2023 WAEC CHEMISTRY PRACTICAL
(d) The usual apparatus and reagents for qualitative work including the following with all reagents appropriately labeled;
(i) Dilute sodium hydroxide solution;
(ii) Dilute hydrochloric acid;
(iii) Dilute trioxonitrate (V) acid;
(iv) Silver trioxonitrate (V) solution;
(v) Acidified potassium dichromate solution;
(vi) Aqueous ammonia;
(vii) Lime water;
(viii) Red and Blue litmus paper;
(ix) Dilute tetraoxosulphate (VI) acid;
(x) Fehlings solution A & B.
(e) Spatula;
(f) Filtration apparatus;
(g) One beaker;
(h) One boiling tube;
(i) Four test tube;
(j) Methyl orange as indicator;
(k) Glass rod;
(l) Wash bottle containing distilled/deionized water;
(m) Burning splint;
(n) Watch glass;
(o) Bunsen burner/source of heat;
(p) Droppers;
(q) Mathematical table/calculator.
(a) 150cm3 of a solution of HCL in a corked flask or bottle labeled ‘An’. These should all be the same containing 8.5cm3 of concentrated HCl per dm³ of solution.
(b) 150cm3 of Na2CO3
.10H2O in a corked flask or bottle labeled ‘Bn’. These should all be the same cointaing 5.0g of the hydrated salt per dm3 of solution.
(c) One spatulaful of glucose in a specimen bottle labeled ‘Cn’. This must be the same for all candidate
(d) One spatulaful of zinc oxide powder in a specimen bottle labeled ‘Dn’. This must be the same for all candidates.
: (1a)
Volume of pipette used, VA = 25.0cm³
Indicator Used = Methyl Orange
[TABULATE]
Burette Reading | Rough titre | 1st titre | 2nd titre |
Final Burette reading (cm³) | 25.60 | 32.80 | 26.40
Initial Burette reading (cm³) | 1.20 | 8.70 | 2.30
Volume of HCL used | 24.40 | 24.10 | 24.10
Average burette reading = (24.10 + 24.10)/2
volume of HCL used = 24.10cm³
Number 3a
(i) If a few drops of NaOH solution are added to 2 cm^3 of a solution of Pb(NO3)2, a white precipitate of lead(II) hydroxide [Pb(OH)2] is formed. The balanced equation for the reaction is as follows:
Pb(NO3)2 + 2NaOH → Pb(OH)2 + 2NaNO3
(ii) If a few drops of NaOH solution are added to 2 cm^3 of a solution of Fe2(SO4)2, a green precipitate of iron(II) hydroxide [Fe(OH)2] is formed. However, on exposure to air, the Fe(OH)2 precipitate can be oxidized by atmospheric oxygen to form a reddish-brown precipitate of iron(III) hydroxide [Fe(OH)3]. The balanced equations for the reaction and the subsequent oxidation are as follows:
Fe2(SO4)2 + 4NaOH → 2Fe(OH)2↓ + Na2SO4
4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3
Therefore, the initial greenish precipitate of Fe(OH)2 will turn reddish-brown in color due to the oxidation to Fe(OH)3.
Another 3b🇳🇬
It is important to ensure that the powder is packed tightly into the tube and that the temperature is increased slowly to prevent the sample from overheating or decomposing. The melting process should be observed carefully to ensure accuracy in recording the melting point range. The melting point should be repeated several times to ensure reproducibility and accuracy.
TAKE NOTE
For question 1
Use your school’s endpoint and use it wherever you see our endpoint (15.15cm³) …and calculate with it
Replace our 15.15 in 1b and do the calculation.
If your school wants to use mine, then the chemistry teacher must submit endpoint similar to ours without much difference…eg 15.15, 15.20, 15.30, etc
STILL SOLVING 💫 💫
2B.*
(b) Test for inorganic compound D:
(i) Heating and cooling:
Observation: Upon heating, compound D may undergo a physical change (e.g., melting, sublimation, decomposition) or remain unchanged. After cooling, the appearance and properties of compound D are observed.
Conclusion: The observation after heating and cooling can provide information about the thermal stability, volatility, and physical properties of compound D.
(ii) Addition of dilute HCl and heating:
Observation: If a reaction occurs, such as effervescence (gas evolution) or formation of a precipitate, it indicates a chemical reaction between compound D and HCl.
