WAEC SSCE - Chemistry - 2008 (Practical)
Pergunta 1 Relatório
Burette readings(initial and final) must be given to two decimal places. Volume of pipette user must also be recorded but on account of experimental procedure is required. All calculations must be done in your answer book. A is O.050 mol dm\(^{-3}\) of acid HX. Bis a solution of NaOH containing 0.025 moles per 250 solutions.
(a) Put A into the burette and titrate it against 20.00 cm\(^{3}\) or 25.00 cm\(^{3}\) portions B using phenolphthalein as indicator. Tabulate your readings and calculate the average volume or A used.
(b) your results and the information provided above, calculate the;
(i) amount of acid in the average
(ii) amount of base in 20.00 cm\(^{3}\) or 25.00 cm\(^{3}\);
(iii) mole ratio of acid to base
(c) Write a balanced chemical equation for the reaction between the acid H\(_y\)X and the base NaOH
(d) State the basicity of the acid H\(_y\)X.
(a) Table of results. Indicator: phenolphthalein. Volume of pipette = 25.00 cm\(^3\). A (acid HX) is in the burette, titrated against 25.00 cm\(^3\) portions of B (NaOH).
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|---|---|---|---|
| Final reading | 22.60 | 22.40 | 22.30 | 22.30 |
| Initial reading | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used | 22.60 | 22.40 | 22.30 | 22.30 |
Average of the concordant titres:
\[V_A=\frac{22.40+22.30+22.30}{3}=\frac{67.00}{3}=22.30\ \text{cm}^3.\](b)(i) Amount of acid in the average titre. A is \(0.050\ \text{mol dm}^{-3}\):
\[n_{acid}=\frac{0.050\times22.30}{1000}=0.0011\ \text{mol}.\](ii) Amount of base in 25.00 cm\(^3\) of B. Concentration of NaOH:
\[[\text{NaOH}]=\frac{0.025}{250/1000}=\frac{0.025}{0.250}=0.10\ \text{mol dm}^{-3}.\]\[n_{base}=\frac{0.10\times25.00}{1000}=0.0025\ \text{mol}.\](iii) Mole ratio of acid to base.
\[\frac{n_{acid}}{n_{base}}=\frac{0.0011}{0.0025}=\frac{1}{2},\qquad\text{acid : base}=1:2.\](c) Balanced equation. The acid combines with two moles of NaOH, so it is H\(_2\)X:
\[H_2X+2NaOH\rightarrow Na_2X+2H_2O.\](d) Basicity of the acid. The acid reacts with NaOH in the ratio 1 : 2, so it has two replaceable hydrogen atoms. Basicity = 2.
Detalhes da Resposta
(a) Table of results. Indicator: phenolphthalein. Volume of pipette = 25.00 cm\(^3\). A (acid HX) is in the burette, titrated against 25.00 cm\(^3\) portions of B (NaOH).
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|---|---|---|---|
| Final reading | 22.60 | 22.40 | 22.30 | 22.30 |
| Initial reading | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used | 22.60 | 22.40 | 22.30 | 22.30 |
Average of the concordant titres:
\[V_A=\frac{22.40+22.30+22.30}{3}=\frac{67.00}{3}=22.30\ \text{cm}^3.\](b)(i) Amount of acid in the average titre. A is \(0.050\ \text{mol dm}^{-3}\):
\[n_{acid}=\frac{0.050\times22.30}{1000}=0.0011\ \text{mol}.\](ii) Amount of base in 25.00 cm\(^3\) of B. Concentration of NaOH:
\[[\text{NaOH}]=\frac{0.025}{250/1000}=\frac{0.025}{0.250}=0.10\ \text{mol dm}^{-3}.\]\[n_{base}=\frac{0.10\times25.00}{1000}=0.0025\ \text{mol}.\](iii) Mole ratio of acid to base.
\[\frac{n_{acid}}{n_{base}}=\frac{0.0011}{0.0025}=\frac{1}{2},\qquad\text{acid : base}=1:2.\](c) Balanced equation. The acid combines with two moles of NaOH, so it is H\(_2\)X:
\[H_2X+2NaOH\rightarrow Na_2X+2H_2O.\](d) Basicity of the acid. The acid reacts with NaOH in the ratio 1 : 2, so it has two replaceable hydrogen atoms. Basicity = 2.
Pergunta 2 Relatório
(a) In the laboratory preparation of crystals of CuSO\(_4\), a green powder Q was added to dilute H\(_2\)SO\(_4\), and stirred. Effervescence occurred and a gas R was given off which turned lime water milky. Excess Q was removed from the mixture. The solution of Cu\(_2\)SO\(_4\) was concentrated to half its original volume and allowed to stand.
(i) What is substance Q?
(ii) Name gas R
(ii) Why was excess Q used?
(iv) How would you know that the reaction is complete?
