Most Common Repeated Waec Chemistry Practical Specimen Questions

Waec Chemistry Practical
Waec Chemistry Practical
Written by Nkedugist

WAEC Chemistry Practical questions to expect. Many WAEC candidates find it difficult to answer the WAEC Practical chemistry correctly due to not understanding the questions. We are going to be releasing the most Common Repeated Waec Chemistry Practical Specimen Questions which are likely to come out.

In this article, we are going to publish most chemistry specimens practical questions you are to expect during your exam day with instructions for the upcoming WAEC examinations for all Waec candidates who are writing.


Waec Chemistry Practical Alternative A

  1. Dilute sodium hydroxide solution
  2. Dilute ammonia solution
  3. Dilute hydrochloric acid
  4. Distilled water
  5. Red and blue litmus paper
  6. Fehlings solution (I II)
  7. Concentrated tetraoxosulphate (vi) solution (H2SO4)
  8. Freshly prepared iron (II) tetraoxosulphate (VI) solution (Fe2SO4), burette (50cm 3), pipette (20cm 3 or 25cm 3), filtration apparatus, methyl orange, boiling tube, test tubes, SALTS: CuO,ZnSO4, Na2CO3
 Measurement Instructions
  • One burette of 50cm3 capacity
  • One pipette, either 20cm3 or 25cm3. All candidates at one centre must use pipettes of the same volume. These should be clean and free from grease

Waec Chemistry Practical Alternative B

(i) Red and blue litmus paper
(ii) Aqueous ammonia
(iii) Dilute hydrochloric acid
(iv) Dilute sodium hydroxide solution
(v) Barium chloride solution
(vi) Dilute trioxonitrate (V) acid
(vii) Silver trioxonitrate (V) acid

(a) One burette of 50cm^3 Capacity
(b) One pipette, either 20cm^3 or 25cm^3. All candidates at one centre must use pipettes of the same volume. These should be clean and free from grease.
(c) The usual apparatus for titration;
(d) The usually apparatus and reagent for qualitative work include the following with all reagents appropriately labelled;


A is a solution containing 6.22 g of an acid H2Y per dm3.
B contains 3.90 g of NaOH per dm3 of solution.
(a)        Put into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portions of using methyl orange as an indicator. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of acid A used.
The equation for the reaction involved in the titration is:
H2Y(aq) + 2NaOH(aq)                 Na2Y(aq)  + 2H2O(l)
[H = 1.00; O = 16.0; Na = 23.0]                                                               [10 marks]

(b)        From your result and the information provided above, calculate the:
(i) concentration of B in moldm-3
(ii) concentration of A in moldm-3;
(iii) molar mass of H2Y.                                                                                         [9 marks] (c)        State whether the pH of each of the following solutions is lower than 7, greater than 7 or equal to 7. The:

  1. solution A before titration;
  2. solution B before titration.

art (a): This question was well attempted by majority of the candidates and the performance was quite good. Candidates correctly tabulated their burette readings and calculated the average volume of A used. Some candidates lost marks as a result of deliberate alteration/cancellation of titre value to agree with those of the supervisor and for using non concordant readings to calculate the average titre.
In question (b)(i), (ii) and (iii), candidates calculates the concentration of B in moldm-3, concentration of A in moldm-3 and the molar mass of H2Y. Some candidates however could not express their answers to three significant figures.
Majority of the candidates did not express the molar mass of H2Y in g mol-1

The expected answers were:
(ii)        Concentration of A in mol dm-3
From the reaction equation
CA VA              =                      1
CB VB                                      2
CA        =                  CB VB
2 VA
CA        =               0.0975 x 25
2 VA
                        =         P mol dm-3                  

Alternative Method:
Amount of acid used = 0.025 x 0.0975 = 0.00244 mol
From the equation of reaction:
2 mol of B require 1 mol of A
0.00244 mol of B requires ½ x 0.00244 mol of A
= 0.00122 mol of A.
i.e VA cm3 contain 0.00122 mol.
1000 cm3 contain 0.00122 x 1000/VA mol.
i.e. Conc of A = 1.22/VA = P mol dm-3
(iii)       Molar mass of H2Y
Molar mass of H2Y =    6.22
=          Q g mol-1

What you see is similiar to what you are going to do during your Waec Specimen Practical Exam Day.

All the repeated Waec Specimen Questions for all Waec candidates are published here. The Answers to all the repeated Waec biology specimens will be released shortly. Kindly check here to see all the answers to the Chemistry specimen.


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