas (carbon(IV) oxide) is evolved which turns limewater milky:
\[2CH_3COOH + K_2CO_3 \rightarrow 2CH_3COOK + H_2O + CO_2\]
Antwortdetails
(i) Lime juice is acidic, so methyl orange turns red (pink).
(ii) The yellow/brown iron(III) chloride solution turns pale green (Fe3+ is reduced to Fe2+) and a pale yellow precipitate of sulphur forms, making the solution turbid:
\[2FeCl_3 + H_2S \rightarrow 2FeCl_2 + 2HCl + S\]
(iii) The purple colour of the KMnO4 is decolourized. Sulphur(IV) oxide is a reducing agent and reduces MnO4- (purple) to Mn2+ (colourless).
(iv) There is effervescence and a colourless gas (carbon(IV) oxide) is evolved which turns limewater milky:
\[2CH_3COOH + K_2CO_3 \rightarrow 2CH_3COOK + H_2O + CO_2\]
Frage 2 Bericht
All your burette readings (initials and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
F is 2 mixture of two inorganic salts. Carry out the following exercises on F. Record your observations and identify any gas(es) evolved. State the conclusions you draw from the result of each test.
(a) Put all of F in a beaker and add about \(10\text{ cm}^3\) of distilled water. Stir well and filter. Keep the filtrate and the residue.
(b)(i) To about \(2\text{cm}^3\) of the filtrate. add \(\mathrm{NaOH}_{(aq)}\) in drops and then in excess.
(ii) To another \(2\text{cm}^3\) portion of the solution, add a few drops of \(\mathrm{NH3}_{(aq)}\) in drops and then in excess.
(c) To about \(2\text{cm}^3\) of the solution, add a few drops of \(\mathrm{HNO}_{3(aq)}\) followed by few drops of the drops of \(\mathrm{AgNO}_{3(aq)}\)
(d)(i) Put all the residue into a clean test-tube and add \(\mathrm{HNO}_{3(aq)}\)
(ii) To a portion of the solution from (d)(i)) add \(\mathrm{NaOH}_{(aq)}\) in drops and then in excess.
Qualitative analysis of F (a mixture of two inorganic salts)
The results of each test on F are recorded below. For each test the observation is stated and the inference (conclusion) drawn from it is given.
| Test | Observation | Inference |
|---|---|---|
| (a) All of F + about 10 cm\(^3\) distilled water; stir and filter. | Slight effervescence; a colourless, odourless gas is evolved. On filtering, a pale-blue (almost colourless) filtrate and a green residue are obtained. | F is a mixture of a soluble salt (in the filtrate) and an insoluble salt (the green residue). A copper(II) compound is indicated by the green residue and pale-blue filtrate. |
| (b)(i) Filtrate + NaOH\(_{(aq)}\) in drops, then in excess. | A white precipitate forms; it is insoluble in excess NaOH\(_{(aq)}\). A pale-blue precipitate is also present which remains insoluble in excess. | White, insoluble precipitate → Ca\(^{2+}\) (or Pb\(^{2+}\)); Ca\(^{2+}\) indicated. Pale-blue precipitate insoluble in excess → Cu\(^{2+}\) also present. |
| (b)(ii) Filtrate + NH\(_3{}_{(aq)}\) in drops, then in excess. | No white precipitate with Ca\(^{2+}\). A pale-blue precipitate forms which dissolves in excess to give a deep-blue solution. | No precipitate with ammonia confirms Ca\(^{2+}\) (not Pb\(^{2+}\)). Deep-blue solution in excess ammonia confirms Cu\(^{2+}\). |
| (c) Filtrate + HNO\(_{3(aq)}\), then AgNO\(_{3(aq)}\). | No gas evolved; a white precipitate forms. | Cl\(^{-}\) present (white AgCl). \[\text{Ag}^{+}_{(aq)} + \text{Cl}^{-}_{(aq)} \rightarrow \text{AgCl}_{(s)}\] |
| (d)(i) All of residue + HNO\(_{3(aq)}\). | Effervescence; a colourless, odourless gas is evolved that turns damp blue litmus paper red and turns lime water milky. The green residue dissolves to give a blue solution. | The gas is CO\(_2\); CO\(_3^{2-}\) (trioxocarbonate(IV)) present. \[\text{CO}_3^{2-} + 2\text{H}^{+} \rightarrow \text{H}_2\text{O} + \text{CO}_{2(g)}\] Blue solution → Cu\(^{2+}\) in the residue. |
| (d)(ii) Solution from (d)(i) + NaOH\(_{(aq)}\) in drops, then in excess. | A blue precipitate forms; it is insoluble in excess NaOH\(_{(aq)}\). | Cu\(^{2+}\) present (blue Cu(OH)\(_2\)). \[\text{Cu}^{2+}_{(aq)} + 2\text{OH}^{-}_{(aq)} \rightarrow \text{Cu(OH)}_{2(s)}\] |
Conclusion: F contains the cations Ca\(^{2+}\) and Cu\(^{2+}\) and the anions Cl\(^{-}\) and CO\(_3^{2-}\). The soluble portion (filtrate) supplies Ca\(^{2+}\) and Cl\(^{-}\) (calcium chloride, CaCl\(_2\)), while the green insoluble residue is a copper(II) trioxocarbonate(IV), CuCO\(_3\), which supplies Cu\(^{2+}\) and CO\(_3^{2-}\).
