WAEC SSCE - Chemistry - 2016 (Practical)
Swali 1 Ripoti
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) What difference in physical properties enable the separation of mixtures by:
(i) simple distillation,
(ii) paper chromatography;
(iii) fractional distillation.
(b) Give a reason for each of the following practices during titration in the laboratory.
(i) White tile is placed under the conical flask.
(ii) Burette readings are always recorded to two decimal places.
(iii) Calculate the volume of 2.5 moldm\(^{-3}\) stock HČI required to prepare 500 cm\(^3\) of 0.20 moldm HCI.
(a) Physical property differences that allow separation:
- (i) Simple distillation: a difference in boiling points, one component being volatile (a liquid) and the other non-volatile (e.g. a dissolved solid), so the liquid boils off and is condensed.
- (ii) Paper chromatography: a difference in solubility and adsorption, i.e. the components dissolve to different extents in the solvent and are adsorbed to different extents on the paper, so they travel at different rates.
- (iii) Fractional distillation: a difference in the boiling points of two (or more) miscible liquids whose boiling points are fairly close together; the fractionating column allows them to be separated.
(b) Reasons for titration practices:
- (i) A white tile is placed under the conical flask so that the colour change at the end point can be seen clearly against a white background.
- (ii) Burette readings are recorded to two decimal places for accuracy/precision, because the burette is graduated to 0.1 cm\(^3\) and readings are estimated to the nearest 0.05 cm\(^3\).
(iii) Volume of 2.5 mol dm\(^{-3}\) stock HCl needed to prepare 500 cm\(^3\) of 0.20 mol dm\(^{-3}\) HCl. Using \(C_1V_1 = C_2V_2\):
\[2.5 \times V_1 = 0.20 \times 500\]
\[V_1 = \frac{0.20 \times 500}{2.5} = \frac{100}{2.5} = 40\,cm^3\]
So 40 cm\(^3\) of the stock acid is required, then made up to 500 cm\(^3\) with distilled water.
Maelezo ya Majibu
(a) Physical property differences that allow separation:
- (i) Simple distillation: a difference in boiling points, one component being volatile (a liquid) and the other non-volatile (e.g. a dissolved solid), so the liquid boils off and is condensed.
- (ii) Paper chromatography: a difference in solubility and adsorption, i.e. the components dissolve to different extents in the solvent and are adsorbed to different extents on the paper, so they travel at different rates.
- (iii) Fractional distillation: a difference in the boiling points of two (or more) miscible liquids whose boiling points are fairly close together; the fractionating column allows them to be separated.
(b) Reasons for titration practices:
- (i) A white tile is placed under the conical flask so that the colour change at the end point can be seen clearly against a white background.
- (ii) Burette readings are recorded to two decimal places for accuracy/precision, because the burette is graduated to 0.1 cm\(^3\) and readings are estimated to the nearest 0.05 cm\(^3\).
(iii) Volume of 2.5 mol dm\(^{-3}\) stock HCl needed to prepare 500 cm\(^3\) of 0.20 mol dm\(^{-3}\) HCl. Using \(C_1V_1 = C_2V_2\):
\[2.5 \times V_1 = 0.20 \times 500\]
\[V_1 = \frac{0.20 \times 500}{2.5} = \frac{100}{2.5} = 40\,cm^3\]
So 40 cm\(^3\) of the stock acid is required, then made up to 500 cm\(^3\) with distilled water.
Swali 2 Ripoti
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 solution of 0.050 moldm\(^{3}\) \(\mathrm{H_2C_2O_4}\) (ethanedioic acid). B is a solution of \(\mathrm{KMnO_4}\), (potassium tetraoxomanganate (VII), of unknown concentration.
