WAEC SSCE - Chemistry - 2001 (Practical)

(a) If you were provided with anhydrous Na\(_2\)CO\(_3\), spatula and stirrer;

(i) list three other materials you would require to prepare a standard solution of Na\(_2\)CO\(_3\)

(ii) state what you would observe on adding diluted H\(_2\)SO\(_4\) to a portion of the Na\(_2\)CO\(_3\)

(b)(i) Describe briefly one chemical test you would perform to distinguish between zinc ions and aluminium ions in solution.

(ii) Mention one laboratory reagent you would use to;

I. produce ammonia from (NH\(_4\))\(_2\)SO\(_4\)

II. differentiate between precipitates of AgCl and Agl

lll. dehydrate ethanol

(c) Give the reason for each of the following laboratory practices

(i) Aqueous solutions of FeSO\(_4\) are freshly prepared when required for use.

(ii) The first jar of hydrogen collected during its preparation is discarded

Jibu

(a) Standard solution of Na₂CO₃

(i) Three other materials required: a chemical (weighing) balance, a volumetric (standard) flask, and a wash bottle of distilled water (a funnel and a beaker may also be used).

(ii) On adding dilute H₂SO₄ to the Na₂CO₃: brisk effervescence occurs and a colourless, odourless gas (CO₂) is evolved which turns lime water milky.

\[ Na_2CO_3 + H_2SO_4 \to Na_2SO_4 + H_2O + CO_2 \]

(b)(i) Distinguishing Zn²⁺ from Al³⁺

Add aqueous ammonia dropwise then in excess to each solution. Both give a white gelatinous precipitate; with Zn²⁺ the precipitate dissolves in excess ammonia (forming a colourless complex), whereas with Al³⁺ the precipitate is insoluble in excess ammonia. This distinguishes them.

(ii) Laboratory reagents:

I. Produce NH₃ from (NH₄)₂SO₄: warm with sodium hydroxide (or calcium hydroxide).
II. Differentiate AgCl from AgI: aqueous ammonia - AgCl dissolves, AgI does not.
III. Dehydrate ethanol: concentrated sulphuric acid (or hot aluminium oxide).

(c) Reasons

(i) FeSO₄ solutions are freshly prepared because the Fe²⁺ ions are readily oxidised by air (oxygen) to Fe³⁺, so a fresh solution is needed to keep the iron as pure Fe²⁺.

(ii) The first jar of hydrogen is discarded because it is mixed with the air already in the apparatus, forming an explosive mixture; discarding it ensures the hydrogen collected afterwards is pure and safe.

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Soma maelezo ya somo kwenye Solutions (WAEC)

Soma maelezo ya somo kwenye Acids, Bases And Salts (Ghana Only) (WAEC)

Maelezo ya Majibu

(a) Standard solution of Na₂CO₃

(i) Three other materials required: a chemical (weighing) balance, a volumetric (standard) flask, and a wash bottle of distilled water (a funnel and a beaker may also be used).

(ii) On adding dilute H₂SO₄ to the Na₂CO₃: brisk effervescence occurs and a colourless, odourless gas (CO₂) is evolved which turns lime water milky.

\[ Na_2CO_3 + H_2SO_4 \to Na_2SO_4 + H_2O + CO_2 \]

(b)(i) Distinguishing Zn²⁺ from Al³⁺

Add aqueous ammonia dropwise then in excess to each solution. Both give a white gelatinous precipitate; with Zn²⁺ the precipitate dissolves in excess ammonia (forming a colourless complex), whereas with Al³⁺ the precipitate is insoluble in excess ammonia. This distinguishes them.

(ii) Laboratory reagents:

I. Produce NH₃ from (NH₄)₂SO₄: warm with sodium hydroxide (or calcium hydroxide).
II. Differentiate AgCl from AgI: aqueous ammonia - AgCl dissolves, AgI does not.
III. Dehydrate ethanol: concentrated sulphuric acid (or hot aluminium oxide).

(c) Reasons

(i) FeSO₄ solutions are freshly prepared because the Fe²⁺ ions are readily oxidised by air (oxygen) to Fe³⁺, so a fresh solution is needed to keep the iron as pure Fe²⁺.

(ii) The first jar of hydrogen is discarded because it is mixed with the air already in the apparatus, forming an explosive mixture; discarding it ensures the hydrogen collected afterwards is pure and safe.

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Soma maelezo ya somo kwenye Solutions (WAEC)

Soma maelezo ya somo kwenye Acids, Bases And Salts (Ghana Only) (WAEC)

Chemistry, Industry And The Environment

Solutions

Acids, Bases And Salts (Ghana Only)

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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 one of the following substances; starch or sucrose or glucose D is a simple salt. 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) Add about 5 cm\(^3\) of distilled water to a portion of C in a test tube. Stir thoroughly and test with litmus

(ii) Add about 2cm\(^2\) of Fehling's solution to the resulting mixture from (a)(i) above the heat.