Antwortdetails
Qualitative analysis of F (a mixture of two inorganic salts)
The results of each test on F are recorded below. For each test the observation is stated and the inference (conclusion) drawn from it is given.
| Test | Observation | Inference |
|---|---|---|
| (a) All of F + about 10 cm\(^3\) distilled water; stir and filter. | Slight effervescence; a colourless, odourless gas is evolved. On filtering, a pale-blue (almost colourless) filtrate and a green residue are obtained. | F is a mixture of a soluble salt (in the filtrate) and an insoluble salt (the green residue). A copper(II) compound is indicated by the green residue and pale-blue filtrate. |
| (b)(i) Filtrate + NaOH\(_{(aq)}\) in drops, then in excess. | A white precipitate forms; it is insoluble in excess NaOH\(_{(aq)}\). A pale-blue precipitate is also present which remains insoluble in excess. | White, insoluble precipitate → Ca\(^{2+}\) (or Pb\(^{2+}\)); Ca\(^{2+}\) indicated. Pale-blue precipitate insoluble in excess → Cu\(^{2+}\) also present. |
| (b)(ii) Filtrate + NH\(_3{}_{(aq)}\) in drops, then in excess. | No white precipitate with Ca\(^{2+}\). A pale-blue precipitate forms which dissolves in excess to give a deep-blue solution. | No precipitate with ammonia confirms Ca\(^{2+}\) (not Pb\(^{2+}\)). Deep-blue solution in excess ammonia confirms Cu\(^{2+}\). |
| (c) Filtrate + HNO\(_{3(aq)}\), then AgNO\(_{3(aq)}\). | No gas evolved; a white precipitate forms. | Cl\(^{-}\) present (white AgCl). \[\text{Ag}^{+}_{(aq)} + \text{Cl}^{-}_{(aq)} \rightarrow \text{AgCl}_{(s)}\] |
| (d)(i) All of residue + HNO\(_{3(aq)}\). | Effervescence; a colourless, odourless gas is evolved that turns damp blue litmus paper red and turns lime water milky. The green residue dissolves to give a blue solution. | The gas is CO\(_2\); CO\(_3^{2-}\) (trioxocarbonate(IV)) present. \[\text{CO}_3^{2-} + 2\text{H}^{+} \rightarrow \text{H}_2\text{O} + \text{CO}_{2(g)}\] Blue solution → Cu\(^{2+}\) in the residue. |
| (d)(ii) Solution from (d)(i) + NaOH\(_{(aq)}\) in drops, then in excess. | A blue precipitate forms; it is insoluble in excess NaOH\(_{(aq)}\). | Cu\(^{2+}\) present (blue Cu(OH)\(_2\)). \[\text{Cu}^{2+}_{(aq)} + 2\text{OH}^{-}_{(aq)} \rightarrow \text{Cu(OH)}_{2(s)}\] |
Conclusion: F contains the cations Ca\(^{2+}\) and Cu\(^{2+}\) and the anions Cl\(^{-}\) and CO\(_3^{2-}\). The soluble portion (filtrate) supplies Ca\(^{2+}\) and Cl\(^{-}\) (calcium chloride, CaCl\(_2\)), while the green insoluble residue is a copper(II) trioxocarbonate(IV), CuCO\(_3\), which supplies Cu\(^{2+}\) and CO\(_3^{2-}\).
Frage 3 Bericht
All your burette readings (initials and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet.