(v) What method was used to remove excess Q? (vi) Why was the solution of CusO\(_4\) not heated to dryness?
(b) Name the reagent(s) used for testing each of the following substances in the laboratory: (i) Water; (i) Primary alkanol.
Preparation of copper(II) sulphate crystals
A green powder Q reacts with dilute H\(_2\)SO\(_4\) with effervescence, giving off a gas R that turns lime water milky (so R is carbon dioxide). This identifies Q as a copper(II) carbonate.
- (i) Substance Q: copper(II) carbonate (basic copper carbonate), CuCO\(_3\). \[CuCO_3 + H_2SO_4 \to CuSO_4 + H_2O + CO_2\uparrow\]
- (ii) Gas R: carbon dioxide, CO\(_2\) (it turns lime water milky).
- (iii) Why excess Q was used: to make sure that all the acid is used up (neutralised), so that no free/unreacted sulphuric acid is left to contaminate the copper(II) sulphate.
- (iv) How you know the reaction is complete: effervescence stops and some of the green solid Q remains undissolved (unreacted), showing the acid has all been used up.
- (v) Method used to remove excess Q: filtration (the undissolved excess Q is left on the filter paper as residue and the copper(II) sulphate solution passes through as filtrate).
- (vi) Why the CuSO\(_4\) solution was not heated to dryness: heating to dryness would drive off the water of crystallisation and could decompose the salt, giving a white anhydrous powder instead of the blue hydrated crystals CuSO\(_4\).5H\(_2\)O. It is therefore evaporated only to the point of crystallisation and allowed to cool so that good crystals form.
(b) Reagents for testing
- (i) Water: anhydrous copper(II) sulphate (turns from white to blue), or anhydrous cobalt(II) chloride paper (turns from blue to pink). A confirmatory test is that pure water boils at 100 degrees Celsius.
- (ii) Primary alkanol (primary alcohol): acidified potassium tetraoxomanganate(VII) or acidified potassium dichromate(VI). On warming, the alkanol is oxidised, so acidified KMnO\(_4\) is decolourised (purple to colourless) and acidified K\(_2\)Cr\(_2\)O\(_7\) turns from orange to green.
Detalhes da Resposta
Preparation of copper(II) sulphate crystals
A green powder Q reacts with dilute H\(_2\)SO\(_4\) with effervescence, giving off a gas R that turns lime water milky (so R is carbon dioxide). This identifies Q as a copper(II) carbonate.
- (i) Substance Q: copper(II) carbonate (basic copper carbonate), CuCO\(_3\). \[CuCO_3 + H_2SO_4 \to CuSO_4 + H_2O + CO_2\uparrow\]
- (ii) Gas R: carbon dioxide, CO\(_2\) (it turns lime water milky).
- (iii) Why excess Q was used: to make sure that all the acid is used up (neutralised), so that no free/unreacted sulphuric acid is left to contaminate the copper(II) sulphate.
- (iv) How you know the reaction is complete: effervescence stops and some of the green solid Q remains undissolved (unreacted), showing the acid has all been used up.
- (v) Method used to remove excess Q: filtration (the undissolved excess Q is left on the filter paper as residue and the copper(II) sulphate solution passes through as filtrate).
- (vi) Why the CuSO\(_4\) solution was not heated to dryness: heating to dryness would drive off the water of crystallisation and could decompose the salt, giving a white anhydrous powder instead of the blue hydrated crystals CuSO\(_4\).5H\(_2\)O. It is therefore evaporated only to the point of crystallisation and allowed to cool so that good crystals form.
(b) Reagents for testing
- (i) Water: anhydrous copper(II) sulphate (turns from white to blue), or anhydrous cobalt(II) chloride paper (turns from blue to pink). A confirmatory test is that pure water boils at 100 degrees Celsius.
- (ii) Primary alkanol (primary alcohol): acidified potassium tetraoxomanganate(VII) or acidified potassium dichromate(VI). On warming, the alkanol is oxidised, so acidified KMnO\(_4\) is decolourised (purple to colourless) and acidified K\(_2\)Cr\(_2\)O\(_7\) turns from orange to green.
Pergunta 3 Relatório
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 centered in your answer book, at the time they are made.
C is a mixture of two salts. Carry outline following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion you draw from the result of each test.
(a) Put C into a beaker and add about \(10\ \text{cm}^3\) of distilled water, stir the mixture, and filter. Test the filtrate with litmus paper. Keep the residue and the filtrate.
(b)(i) To about \(2\ \text{cm}^3\) of the filtrate, add few drops of aqueous \(\mathrm{HNO_3}\) followed by \(\mathrm{AgNO_{3(aq)}}\)
(ii) Add excess \(\mathrm{NH_3}\) solution to the resulting mixture.