(b)(i) Heat a portion of D strongly in a test tube

(ii) Put the rest of D in a boiling tube and add about 10 cm\(^3\) of distilled water. Shake the mixture

(iii) Put about 2 cm\(^3\) of the mixture from (b)(ii)) in a test tube. Add aqueous ammonia in drops and then in excess

Jibu

Test	Observation	Inference
(a)(i) Add distilled water to C, stir and test with litmus paper.	C dissolves to give a colourless solution. There is no change in either red or blue litmus paper.	C is soluble in water and is neutral. It is a carbohydrate (sugar).
(a)(ii) Add Fehling's solution to the mixture and heat.	A brick-red precipitate is formed.	C is a reducing sugar. Therefore, C is glucose.
(b)(i) Heat D strongly in a dry test tube.	A reddish-brown gas is evolved and a black residue remains.	The gas is nitrogen(IV) oxide, \(\mathrm{NO_2}\), showing the presence of nitrate ions, \(\mathrm{NO_3^-}\). The black residue is copper(II) oxide, \(\mathrm{CuO}\).
(b)(ii) Add distilled water to the remaining D and shake.	D dissolves to give a light-blue solution.	D is soluble and copper(II) ions, \(\mathrm{Cu^{2+}}\), are suspected.
(b)(iii) Add aqueous ammonia dropwise to the solution of D, then add it in excess.	A pale-blue precipitate is formed. The precipitate dissolves in excess aqueous ammonia to give a deep-blue solution.	Copper(II) ions, \(\mathrm{Cu^{2+}}\), are confirmed.

Test	Observation	Inference
(a)(i) Add distilled water to C, stir and test with litmus paper.	C dissolves to give a colourless solution. There is no change in either red or blue litmus paper.	C is soluble in water and is neutral. It is a carbohydrate (sugar).
(a)(ii) Add Fehling's solution to the mixture and heat.	A brick-red precipitate is formed.	C is a reducing sugar. Therefore, C is glucose.
(b)(i) Heat D strongly in a dry test tube.	A reddish-brown gas is evolved and a black residue remains.	The gas is nitrogen(IV) oxide, \(\mathrm{NO_2}\), showing the presence of nitrate ions, \(\mathrm{NO_3^-}\). The black residue is copper(II) oxide, \(\mathrm{CuO}\).
(b)(ii) Add distilled water to the remaining D and shake.	D dissolves to give a light-blue solution.	D is soluble and copper(II) ions, \(\mathrm{Cu^{2+}}\), are suspected.
(b)(iii) Add aqueous ammonia dropwise to the solution of D, then add it in excess.	A pale-blue precipitate is formed. The precipitate dissolves in excess aqueous ammonia to give a deep-blue solution.	Copper(II) ions, \(\mathrm{Cu^{2+}}\), are confirmed.

Solubility

Acids, Bases And Salts

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Swali 3 Ripoti

All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experiment procedure is required. All calculations must be done in your answer book.

A is mol dm HCI. B is a solution containing 15.0 g dm of a mixture of NaCl and KHCO\(_3\).

(a) Put A 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:

\[ \mathrm{HCl}_{aq} + \mathrm{KHCO}_{3(aq)} \to \mathrm{KCl}_{(aq)} + \mathrm{CO}_{2(g)} \]

(b) From your results and the information provided above, calculate the:

(i) concentration of KHCO\(_3\), in \(\text{mol dm}^{-3}\) in B;

(ii) mass of KHCO\(_3\), in \(\text{g dm}^{-3}\) in B

(ii) Percentage by mass of KHCO\(_3\) in the mixture, [H=1; C = 12; O = 16; K = 39]

(iv) mass of NaCl in the mixture.

Jibu

Titration results

Use a 25.0 cm³ pipette to measure solution B. The rough titre is not included in the average. The concordant titres are 26.00 cm³, 26.05 cm³ and 25.95 cm³.

Reading	Rough	1st accurate	2nd accurate	3rd accurate
Final burette reading / cm³	27.00	36.00	26.05	40.95
Initial burette reading / cm³	0.00	10.00	0.00	15.00
Volume of A used / cm³	27.00	26.00	26.05	25.95

Volume of pipette \(=25.0\ \text{cm}^3\).

Average volume of A used:

\[ \frac{26.00+26.05+25.95}{3}=26.00\ \text{cm}^3 \]

The equation must include water, since hydrogen carbonate reacts with acid to form carbon dioxide and water:

\[ \mathrm{HCl_{(aq)}+KHCO_{3(aq)}\rightarrow KCl_{(aq)}+CO_{2(g)}+H_2O_{(l)}} \]

Only \(\mathrm{KHCO_3}\) reacts with hydrochloric acid. Sodium chloride does not react, so it does not affect the titre. The mole ratio of \(\mathrm{HCl}\) to \(\mathrm{KHCO_3}\) is \(1:1\).