(a) Put B into the burette. Pipette 20.0 cm\(^{3}\) or 25.0 cm\(^{3}\) of A into a Conical flask and add about 10.0 cm\(^{3}\) of dilute \(\mathrm{H_2SO_4}\), Heat the mixture to about 40°C - 50°C and titrate it while still hot with B. Repeat the titration to obtain consistent titre values. Tabulate your results and calculate the average volume of B used. The equation of reaction is;
\[ \mathrm{2MnO}_{4(aq)}^{-} + \mathrm{5C_2O}_{4(aq)}^{2-} + \mathrm{16H}^{+}_{(aq)} \to \mathrm{2MnH}^{2+}_{(aq)} + \mathrm{8H_2O}_{(l)} + \mathrm{10CO}_{2(g)} \](b) From your results and the information provided, calculate the:
(i) concentration of \(\mathrm{MnO_2^-}\) in B in moldm\(^{-1}\)
(ii) concentration of \(\mathrm{KMnO_4^-}\) in B in gdm\(^{-3}\)
(iii) volume of \(\mathrm{CO_2}\) evolved at s.t.p when 25.0 cm\(^{3}\) of \(\mathrm{H_2C_2O_4}\) reacted completely. [0 = 16.0, K= 39.0, Mn = 55.0, Molar volume of gas at s.t.p.= 22.4 dm\(^{3}\) mol\(^{-1}\)]
Credit will be given for strict adherence to the instructions. for observations precisely recorded and jor accurate inferences. All tests, observations and inferences must be clearly entered in your answer book in ink, at the time they are made
This is a redox titration of standard ethanedioic acid A (\(0.050\ \text{mol dm}^{-3}\)) against potassium tetraoxomanganate(VII) B of unknown concentration, using a 25.0 cm\(^3\) pipette of A.
\[2MnO_4^-+5C_2O_4^{2-}+16H^+\rightarrow 2Mn^{2+}+8H_2O+10CO_2\](a) Table of results.
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|---|---|---|---|
| Final reading | 36.50 | 30.40 | 41.70 | 26.80 |
| Initial reading | 12.30 | 05.90 | 17.10 | 01.90 |
| Volume of B used | 24.20 | 24.50 | 24.60 | 24.90 |
Average of the two concordant titres:
\[V_B=\frac{24.50+24.60}{2}=24.55\ \text{cm}^3.\](b)(i) Concentration of \(MnO_4^-\) in B. Using the mole ratio \(n(MnO_4^-):n(C_2O_4^{2-})=2:5\):
\[\frac{C(MnO_4^-)\times24.55}{0.050\times25.0}=\frac{2}{5}\]\[C(MnO_4^-)=\frac{0.050\times25.0\times2}{24.55\times5}=0.0204\ \text{mol dm}^{-3}.\](ii) Concentration of \(KMnO_4\) in g dm\(^{-3}\). Molar mass of \(KMnO_4=39.0+55.0+(4\times16.0)=158\ \text{g mol}^{-1}\):
\[C=0.0204\times158=3.22\ \text{g dm}^{-3}.\](iii) Volume of \(CO_2\) at s.t.p. from 25.0 cm\(^3\) of A.
\[n(C_2O_4^{2-})=0.050\times\frac{25.0}{1000}=0.00125\ \text{mol}.\]From the equation, 5 mol \(C_2O_4^{2-}\) give 10 mol \(CO_2\):
\[n(CO_2)=0.00125\times\frac{10}{5}=0.00250\ \text{mol}.\]\[V(CO_2)=0.00250\times22.4=0.056\ \text{dm}^3=56.0\ \text{cm}^3.\]Maelezo ya Majibu
This is a redox titration of standard ethanedioic acid A (\(0.050\ \text{mol dm}^{-3}\)) against potassium tetraoxomanganate(VII) B of unknown concentration, using a 25.0 cm\(^3\) pipette of A.
\[2MnO_4^-+5C_2O_4^{2-}+16H^+\rightarrow 2Mn^{2+}+8H_2O+10CO_2\](a) Table of results.
