WAEC SSCE - Chemistry - 2005 (Practical)
Swali 1 Ripoti
All your burette readings (initial 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 book.
A is a solution of H\(_2\)SO\(_4\) containing 4.9 gdm-3, B is a solution containing X g dm\(^{-3}\) of Na\(_2\)CO\(_3\).
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methyl orange as an 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\(_{4(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) \(\to\) Na\(_{2}\)SO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(b) From your results and information provided above, calculate the:
(i) Concentration of A In mol dm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) mass of salt formed when 500 cm\(^3\) of B is Completely neutralized by A.
(v) volume of carbon (IV) oxide liberated in (b) (ii) above at s.t.p. [O = 16, Na = 23, S = 32, 1 mole or a gas occupies 22.4 dm\(^3\) at s.t.p.]
(a) Burette readings and average titre
Volume of pipette used = 25.00 cm3 of B.
| Burette reading (cm3) | Rough | 1st titre | 2nd titre | 3rd titre |
|---|---|---|---|---|
| Final reading | 22.80 | 22.70 | 32.70 | 22.70 |
| Initial reading | 0.00 | 0.00 | 10.00 | 0.00 |
| Volume of acid used | 22.00 | 22.70 | 22.70 | 22.70 |
Average volume of A used (concordant titres):
\[ V_A = \frac{22.70 + 22.70 + 22.70}{3} = \frac{68.10}{3} = 22.70\ \text{cm}^3. \]
(b)(i) Concentration of A in mol dm-3
Molar mass of H2SO4 = \(2(1) + 32 + 4(16) = 98\ \text{g mol}^{-1}\).
\[ C_A = \frac{4.9}{98} = 0.05\ \text{mol dm}^{-3}. \]
(b)(ii) Concentration of B in mol dm-3
From the equation the acid : base mole ratio is 1 : 1, so \(\dfrac{C_A V_A}{C_B V_B} = \dfrac{1}{1}\):
\[ C_B = \frac{C_A V_A}{V_B} = \frac{0.05 \times 22.70}{25.0} = 0.0454 \approx 0.05\ \text{mol dm}^{-3}. \]
(b)(iii) Mass of salt (Na2SO4) formed by 500 cm3 of B
Moles of Na2CO3 in 500 cm3 of B \(= \dfrac{0.05 \times 500}{1000} = 0.025\ \text{mol}\).
From the equation, moles of Na2SO4 formed = 0.025 mol. Molar mass of Na2SO4 \(= 2(23) + 32 + 4(16) = 142\ \text{g mol}^{-1}\).
\[ \text{Mass of Na}_2\text{SO}_4 = 0.025 \times 142 = 3.55\ \text{g}. \]
(b)(v) Volume of CO2 liberated at s.t.p.
Moles of CO2 = moles of Na2CO3 reacted = 0.025 mol.
\[ V_{CO_2} = 0.025 \times 22.4 = 0.56\ \text{dm}^3\ \text{at s.t.p.} \]
Maelezo ya Majibu
(a) Burette readings and average titre
Volume of pipette used = 25.00 cm3 of B.
| Burette reading (cm3) | Rough | 1st titre | 2nd titre | 3rd titre |
|---|---|---|---|---|
| Final reading | 22.80 | 22.70 | 32.70 | 22.70 |
| Initial reading | 0.00 | 0.00 | 10.00 | 0.00 |
| Volume of acid used | 22.00 | 22.70 | 22.70 | 22.70 |
Average volume of A used (concordant titres):
\[ V_A = \frac{22.70 + 22.70 + 22.70}{3} = \frac{68.10}{3} = 22.70\ \text{cm}^3. \]
(b)(i) Concentration of A in mol dm-3
Molar mass of H2SO4 = \(2(1) + 32 + 4(16) = 98\ \text{g mol}^{-1}\).
\[ C_A = \frac{4.9}{98} = 0.05\ \text{mol dm}^{-3}. \]
(b)(ii) Concentration of B in mol dm-3
From the equation the acid : base mole ratio is 1 : 1, so \(\dfrac{C_A V_A}{C_B V_B} = \dfrac{1}{1}\):
\[ C_B = \frac{C_A V_A}{V_B} = \frac{0.05 \times 22.70}{25.0} = 0.0454 \approx 0.05\ \text{mol dm}^{-3}. \]
(b)(iii) Mass of salt (Na2SO4) formed by 500 cm3 of B
Moles of Na2CO3 in 500 cm3 of B \(= \dfrac{0.05 \times 500}{1000} = 0.025\ \text{mol}\).