(i) Concentration of \(\mathrm{KHCO_3}\) in B

Using the stated concentration of A, \(0.100\ \text{mol dm}^{-3}\):

\[ n(\mathrm{HCl})=CV \]

\[ =0.100\times\frac{26.00}{1000} =2.60\times10^{-3}\ \text{mol} \]

Therefore, by the \(1:1\) ratio:

\[ n(\mathrm{KHCO_3})=2.60\times10^{-3}\ \text{mol} \]

This amount is present in \(25.0\ \text{cm}^3=0.0250\ \text{dm}^3\) of B:

\[ [\mathrm{KHCO_3}] =\frac{2.60\times10^{-3}}{0.0250} =\boxed{0.104\ \text{mol dm}^{-3}} \]

(ii) Mass concentration of \(\mathrm{KHCO_3}\) in B

\[ M_r(\mathrm{KHCO_3})=39+1+12+(3\times16)=100 \]

\[ \text{Mass concentration} =0.104\times100 =\boxed{10.4\ \text{g dm}^{-3}} \]

(iii) Percentage by mass of \(\mathrm{KHCO_3}\) in the mixture

The total mass concentration of the mixture is \(15.0\ \text{g dm}^{-3}\), of which \(10.4\ \text{g dm}^{-3}\) is \(\mathrm{KHCO_3}\):

\[ \%\,\mathrm{KHCO_3} =\frac{10.4}{15.0}\times100 =\boxed{69.3\%} \]

(iv) Mass of \(\mathrm{NaCl}\) in the mixture

\[ \text{Mass concentration of NaCl} =15.0-10.4 =\boxed{4.6\ \text{g dm}^{-3}} \]

The percentage calculation must use \(\frac{\text{mass of KHCO}_3}{\text{total mass of mixture}}\times100\). Using \(15.0-10.4\) in this fraction would calculate the percentage of sodium chloride instead, not the percentage of potassium hydrogencarbonate.

Soma maelezo ya somo kwenye Chemical Equilibria (JAMB)

Maelezo ya Majibu

Titration results

Use a 25.0 cm³ pipette to measure solution B. The rough titre is not included in the average. The concordant titres are 26.00 cm³, 26.05 cm³ and 25.95 cm³.

Reading	Rough	1st accurate	2nd accurate	3rd accurate
Final burette reading / cm³	27.00	36.00	26.05	40.95
Initial burette reading / cm³	0.00	10.00	0.00	15.00
Volume of A used / cm³	27.00	26.00	26.05	25.95

Volume of pipette \(=25.0\ \text{cm}^3\).

Average volume of A used:

\[ \frac{26.00+26.05+25.95}{3}=26.00\ \text{cm}^3 \]

The equation must include water, since hydrogen carbonate reacts with acid to form carbon dioxide and water:

\[ \mathrm{HCl_{(aq)}+KHCO_{3(aq)}\rightarrow KCl_{(aq)}+CO_{2(g)}+H_2O_{(l)}} \]

Only \(\mathrm{KHCO_3}\) reacts with hydrochloric acid. Sodium chloride does not react, so it does not affect the titre. The mole ratio of \(\mathrm{HCl}\) to \(\mathrm{KHCO_3}\) is \(1:1\).

(i) Concentration of \(\mathrm{KHCO_3}\) in B

Using the stated concentration of A, \(0.100\ \text{mol dm}^{-3}\):

\[ n(\mathrm{HCl})=CV \]

\[ =0.100\times\frac{26.00}{1000} =2.60\times10^{-3}\ \text{mol} \]

Therefore, by the \(1:1\) ratio:

\[ n(\mathrm{KHCO_3})=2.60\times10^{-3}\ \text{mol} \]

This amount is present in \(25.0\ \text{cm}^3=0.0250\ \text{dm}^3\) of B:

\[ [\mathrm{KHCO_3}] =\frac{2.60\times10^{-3}}{0.0250} =\boxed{0.104\ \text{mol dm}^{-3}} \]

(ii) Mass concentration of \(\mathrm{KHCO_3}\) in B

\[ M_r(\mathrm{KHCO_3})=39+1+12+(3\times16)=100 \]

\[ \text{Mass concentration} =0.104\times100 =\boxed{10.4\ \text{g dm}^{-3}} \]

(iii) Percentage by mass of \(\mathrm{KHCO_3}\) in the mixture

The total mass concentration of the mixture is \(15.0\ \text{g dm}^{-3}\), of which \(10.4\ \text{g dm}^{-3}\) is \(\mathrm{KHCO_3}\):

\[ \%\,\mathrm{KHCO_3} =\frac{10.4}{15.0}\times100 =\boxed{69.3\%} \]

(iv) Mass of \(\mathrm{NaCl}\) in the mixture

\[ \text{Mass concentration of NaCl} =15.0-10.4 =\boxed{4.6\ \text{g dm}^{-3}} \]

The percentage calculation must use \(\frac{\text{mass of KHCO}_3}{\text{total mass of mixture}}\times100\). Using \(15.0-10.4\) in this fraction would calculate the percentage of sodium chloride instead, not the percentage of potassium hydrogencarbonate.

Soma maelezo ya somo kwenye Chemical Equilibria (JAMB)

Chemical Equilibria

Tazama Jibu

WAEC SSCE - Chemistry - 2001 (Practical)

Tests on C and D

Tests on C and D






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