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|---|---|---|---|
| Final reading | 36.50 | 30.40 | 41.70 | 26.80 |
| Initial reading | 12.30 | 05.90 | 17.10 | 01.90 |
| Volume of B used | 24.20 | 24.50 | 24.60 | 24.90 |
Average of the two concordant titres:
\[V_B=\frac{24.50+24.60}{2}=24.55\ \text{cm}^3.\](b)(i) Concentration of \(MnO_4^-\) in B. Using the mole ratio \(n(MnO_4^-):n(C_2O_4^{2-})=2:5\):
\[\frac{C(MnO_4^-)\times24.55}{0.050\times25.0}=\frac{2}{5}\]\[C(MnO_4^-)=\frac{0.050\times25.0\times2}{24.55\times5}=0.0204\ \text{mol dm}^{-3}.\](ii) Concentration of \(KMnO_4\) in g dm\(^{-3}\). Molar mass of \(KMnO_4=39.0+55.0+(4\times16.0)=158\ \text{g mol}^{-1}\):
\[C=0.0204\times158=3.22\ \text{g dm}^{-3}.\](iii) Volume of \(CO_2\) at s.t.p. from 25.0 cm\(^3\) of A.
\[n(C_2O_4^{2-})=0.050\times\frac{25.0}{1000}=0.00125\ \text{mol}.\]From the equation, 5 mol \(C_2O_4^{2-}\) give 10 mol \(CO_2\):
\[n(CO_2)=0.00125\times\frac{10}{5}=0.00250\ \text{mol}.\]\[V(CO_2)=0.00250\times22.4=0.056\ \text{dm}^3=56.0\ \text{cm}^3.\]Swali 3 Ripoti
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
C is a mıxture of two salts, containing one cation and two anions. Carry out the 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 all of C in a beaker and add about \(10\ \text{cm}^3\) of distilled water. Stir well and filter. Keep the filtrate and the residue.
(b) To about \(2\ \text{cm}^3\) of the filtrate, add few drops of \(\mathrm{AgNO}_{(aq)}\), followed by \(\mathrm{HNO}_{3(g)}\). Add excess \(\mathrm{NH}_{3(aq)}\) to the resulting mixture.
(c)(i) Put all of the residue into a clean test tube and add about 5 cm of \(\mathrm{HNO}_{3(aq)}\)
(ii) To about \(2\ \text{cm}^3\) of the solution from 2(c)(i), add \(\mathrm{NaOH}_{(aq)}\) in drops and then in excess
(iii) To another \(2\ \text{cm}^3\) of the solution from 2(c)(i), add \(\mathrm{NH}_{3(aq)}\) in drops and then in excess.
Specimen C is a mixture of two salts sharing one cation, Zn2+, and containing two anions: the soluble part supplies chloride, Cl−, and the insoluble residue supplies carbonate, CO32−. That is, C is a mixture of zinc chloride and zinc carbonate.
The observations and inferences for each test are recorded below.
| Test | Observation | Inference |
|---|---|---|
| (a) All of C placed in a beaker, about 10 cm3 distilled water added, stirred and filtered. | C partly dissolves; a colourless filtrate is obtained and a residue remains on the filter paper. | C is a mixture of a soluble salt (in the filtrate) and an insoluble salt (the residue). |
| (b) To about 2 cm3 of the filtrate, a few drops of AgNO3(aq) are added, followed by dilute HNO3(aq), then excess NH3(aq). | A white precipitate forms; it is insoluble in dilute HNO3; it dissolves completely in excess aqueous ammonia. | Cl− present in the soluble salt. \(\text{Ag}^{+} + \text{Cl}^{-} \rightarrow \text{AgCl}\downarrow\) (white), and \(\text{AgCl} + 2\text{NH}_3 \rightarrow [\text{Ag(NH}_3)_2]^{+} + \text{Cl}^{-}\). |
| (c)(i) All of the residue is put into a clean test tube and about 5 cm3 of dilute HNO3(aq) is added. | Brisk effervescence of a colourless, odourless gas that turns lime water milky; the residue dissolves to a colourless solution. | Gas is CO2; CO32− present in the insoluble salt. \(\text{CO}_3^{2-} + 2\text{H}^{+} \rightarrow \text{H}_2\text{O} + \text{CO}_2\uparrow\). |
| (c)(ii) To about 2 cm3 of the solution from (c)(i), NaOH(aq) is added in drops and then in excess. | A white gelatinous precipitate forms which dissolves in excess NaOH to give a colourless solution. | Zn2+ or Al3+ present. \(\text{Zn}^{2+} + 2\text{OH}^{-} \rightarrow \text{Zn(OH)}_2\downarrow\), soluble in excess as \([\text{Zn(OH)}_4]^{2-}\). |
| (c)(iii) To a further 2 cm3 of the solution from (c)(i), NH3(aq) is added in drops and then in excess. | A white gelatinous precipitate forms which dissolves in excess aqueous ammonia to give a colourless solution. | Zn2+ confirmed; Al3+ excluded, since Al(OH)3 is insoluble in excess NH3. \(\text{Zn(OH)}_2 + 4\text{NH}_3 \rightarrow [\text{Zn(NH}_3)_4]^{2+} + 2\text{OH}^{-}\). |
Conclusion: The single cation in C is Zn2+ (given by (c)(ii) and confirmed by its dissolving in excess NH3 in (c)(iii)). The two anions are chloride, Cl−, in the soluble portion (test (b)) and carbonate, CO32−, in the insoluble residue (test (c)(i)). Therefore C is a mixture of zinc chloride, ZnCl2, and zinc carbonate, ZnCO3.
