QUANTITATIVE ANALYSIS — PART 3

Final WAEC quantitative analysis mastery class: full model practical papers, independent titration problem-solving, examiner-style marking schemes, advanced errors, graphical reasoning and real practical strategies for accurate values.

Part 3 turns students from learners into exam-ready practical candidates. It combines theory, laboratory procedure, accurate titre simulation and full WAEC marking style.

1. Final Quantitative Analysis Mindset

At WAEC level, quantitative analysis is not only about titration. It is about proving that you can measure accurately, record neatly, calculate logically and explain the chemistry behind the values obtained.

Accurate Recording

Record initial and final burette readings to two decimal places. Use only concordant titres.

Mole Ratio First

Every titration calculation begins with the balanced equation and mole ratio.

Examiner Style

WAEC rewards method, units, correct substitution, significant figures and final answer.

FINAL WARNING: A correct numerical answer with no working may lose marks. WAEC practical chemistry rewards traceable calculation steps.

2. Model Paper 1 — Strong Acid vs Strong Alkali

Question

A is dilute tetraoxosulphate(VI) acid, \(H_2SO_4\). B is sodium hydroxide solution containing \(4.0g\,dm^{-3}\) of NaOH. \(25.0cm^3\) of B was titrated against A using phenolphthalein indicator.

TitrationRough1st2nd3rd
Final burette reading / cm³12.4012.1512.1012.15
Initial burette reading / cm³0.000.000.000.00
Volume of A used / cm³12.4012.1512.1012.15
  1. Calculate the average titre.
  2. Calculate concentration of B in \(mol\,dm^{-3}\).
  3. Calculate concentration of A in \(mol\,dm^{-3}\).
  4. Calculate mass concentration of A in \(g\,dm^{-3}\).
  5. Calculate moles of \(H^+\) in the average titre.

Teacher

This is the exact type of calculation that can appear in WAEC. The secret is to respect the mole ratio \(1:2\).

\[ Average=\frac{12.15+12.10+12.15}{3}=12.13cm^3 \] \[ Molar\ mass\ of\ NaOH=23+16+1=40g\,mol^{-1} \] \[ C_B=\frac{4.0}{40}=0.100mol\,dm^{-3} \] \[ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O \] \[ H_2SO_4:NaOH=1:2 \] \[ \frac{C_A\times12.13}{0.100\times25.0}=\frac{1}{2} \] \[ C_A=\frac{0.100\times25.0}{2\times12.13}=0.103mol\,dm^{-3} \] \[ Mass\ concentration=0.103\times98=10.1g\,dm^{-3} \] \[ n(H_2SO_4)=\frac{0.103\times12.13}{1000}=1.25\times10^{-3}mol \] \[ n(H^+)=2.50\times10^{-3}mol \]
Endpoint with phenolphthalein: pink to colourless.

3. Model Paper 2 — Methyl Orange Variant

Question

A is dilute \(H_2SO_4\). B is \(0.100mol\,dm^{-3}\) NaOH solution. \(20.0cm^3\) of B was pipetted into a conical flask and titrated with A using methyl orange indicator.

TitrationRough1st2nd3rd
Final reading / cm³10.209.809.759.80
Initial reading / cm³0.000.000.000.00
Titre / cm³10.209.809.759.80
  1. Calculate the average titre.
  2. Calculate the concentration of A.
  3. Calculate the mass concentration of A.
  4. State the colour change of methyl orange.

Teacher

The lower titre occurs because only \(20.0cm^3\) of NaOH was used instead of \(25.0cm^3\). Always connect titre size to volume used.

\[ Average=\frac{9.80+9.75+9.80}{3}=9.78cm^3 \] \[ \frac{C_A\times9.78}{0.100\times20.0}=\frac{1}{2} \] \[ C_A=\frac{0.100\times20.0}{2\times9.78}=0.102mol\,dm^{-3} \] \[ Mass\ concentration=0.102\times98=10.0g\,dm^{-3} \]
Methyl orange colour change: yellow in alkali to orange/pink at endpoint.