From the equation, moles of Na2SO4 formed = 0.025 mol. Molar mass of Na2SO4 \(= 2(23) + 32 + 4(16) = 142\ \text{g mol}^{-1}\).
\[ \text{Mass of Na}_2\text{SO}_4 = 0.025 \times 142 = 3.55\ \text{g}. \]
(b)(v) Volume of CO2 liberated at s.t.p.
Moles of CO2 = moles of Na2CO3 reacted = 0.025 mol.
\[ V_{CO_2} = 0.025 \times 22.4 = 0.56\ \text{dm}^3\ \text{at s.t.p.} \]
Swali 2 Ripoti
Credit will be given for strict adherence to 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 time they are made.
C Is an inorganic salt. D is an organic compound. Carry out the following exercises on C and D. Record your observations and identify any gases evolved. State the conclusion you draw from the result of each test.
(a) (i) Put C in a test tube and add about 10 cm\(^3\) of distilled water. Stir well.
(ii) Divide the resulting solution into two portions. To the first portion add sodium hydroxide solution in drops and then in excess.
(iii) To the second portion add dilute trioxonitrate (V) acid. Then add silver trioxonitrate (V) solution followed by aqueous ammonia in excess. tube and odd about 2- 5 cm\(^3\) of 'Xn'.
(ii) Identify the functional group present in D.
Identification: Specimen C is calcium chloride, \(\text{CaCl}_2\) (supplying \(\text{Ca}^{2+}\) and \(\text{Cl}^{-}\) ions), and specimen D is a carboxylic (alkanoic) acid carrying the carboxyl functional group, \(-\text{COOH}\).
| Test | Observation | Inference |
|---|---|---|
| (a)(i) C put in a test tube; about \(10\ \text{cm}^3\) of distilled water added and stirred well. | C dissolves completely to give a colourless solution. | C is a soluble salt. |
| (a)(ii) First portion + sodium hydroxide solution, \(\text{NaOH}_{(aq)}\), in drops and then in excess. | A white precipitate forms which is insoluble in excess \(\text{NaOH}\). | \(\text{Ca}^{2+}\) is present. \(\text{Ca}^{2+}_{(aq)} + 2\text{OH}^{-}_{(aq)} \rightarrow \text{Ca(OH)}_2{}_{(s)}\) |
| (a)(iii) Second portion + dilute trioxonitrate(V) acid, \(\text{HNO}_{3(aq)}\), then silver trioxonitrate(V) solution, \(\text{AgNO}_{3(aq)}\), then aqueous ammonia, \(\text{NH}_{3(aq)}\), in excess. | No reaction with the acid; on adding \(\text{AgNO}_{3(aq)}\) a white precipitate forms, which dissolves in excess aqueous ammonia to give a colourless solution. | \(\text{Cl}^{-}\) is present; \(\text{Cl}^{-}\) confirmed. \(\text{Ag}^{+}_{(aq)} + \text{Cl}^{-}_{(aq)} \rightarrow \text{AgCl}_{(s)}\) Together with (ii), C is calcium chloride. |
| (b) D + sodium hydrogentrioxocarbonate(IV) solution, \(\text{NaHCO}_{3(aq)}\). | Brisk effervescence of a colourless gas which turns lime water milky. | The gas is carbon(IV) oxide, \(\text{CO}_2\); D is an organic (alkanoic) acid. The functional group present is the carboxyl group, \(-\text{COOH}\). \(\text{RCOOH} + \text{NaHCO}_3 \rightarrow \text{RCOONa} + \text{H}_2\text{O} + \text{CO}_2\) |
Conclusion: C is calcium chloride, containing the \(\text{Ca}^{2+}\) cation and the \(\text{Cl}^{-}\) anion. D is a carboxylic (alkanoic) acid, and the functional group present in D is the carboxyl group, \(-\text{COOH}\).