Maelezo ya Majibu
Specimen C is a mixture of two salts sharing one cation, Zn2+, and containing two anions: the soluble part supplies chloride, Cl−, and the insoluble residue supplies carbonate, CO32−. That is, C is a mixture of zinc chloride and zinc carbonate.
The observations and inferences for each test are recorded below.
| Test | Observation | Inference |
|---|---|---|
| (a) All of C placed in a beaker, about 10 cm3 distilled water added, stirred and filtered. | C partly dissolves; a colourless filtrate is obtained and a residue remains on the filter paper. | C is a mixture of a soluble salt (in the filtrate) and an insoluble salt (the residue). |
| (b) To about 2 cm3 of the filtrate, a few drops of AgNO3(aq) are added, followed by dilute HNO3(aq), then excess NH3(aq). | A white precipitate forms; it is insoluble in dilute HNO3; it dissolves completely in excess aqueous ammonia. | Cl− present in the soluble salt. \(\text{Ag}^{+} + \text{Cl}^{-} \rightarrow \text{AgCl}\downarrow\) (white), and \(\text{AgCl} + 2\text{NH}_3 \rightarrow [\text{Ag(NH}_3)_2]^{+} + \text{Cl}^{-}\). |
| (c)(i) All of the residue is put into a clean test tube and about 5 cm3 of dilute HNO3(aq) is added. | Brisk effervescence of a colourless, odourless gas that turns lime water milky; the residue dissolves to a colourless solution. | Gas is CO2; CO32− present in the insoluble salt. \(\text{CO}_3^{2-} + 2\text{H}^{+} \rightarrow \text{H}_2\text{O} + \text{CO}_2\uparrow\). |
| (c)(ii) To about 2 cm3 of the solution from (c)(i), NaOH(aq) is added in drops and then in excess. | A white gelatinous precipitate forms which dissolves in excess NaOH to give a colourless solution. | Zn2+ or Al3+ present. \(\text{Zn}^{2+} + 2\text{OH}^{-} \rightarrow \text{Zn(OH)}_2\downarrow\), soluble in excess as \([\text{Zn(OH)}_4]^{2-}\). |
| (c)(iii) To a further 2 cm3 of the solution from (c)(i), NH3(aq) is added in drops and then in excess. | A white gelatinous precipitate forms which dissolves in excess aqueous ammonia to give a colourless solution. | Zn2+ confirmed; Al3+ excluded, since Al(OH)3 is insoluble in excess NH3. \(\text{Zn(OH)}_2 + 4\text{NH}_3 \rightarrow [\text{Zn(NH}_3)_4]^{2+} + 2\text{OH}^{-}\). |
Conclusion: The single cation in C is Zn2+ (given by (c)(ii) and confirmed by its dissolving in excess NH3 in (c)(iii)). The two anions are chloride, Cl−, in the soluble portion (test (b)) and carbonate, CO32−, in the insoluble residue (test (c)(i)). Therefore C is a mixture of zinc chloride, ZnCl2, and zinc carbonate, ZnCO3.