A is \(0.200\ \mathrm{moldm^3}\) of HCl. C is a solution containing \(14.3\mathrm{g}\) of \(\mathrm{Na_2CO_3\ .\ xH_2O}\) in \(500\ \mathrm{cm^3}\) of solution.
a) Put A into the burette and titrate it against \(20.0\ \mathrm{cm^3}\) or \(25.0\mathrm{cm^3}\) portions of C using methyl orange as indicator. Repeat the titration to obtain Consistent titre values. Tabulate vour results and calculate the average volume of A used. The equation for the reaction is;
\[ \mathrm{Na_2CO_3\ .\ xH_2O + 2HCl_{(aq)} \to 2NaCl_{(aq)} + CO_{2(g)} + (x + 1)_3H_2O_{(l)}} \]
(b) From your results and the information provided. calculate the:
(i) concentration of C in \(\mathrm{moldm^{-3}}\)
(ii) concentration of C in \(\mathrm{gdm^{-3}}\)
(iii) molar mass of \(\mathrm{Na_2CO_3\ .\ xH_2O}\)
(iv) the value of x in \(\mathrm{Na_2CO_3\ .\ xH_2O}\). [H =1.0; C = 12.0; O = 16.0; Na = 23.0]
Credit will be given for strict adherence to the instruction, for observations precisely recorded and for accurale references. All tests. obsenations and influences must be cleary entered in the booklet in ink at the same time they are made.
Indicator: methyl orange. Volume of C (base) pipetted: 25.00 cm3.
(a) Burette readings and average titre
| Titration | Rough | 1st | 2nd | 3rd |
|---|---|---|---|---|
| Final burette reading / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
| Initial burette reading / cm3 | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
Using the three consistent titres:
\[ \text{Average titre} = \frac{24.80 + 24.70 + 24.90}{3} = 24.80\ \text{cm}^3 \](b)(i) Concentration of C in mol dm-3
Equation: \(Na_2CO_3\cdot xH_2O + 2HCl \rightarrow 2NaCl + CO_2 + (x+1)H_2O\), so \(\dfrac{C_AV_A}{C_BV_B} = \dfrac{n_A}{n_B} = \dfrac{2}{1}\).
With \(C_A = 0.200\ \text{mol dm}^{-3}\), \(V_A = 24.80\ \text{cm}^3\), \(V_B = 25.00\ \text{cm}^3\):
\[ \frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1} \] \[ C_B = \frac{1 \times 0.200 \times 24.80}{2 \times 25.00} = 0.0992\ \text{mol dm}^{-3} \](ii) Concentration of C in g dm-3
C contains 14.3 g in 500 cm3, so
\[ \frac{14.3}{500} \times 1000 = 28.6\ \text{g dm}^{-3} \](iii) Molar mass of Na2CO3·xH2O
\[ M = \frac{\text{concentration in g dm}^{-3}}{\text{concentration in mol dm}^{-3}} = \frac{28.6}{0.0992} = 288\ \text{g mol}^{-1} \](iv) Value of x
\[ 2(23.0) + 12.0 + 3(16.0) + x(2(1.0)+16.0) = 288 \] \[ 46 + 12 + 48 + 18x = 288 \] \[ 106 + 18x = 288 \] \[ 18x = 182,\qquad x = \frac{182}{18} = 10.11 \approx 10 \]The salt is Na2CO3·10H2O.
Antwortdetails
Indicator: methyl orange. Volume of C (base) pipetted: 25.00 cm3.
(a) Burette readings and average titre
| Titration | Rough | 1st | 2nd | 3rd |
|---|---|---|---|---|
| Final burette reading / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
| Initial burette reading / cm3 | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
Using the three consistent titres:
\[ \text{Average titre} = \frac{24.80 + 24.70 + 24.90}{3} = 24.80\ \text{cm}^3 \](b)(i) Concentration of C in mol dm-3
Equation: \(Na_2CO_3\cdot xH_2O + 2HCl \rightarrow 2NaCl + CO_2 + (x+1)H_2O\), so \(\dfrac{C_AV_A}{C_BV_B} = \dfrac{n_A}{n_B} = \dfrac{2}{1}\).
With \(C_A = 0.200\ \text{mol dm}^{-3}\), \(V_A = 24.80\ \text{cm}^3\), \(V_B = 25.00\ \text{cm}^3\):
\[ \frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1} \] \[ C_B = \frac{1 \times 0.200 \times 24.80}{2 \times 25.00} = 0.0992\ \text{mol dm}^{-3} \](ii) Concentration of C in g dm-3
C contains 14.3 g in 500 cm3, so
\[ \frac{14.3}{500} \times 1000 = 28.6\ \text{g dm}^{-3} \](iii) Molar mass of Na2CO3·xH2O
\[ M = \frac{\text{concentration in g dm}^{-3}}{\text{concentration in mol dm}^{-3}} = \frac{28.6}{0.0992} = 288\ \text{g mol}^{-1} \](iv) Value of x
\[ 2(23.0) + 12.0 + 3(16.0) + x(2(1.0)+16.0) = 288 \] \[ 46 + 12 + 48 + 18x = 288 \] \[ 106 + 18x = 288 \] \[ 18x = 182,\qquad x = \frac{182}{18} = 10.11 \approx 10 \]The salt is Na2CO3·10H2O.
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