Maelezo ya Majibu
Identification: Specimen C is calcium chloride, \(\text{CaCl}_2\) (supplying \(\text{Ca}^{2+}\) and \(\text{Cl}^{-}\) ions), and specimen D is a carboxylic (alkanoic) acid carrying the carboxyl functional group, \(-\text{COOH}\).
| Test | Observation | Inference |
|---|---|---|
| (a)(i) C put in a test tube; about \(10\ \text{cm}^3\) of distilled water added and stirred well. | C dissolves completely to give a colourless solution. | C is a soluble salt. |
| (a)(ii) First portion + sodium hydroxide solution, \(\text{NaOH}_{(aq)}\), in drops and then in excess. | A white precipitate forms which is insoluble in excess \(\text{NaOH}\). | \(\text{Ca}^{2+}\) is present. \(\text{Ca}^{2+}_{(aq)} + 2\text{OH}^{-}_{(aq)} \rightarrow \text{Ca(OH)}_2{}_{(s)}\) |
| (a)(iii) Second portion + dilute trioxonitrate(V) acid, \(\text{HNO}_{3(aq)}\), then silver trioxonitrate(V) solution, \(\text{AgNO}_{3(aq)}\), then aqueous ammonia, \(\text{NH}_{3(aq)}\), in excess. | No reaction with the acid; on adding \(\text{AgNO}_{3(aq)}\) a white precipitate forms, which dissolves in excess aqueous ammonia to give a colourless solution. | \(\text{Cl}^{-}\) is present; \(\text{Cl}^{-}\) confirmed. \(\text{Ag}^{+}_{(aq)} + \text{Cl}^{-}_{(aq)} \rightarrow \text{AgCl}_{(s)}\) Together with (ii), C is calcium chloride. |
| (b) D + sodium hydrogentrioxocarbonate(IV) solution, \(\text{NaHCO}_{3(aq)}\). | Brisk effervescence of a colourless gas which turns lime water milky. | The gas is carbon(IV) oxide, \(\text{CO}_2\); D is an organic (alkanoic) acid. The functional group present is the carboxyl group, \(-\text{COOH}\). \(\text{RCOOH} + \text{NaHCO}_3 \rightarrow \text{RCOONa} + \text{H}_2\text{O} + \text{CO}_2\) |
Conclusion: C is calcium chloride, containing the \(\text{Ca}^{2+}\) cation and the \(\text{Cl}^{-}\) anion. D is a carboxylic (alkanoic) acid, and the functional group present in D is the carboxyl group, \(-\text{COOH}\).
Swali 3 Ripoti
(a) State what would be observed when:
(i) Chlorine is passed through a freshly prepared solution of FeCl\(_2\).
(ii) SO\(_2\) is bubbled into a solution of FeCl\(_3\);
(iii) A few drops of water Is added to sodium hydroxide pellets in a test tube;
(iv) Dilute H\(_2\)SO\(_4\) is added to CaCO\(_{3(s)}\)
(b) Three test tubes contain solutions of SO\(_3^{2-}\), CO\(^{2-}_3\) and SO\(_4^{2-}\) respectively. Describe one chemical method that you would use to identify the solution containing SO\(_4^{2-}\)
(c)(i) Draw and label a diagram to illustrate the separation of a mixture of petrol and water
(ii) Which of the following will dissolve faster? 10g of NaOH pellets in 100 cm\(^3\) of water; 10g of NaOH powder in 50 cm\(^3\) of water. Give the reason for your answer.
(a) Observations
(i) Chlorine passed through freshly prepared FeCl\(_2\) solution: the pale green solution turns yellow-brown. Chlorine (an oxidising agent) oxidises iron(II) ions to iron(III) ions.
\[2\text{FeCl}_2 + \text{Cl}_2 \rightarrow 2\text{FeCl}_3\](ii) SO\(_2\) bubbled into FeCl\(_3\) solution: the yellow-brown solution turns pale green. Sulphur(IV) oxide (a reducing agent) reduces iron(III) ions to iron(II) ions.
\[2\text{FeCl}_3 + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{FeCl}_2 + \text{H}_2\text{SO}_4 + 2\text{HCl}\](iii) A few drops of water added to sodium hydroxide pellets: the pellets dissolve and a large amount of heat is evolved (the process is strongly exothermic), so the test tube becomes warm/hot.