(c) To about half of the residue from (a) above, add about \(5\text{cm}^3\) of dilute \(\mathrm{HNO_3}\) in drops. Divide the resulting solution into two equal portions.
(d)(i) To the first portion add ammonia solution in drops and then in excess.
(ii) To the second portion add dilute hydrochloric acid.
Tests on C
C is a mixture of two salts.
| Test | Observation | Inference / conclusion |
|---|---|---|
| (a) Add distilled water to C, stir and filter. Test the filtrate with litmus paper. | C partly dissolves, giving a colourless filtrate and a white residue. The filtrate has no effect on either red or blue litmus paper. | C contains a soluble salt and an insoluble salt. The soluble salt gives a neutral solution. |
| (b)(i) To the filtrate add dilute nitric acid, followed by aqueous silver nitrate. | No visible change occurs on adding nitric acid. A white precipitate forms on adding silver nitrate solution. | Chloride ions, \(\mathrm{Cl^-}\), may be present. |
| (b)(ii) Add excess aqueous ammonia to the mixture in (b)(i). | The white precipitate dissolves in excess ammonia solution to give a colourless solution. | \(\mathrm{Cl^-}\) is confirmed. The precipitate is silver chloride, \(\mathrm{AgCl}\). |
| (c) Add dilute nitric acid dropwise to the residue. | The white residue dissolves with effervescence. The gas evolved turns limewater milky. | The gas is carbon dioxide, \(\mathrm{CO_2}\). Carbonate ions, \(\mathrm{CO_3^{2-}}\), are present. |
| (d)(i) To the first portion of the solution from (c), add aqueous ammonia dropwise and then in excess. | A white precipitate forms. It is insoluble in excess ammonia solution. | \(\mathrm{Pb^{2+}}\) may be present. |
| (d)(ii) To the second portion of the solution from (c), add dilute hydrochloric acid. | A white chalky precipitate is formed. | \(\mathrm{Pb^{2+}}\) is confirmed, due to formation of lead(II) chloride, \(\mathrm{PbCl_2}\). |
Gas evolved: Carbon dioxide, \(\mathrm{CO_2}\).
Overall conclusion: C contains a soluble neutral chloride, represented by sodium chloride, \(\mathrm{NaCl}\), and insoluble lead(II) trioxocarbonate(IV), \(\mathrm{PbCO_3}\).
Relevant ionic equations:
\[\mathrm{Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)}\]
\[\mathrm{PbCO_3(s)+2H^+(aq)\rightarrow Pb^{2+}(aq)+CO_2(g)+H_2O(l)}\]
\[\mathrm{Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)}\]
Detalhes da Resposta
Tests on C
C is a mixture of two salts.
| Test | Observation | Inference / conclusion |
|---|---|---|
| (a) Add distilled water to C, stir and filter. Test the filtrate with litmus paper. | C partly dissolves, giving a colourless filtrate and a white residue. The filtrate has no effect on either red or blue litmus paper. | C contains a soluble salt and an insoluble salt. The soluble salt gives a neutral solution. |
| (b)(i) To the filtrate add dilute nitric acid, followed by aqueous silver nitrate. | No visible change occurs on adding nitric acid. A white precipitate forms on adding silver nitrate solution. | Chloride ions, \(\mathrm{Cl^-}\), may be present. |
| (b)(ii) Add excess aqueous ammonia to the mixture in (b)(i). | The white precipitate dissolves in excess ammonia solution to give a colourless solution. | \(\mathrm{Cl^-}\) is confirmed. The precipitate is silver chloride, \(\mathrm{AgCl}\). |
| (c) Add dilute nitric acid dropwise to the residue. | The white residue dissolves with effervescence. The gas evolved turns limewater milky. | The gas is carbon dioxide, \(\mathrm{CO_2}\). Carbonate ions, \(\mathrm{CO_3^{2-}}\), are present. |
| (d)(i) To the first portion of the solution from (c), add aqueous ammonia dropwise and then in excess. | A white precipitate forms. It is insoluble in excess ammonia solution. | \(\mathrm{Pb^{2+}}\) may be present. |
| (d)(ii) To the second portion of the solution from (c), add dilute hydrochloric acid. | A white chalky precipitate is formed. | \(\mathrm{Pb^{2+}}\) is confirmed, due to formation of lead(II) chloride, \(\mathrm{PbCl_2}\). |
Gas evolved: Carbon dioxide, \(\mathrm{CO_2}\).
Overall conclusion: C contains a soluble neutral chloride, represented by sodium chloride, \(\mathrm{NaCl}\), and insoluble lead(II) trioxocarbonate(IV), \(\mathrm{PbCO_3}\).
Relevant ionic equations:
\[\mathrm{Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)}\]
\[\mathrm{PbCO_3(s)+2H^+(aq)\rightarrow Pb^{2+}(aq)+CO_2(g)+H_2O(l)}\]
\[\mathrm{Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)}\]