4. Hard Calculation Masterclass

Task A: Find Mass in a Given Volume

If \(A=0.103mol\,dm^{-3}\) \(H_2SO_4\), calculate the mass of \(H_2SO_4\) in \(250cm^3\) of A.

\[ n=CV=\frac{0.103\times250}{1000}=0.02575mol \] \[ mass=n\times molar\ mass=0.02575\times98=2.52g \]

Task B: Find Volume Needed for Neutralization

Calculate the volume of \(0.103mol\,dm^{-3}\) \(H_2SO_4\) needed to neutralize \(20.0cm^3\) of \(0.100mol\,dm^{-3}\) NaOH.

\[ \frac{C_A V_A}{C_B V_B}=\frac{1}{2} \] \[ V_A=\frac{0.100\times20.0}{2\times0.103}=9.71cm^3 \]

Task C: Calculate Percentage Error

A student obtained \(12.50cm^3\) instead of the correct \(12.13cm^3\). Calculate percentage error.

\[ Error=12.50-12.13=0.37cm^3 \] \[ \% error=\frac{0.37}{12.13}\times100=3.05\% \]

Teacher

WAEC may not always stop at concentration. You should be ready for mass, volume, percentage error and moles of ions. The same data can generate many questions.

Student

Sir, why do we divide cm³ by 1000?

Teacher

Because concentration in \(mol\,dm^{-3}\) works with volume in \(dm^3\). Since \(1000cm^3=1dm^3\), divide by 1000.

5. Result Analysis and Graphical Reasoning

Although WAEC chemistry titration usually relies on tables, students must understand how repeated readings show accuracy. Closely grouped titres show good technique.

StudentConcordant Titres / cm³Comment
A12.15, 12.10, 12.15Excellent concordance
B12.10, 12.70, 13.20Poor technique; likely endpoint error
C11.90, 12.00, 12.05Close, but may have stopped slightly early
Correct burette reading at the meniscus Teacher guide: Read the bottom of the meniscus at eye level and record to two decimal places. Good titre values cluster closely; poor titres scatter widely.
WAEC PRACTICAL SKILL: Concordance is evidence of repeatability. A student who repeats carefully earns more confidence in the result.

6. Advanced Errors and Their Effects

ErrorEffect on TitreEffect on Calculated Acid Concentration
Overshooting endpointToo highCalculated acid concentration becomes too low
Stopping before endpointToo lowCalculated acid concentration becomes too high
Burette contains water before adding acidMay require larger titreAcid appears less concentrated
Pipette contains water before measuring alkaliMay require smaller titreAcid appears more concentrated
Air bubble in burette jetUnreliable; often too highAcid concentration may be underestimated
Reading top of meniscus instead of bottomWrong readingWrong concentration

Teacher

If titre is too high, your formula often makes the calculated acid concentration lower because the same amount of alkali appears to need more acid volume.

Student

Sir, which is worse: wrong endpoint or wrong mole ratio?

Teacher

Wrong mole ratio is more dangerous because it affects every later calculation. Endpoint error affects the titre, but wrong ratio destroys the chemistry.

7. Final Quantitative Analysis Exam-Ready Summary

Step 1

Write the balanced chemical equation.

Step 2

Extract the mole ratio correctly.

Step 3

Calculate average titre using only concordant titres.

Step 4

Convert g dm⁻³ to mol dm⁻³ where necessary.

Step 5

Substitute carefully and attach units.

Step 6

Check whether your answer is chemically reasonable.

Final target: For \(25.0cm^3\) of \(0.100mol\,dm^{-3}\) NaOH, a titre around \(12.1cm^3\) of about \(0.103mol\,dm^{-3}\) \(H_2SO_4\) is realistic because \(H_2SO_4\) is dibasic.
FINAL WAEC WARNING: Never ignore units, never average rough titre, never use an unbalanced equation, and never forget the \(1:2\) acid-alkali ratio.