QUANTITATIVE ANALYSIS — PART 2

Advanced WAEC titration masterclass: mole calculations, dilution, mass concentration, percentage purity, water of crystallization, error control and full examiner-style problem solving.

Part 2 moves beyond basic titration table work into the deeper calculations WAEC may ask after the titre has been obtained.

1. What Makes Quantitative Analysis Advanced?

In Part 1, you learnt how to prepare the titration table, obtain concordant titres and calculate concentration. In Part 2, you will learn how WAEC extends the same titration data into deeper questions.

Calculation Extension

WAEC may ask for concentration in mol dm⁻³, g dm⁻³, mass in a given volume, or moles of ions.

Dilution

You may be asked to prepare a weaker solution from a stronger one using \(C_1V_1=C_2V_2\).

Purity

You may be asked to determine percentage purity of an impure acid, alkali or carbonate sample.

WAEC WARNING: Do not jump into substitution. First write the balanced equation and identify the mole ratio.

2. Formula Bank for WAEC Quantitative Analysis

FormulaMeaningUse
\(n=CV\)Moles = concentration × volume in dm³Finding amount of reacting substance
\(C=\frac{n}{V}\)Concentration = moles ÷ volumeFinding mol dm⁻³
\(C=\frac{mass\ concentration}{molar\ mass}\)Convert g dm⁻³ to mol dm⁻³When solution strength is given in g dm⁻³
\(C_1V_1=C_2V_2\)Dilution formulaPreparing diluted solution
\(\%\ purity=\frac{pure\ mass}{impure\ mass}\times100\)Percentage purityImpure sample questions
\(mass=n \times molar\ mass\)Mass from molesMass calculation
\[ H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l) \] \[ H_2SO_4:NaOH=1:2 \]

Teacher

Students, the formula is not the beginning. The equation is the beginning. The formula only works correctly when the mole ratio is correctly taken from the balanced equation.

Student

Sir, must I convert cm³ to dm³?

Teacher

Yes, when using \(n=CV\), volume must be in dm³. Since \(1000cm^3=1dm^3\), divide cm³ by 1000.

3. Dilution Calculations

Example

Calculate the volume of \(2.00mol\,dm^{-3}\) HCl required to prepare \(500cm^3\) of \(0.200mol\,dm^{-3}\) HCl.

\[ C_1V_1=C_2V_2 \] \[ 2.00\times V_1=0.200\times500 \] \[ V_1=\frac{0.200\times500}{2.00}=50.0cm^3 \]
Volume of concentrated acid required = \(50.0cm^3\). Volume of water needed = \(500-50=450cm^3\).

Teacher

In the laboratory, you measure 50.0 cm³ of the stronger acid and make it up to 500 cm³ in a volumetric flask. You do not simply add 450 cm³ carelessly in any container if precision is required.

Diagram Guide

Dilution using a volumetric flask Teacher guide: Label the pipette or measuring cylinder, concentrated solution, volumetric flask, calibration mark and distilled water. The final volume must be made exactly to the mark.

4. Percentage Purity from Titration

WAEC-Style Example

An impure sample of sodium carbonate was dissolved and made up to \(250cm^3\). \(25.0cm^3\) of the solution required \(24.80cm^3\) of \(0.100mol\,dm^{-3}\) HCl. Calculate the mass of pure sodium carbonate in the \(250cm^3\) solution and the percentage purity if the original impure sample weighed \(1.50g\).

\[ Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2 \] \[ n(HCl)=\frac{0.100\times24.80}{1000}=0.00248mol \] \[ n(Na_2CO_3)=\frac{0.00248}{2}=0.00124mol \] \[ Mass\ in\ 25.0cm^3=0.00124\times106=0.13144g \] \[ Mass\ in\ 250cm^3=0.13144\times10=1.3144g \] \[ \%\ purity=\frac{1.3144}{1.50}\times100=87.6\% \]

Teacher

The most common mistake here is forgetting that only 25.0 cm³ was titrated, while the whole solution is 250 cm³. Since \(250/25=10\), multiply the mass in the aliquot by 10.

Student

Sir, why did we divide HCl moles by 2?

Teacher

Because the equation shows that 2 moles of HCl react with 1 mole of sodium carbonate. The carbonate moles are half the HCl moles.

5. Water of Crystallization

Concept

Some salts contain water molecules in their crystal structure. WAEC may ask you to determine \(x\) in a formula such as \(Na_2CO_3\cdot xH_2O\).

Example

A hydrated sodium carbonate sample has molar mass \(286g\,mol^{-1}\). Find \(x\) in \(Na_2CO_3\cdot xH_2O\).

\[ Molar\ mass\ of\ Na_2CO_3=106 \] \[ Mass\ of\ water=286-106=180 \] \[ x=\frac{180}{18}=10 \] \[ Formula=Na_2CO_3\cdot10H_2O \]

Teacher

Water of crystallization questions are easy when you separate the salt mass from the water mass. The water part is always divided by 18 because \(H_2O=18\).

6. Full WAEC-Style Quantitative Analysis Task

Question

A is dilute \(H_2SO_4\). B is NaOH solution containing \(4.0g\,dm^{-3}\). \(25.0cm^3\) of B required the following volumes of A:

TitrationRough1st2nd3rd
Final reading / cm³12.4012.1512.1012.15
Initial reading / cm³0.000.000.000.00
Titre / 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.
  6. Calculate volume of A needed for \(20.0cm^3\) of B.

WAEC Solution

\[ Average=\frac{12.15+12.10+12.15}{3}=12.13cm^3 \] \[ C_B=\frac{4.0}{40}=0.100mol\,dm^{-3} \] \[ \frac{C_A\times12.13}{0.100\times25.0}=\frac{1}{2} \] \[ C_A=\frac{0.100\times25.0}{2\times12.13}=0.103mol\,dm^{-3} \] \[ Mass\ concentration=0.103\times98=10.1g\,dm^{-3} \] \[ n(H_2SO_4)=\frac{0.103\times12.13}{1000}=1.25\times10^{-3}mol \] \[ n(H^+)=2.50\times10^{-3}mol \] \[ V_A=\frac{0.100\times20.0}{2\times0.103}=9.71cm^3 \]

7. Accuracy, Error Control and WAEC Marking

ProblemResultHow to Avoid It
Using rough titre in averageWrong averageUse only concordant titres
Wrong mole ratioWrong concentrationWrite balanced equation first
Not converting cm³ to dm³Wrong molesDivide volume by 1000
Endpoint overshotTitre too highAdd solution dropwise near endpoint
Endpoint stopped too earlyTitre too lowSwirl well and wait for permanent colour change
No unitLoss of markAttach \(mol\,dm^{-3}\), \(g\,dm^{-3}\), cm³ or mol
WAEC STRATEGY: Your calculation must be readable. Arrange work in steps: equation, mole ratio, given data, substitution, answer and unit.

8. Part 2 Exam-Ready Summary

Dilution

Use \(C_1V_1=C_2V_2\).

Purity

\(\% purity=\frac{pure\ mass}{impure\ mass}\times100\).

Hydrated Salt

Subtract anhydrous mass from hydrated mass, then divide water mass by 18.

Titration

Always begin with the balanced equation and mole ratio.

Average Titre

Average only concordant titres. Ignore rough titre.

Final Answer

Always include units and suitable significant figures.

You are ready for Quantitative Analysis Part 3: complete WAEC practical model papers and independent calculation mastery.