Conclusion: The reaction with HCl can provide information about the acidic or basic nature of compound D, as well as the possibility of it being a carbonate or bicarbonate.
(iii) Addition of NaOH(aq):
Observation: If a reaction occurs, such as the formation of a precipitate or change in color, it indicates a chemical reaction between compound D and NaOH.
Conclusion: The reaction with NaOH can provide information about the acidic or basic nature of compound D and help identify the presence of certain metal ions or hydroxide compounds.
(iv) Addition of aqueous NH3:
Observation: If a reaction occurs, such as the formation of a precipitate or change in color, it indicates a chemical reaction between compound D and aqueous NH3 (ammonia solution).
Conclusion: The reaction with NH3 can provide information about the acidic or basic nature of compound D and help identify the presence of certain metal ions or complexes.
Number 1 (c) 🇳🇬
We are given that it contains 5.0 g dm^-3 of the hydrated salt. Let’s say the formula of the hydrated salt is XCO3. YH2O.
The molar mass of XCO3. YH2O is:
Molar mass = molar mass of XCO3 + molar mass of YH2O
Molar mass = (40.01 + 3×16.00) + Y(2×1.01 + 16.00)
Where Y is the number of moles of water of crystallisation.
We are given that 5.0 g dm^-3 of the hydrated salt is present, which means that in 1 dm^3 (1000 cm^3) of B, there are 5.0 g of the hydrated salt.
Number of moles of hydrated salt in 1 dm^3 = mass/molar mass
Number of moles of hydrated salt in 1 dm^3 = 5.0 g/(40.01 + 3×16.00 + Y(2×1.01 + 16.00)) mol^-1 dm^3
Number of moles of anhydrous salt in 1 dm^3 = 5.0 g/(40.01 + 3×16.00 + Y(2×1.01 + 16.00)) mol^-1 dm^3
Mass of anhydrous salt = concentration x volume x molar mass
For 20.0cm^3: 0.030 x 20.0cm^3 x (40.01 + 3×16.00 + Y(2×1.01 + 16.00)) g dm^-3
For 25.0cm^3: 0.030 x 25.0cm^3 x (40.01 + 3×16.00 + Y(2×1.01 + 16.00)) g dm^-3
We can equate this mass to the mass of anhydrous salt in 1 dm^3 and solve for Y:
Mass of anhydrous salt in 1 dm^3 = Mass of anhydrous salt in
COMPLETE NUMBER 3
3a.
When a few drops of NaOH solution are added to a 2 cm³ solution of Pb(NO3)2 (lead nitrate), the following observation would be made:
A white precipitate of lead(II) hydroxide (Pb(OH)2) would form. Lead(II) hydroxide is insoluble in water and appears as a white solid. The reaction can be represented as:
Pb(NO3)2 + 2NaOH -> Pb(OH)2 + 2NaNO3
The formation of a white precipitate indicates the presence of lead(II) ions in the solution, and the reaction between Pb(NO3)2 and NaOH results in the formation of lead(II) hydroxide.
3aii.
When a few drops of NaOH solution are added to a 2 cm³ solution of Fe2(SO4)3 (iron(III) sulfate), the following observation would be made:
A reddish-brown precipitate of iron(III) hydroxide (Fe(OH)3) would form. Iron(III) hydroxide is insoluble in water and appears as a reddish-brown solid. The reaction can be represented as:
Fe2(SO4)3 + 6NaOH -> 2Fe(OH)3 + 3Na2SO4
3b
To determine the melting point of benzoic acid in the laboratory, the following procedure can be followed:
By following this procedure, the melting point of benzoic acid can be determined in the laboratory.
CHEMISTRY SOLUTION
2A.)
a) Test for organic compound C:
(i) Litmus paper test:
Observation: The litmus paper remains unchanged.
Conclusion: No significant change in the litmus paper indicates that compound C is likely neutral or non-acidic.
(ii) Fehling’s test:
Observation: No significant reaction or color change occurs upon heating.
Conclusion: The lack of reaction with Fehling’s solution indicates that compound C does not contain a reducing sugar. It suggests that compound C is not a carbohydrate or does not possess aldehyde or ketone functional groups.
(b) Test for inorganic compound D:
2B.