(iv) Dilute H\(_2\)SO\(_4\) added to CaCO\(_{3(s)}\): there is effervescence (brisk at first) and a colourless, odourless gas, CO\(_2\), is evolved which turns lime water milky. The reaction soon slows because insoluble calcium tetraoxosulphate(VI) coats the marble.
\[\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} + \text{CO}_2\](b) Identifying the SO\(_4^{2-}\) solution
To each solution add dilute hydrochloric acid, then barium chloride solution. The carbonate and the sulphite react with the acid and give no lasting precipitate (the carbonate effervesces, evolving CO\(_2\); the sulphite evolves SO\(_2\), which has a choking smell). Only the tetraoxosulphate(VI) still gives a white precipitate of barium tetraoxosulphate(VI) that is insoluble in the dilute acid. The solution that gives the acid-insoluble white precipitate is the one containing SO\(_4^{2-}\).
\[\text{Ba}^{2+}_{(aq)} + \text{SO}_4^{2-}_{(aq)} \rightarrow \text{BaSO}_{4(s)}\](c)(i) Separation of a mixture of petrol and water
Petrol and water are immiscible liquids that form two distinct layers, so they are separated using a separating funnel. The mixture is poured into the funnel and allowed to settle: the denser water sinks to the bottom and the less dense petrol floats on top. The tap (stopcock) is opened to run the lower water layer into a beaker, then closed the moment the petrol reaches the tap, so that the petrol is left in the funnel.
(c)(ii) The 10 g of NaOH powder in 50 cm\(^3\) of water dissolves faster. The powder is broken into much finer particles than the pellets, so it exposes a far larger surface area to the water; more particles are attacked by the water molecules at the same time, giving a faster rate of dissolution. (The smaller volume of water also brings the particles into contact with the solvent more readily.)
Maelezo ya Majibu
(a) Observations
(i) Chlorine passed through freshly prepared FeCl\(_2\) solution: the pale green solution turns yellow-brown. Chlorine (an oxidising agent) oxidises iron(II) ions to iron(III) ions.
\[2\text{FeCl}_2 + \text{Cl}_2 \rightarrow 2\text{FeCl}_3\](ii) SO\(_2\) bubbled into FeCl\(_3\) solution: the yellow-brown solution turns pale green. Sulphur(IV) oxide (a reducing agent) reduces iron(III) ions to iron(II) ions.
\[2\text{FeCl}_3 + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{FeCl}_2 + \text{H}_2\text{SO}_4 + 2\text{HCl}\](iii) A few drops of water added to sodium hydroxide pellets: the pellets dissolve and a large amount of heat is evolved (the process is strongly exothermic), so the test tube becomes warm/hot.
(iv) Dilute H\(_2\)SO\(_4\) added to CaCO\(_{3(s)}\): there is effervescence (brisk at first) and a colourless, odourless gas, CO\(_2\), is evolved which turns lime water milky. The reaction soon slows because insoluble calcium tetraoxosulphate(VI) coats the marble.
\[\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} + \text{CO}_2\](b) Identifying the SO\(_4^{2-}\) solution
To each solution add dilute hydrochloric acid, then barium chloride solution. The carbonate and the sulphite react with the acid and give no lasting precipitate (the carbonate effervesces, evolving CO\(_2\); the sulphite evolves SO\(_2\), which has a choking smell). Only the tetraoxosulphate(VI) still gives a white precipitate of barium tetraoxosulphate(VI) that is insoluble in the dilute acid. The solution that gives the acid-insoluble white precipitate is the one containing SO\(_4^{2-}\).
\[\text{Ba}^{2+}_{(aq)} + \text{SO}_4^{2-}_{(aq)} \rightarrow \text{BaSO}_{4(s)}\](c)(i) Separation of a mixture of petrol and water
Petrol and water are immiscible liquids that form two distinct layers, so they are separated using a separating funnel. The mixture is poured into the funnel and allowed to settle: the denser water sinks to the bottom and the less dense petrol floats on top. The tap (stopcock) is opened to run the lower water layer into a beaker, then closed the moment the petrol reaches the tap, so that the petrol is left in the funnel.
(c)(ii) The 10 g of NaOH powder in 50 cm\(^3\) of water dissolves faster. The powder is broken into much finer particles than the pellets, so it exposes a far larger surface area to the water; more particles are attacked by the water molecules at the same time, giving a faster rate of dissolution. (The smaller volume of water also brings the particles into contact with the solvent more readily.)