(b) Test for inorganic compound D:
(i) Heating and cooling:
Observation: Upon heating, compound D may undergo a physical change (e.g., melting, sublimation, decomposition) or remain unchanged. After cooling, the appearance and properties of compound D are observed.
Conclusion: The observation after heating and cooling can provide information about the thermal stability, volatility, and physical properties of compound D.
(ii) Addition of dilute HCl and heating:
Observation: If a reaction occurs, such as effervescence (gas evolution) or formation of a precipitate, it indicates a chemical reaction between compound D and HCl.
Conclusion: The reaction with HCl can provide information about the acidic or basic nature of compound D, as well as the possibility of it being a carbonate or bicarbonate.
(iii) Addition of NaOH(aq):
Observation: If a reaction occurs, such as the formation of a precipitate or change in color, it indicates a chemical reaction between compound D and NaOH.
Conclusion: The reaction with NaOH can provide information about the acidic or basic nature of compound D and help identify the presence of certain metal ions or hydroxide compounds.
(iv) Addition of aqueous NH3:
Observation: If a reaction occurs, such as the formation of a precipitate or change in color, it indicates a chemical reaction between compound D and aqueous NH3 (ammonia solution).
Conclusion: The reaction with NH3 can provide information about the acidic or basic nature of compound D and help identify the presence of certain metal ions or complexes.
COMPLETED
✅✅
Number 1 (c) 🇳🇬
We are given that it contains 5.0 g dm^-3 of the hydrated salt. Let’s say the formula of the hydrated salt is XCO3. YH2O.
The molar mass of XCO3. YH2O is:
Molar mass = molar mass of XCO3 + molar mass of YH2O
Molar mass = (40.01 + 3×16.00) + Y(2×1.01 + 16.00)
Where Y is the number of moles of water of crystallisation.
We are given that 5.0 g dm^-3 of the hydrated salt is present, which means that in 1 dm^3 (1000 cm^3) of B, there are 5.0 g of the hydrated salt.
Number of moles of hydrated salt in 1 dm^3 = mass/molar mass
Number of moles of hydrated salt in 1 dm^3 = 5.0 g/(40.01 + 3×16.00 + Y(2×1.01 + 16.00)) mol^-1 dm^3
Number of moles of anhydrous salt in 1 dm^3 = 5.0 g/(40.01 + 3×16.00 + Y(2×1.01 + 16.00)) mol^-1 dm^3
Mass of anhydrous salt = concentration x volume x molar mass
For 20.0cm^3: 0.030 x 20.0cm^3 x (40.01 + 3×16.00 + Y(2×1.01 + 16.00)) g dm^-3
For 25.0cm^3: 0.030 x 25.0cm^3 x (40.01 + 3×16.00 + Y(2×1.01 + 16.00)) g dm^-3
We can equate this mass to the mass of anhydrous salt in 1 dm^3 and solve for Y:
Mass of anhydrous salt in 1 dm^3 = Mass of anhydrous salt in
(2a)
TEST; C + distilled water + shake
OBSERVATIONS; It dissolves to give a colourless or clear solution
INFERENCE; Soluble salt suspected
(2ai)
TEST; Solution C + litmus paper
OBSERVATIONS; It has no effect on both blue and red litmus paper
INFERENCE; Neutral solution
(2aii)
TEST; solution C + Fehling solution A and B + heat
OBSERVATIONS; It gives a black red colour
INFERENCE; Reading agent confirmed
(2bi)
TEST; D + heat
OBSERVATIONS; It turns yellow on heating and white on cooling
INFERENCE; ZnO is present
(2bii)
TEST; D + dilute Hcl + heat
OBSERVATIONS; It dissolves completely to give a clear or colourless solution
INFERENCE; Solution chlorides of Zn³+, Al³+ is present
(2biii)
TEST; Solution D + NaOH in drops and then in excess
OBSERVATIONS; It gives white gelatinous precipitate. The precipitate is soluble in excess.
INFERENCE; Al³+, Zn²+ present.
Al³+, Zn²+ present .
(2biv)
TEST; Solution D + NH3 in drops and then in excess
OBSERVATIONS; It gives white gelatinous precipitate. The precipitate is soluble in excess NH3
INFERENCE; Al³+, Zn²+ is present.
Zn²+ confirmed.
Kleff 💫
Adeyinka is the founder and content creator at Career Acada. He’s a technology expert and web developer. He holds a degree in Genetics and loves impacting life for a better society.
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