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SubjectFree lesson

Quantitative and qualitative analysis

ClassNotes Team 9 MIN READUPDATED 5 JUL 2026

CHEMISTRY SSS3 Second Term

WEEK 2

Quantitative and qualitative analysis

Performance objectives

Students should be able to:

  1. Solve calculations on acid and base titration.
  2. Solve calculations on redox reaction.
  3. Explain tests for oxidants and reductants.

Content

Calculations on acid base titration

Worked example

  1. The volume of 0.20M H2SO4 that will exactly neutralize 25cm3 of 0.05M of NaOH is?

Solution

The equation of reaction is

H2SO4 + 2NaOH -------à Na2SO4 + 2H2O

1mole      2moles

Mole ratio of the acid to base is 1:2

Concentration of acid CA = 0.20M

Concentration of base CB = 0.05M

Volume of acid VA =?

Volume of base V­B = 25cm3

By using CA VA/CB VB = nA/nB

=0.20 x VA/0.05 x 25 =1/2

VA=0.05 x 25 x 1/0.20 x2 =3.125 cm3

  1. A is a dilute solution of tetraoxosulphate (vi) acid. B is a solution of 2.0g per 250 cm3 of sodium hydroxide. 25 cm3 of B required 20 cm3 of A for complete neutralization. Calculate
  1. Concentration of B in mol dm3
  2. Concentration of A in mol dm3

(NaOH = 40g/mol, H2S04=98g/mol)

Solution

Equation of reaction

H2SO4 + 2NaOH -------à Na2SO4 + 2H2O

1mole      2moles

Mole ratio of the acid to base is 1:2

  1. Mass concentration of B = mass/ volume(dm3)

=2.0/0.250= 8gdm3

Concentration of B in mol dm3 = mass conc/ molar mass of B

= 8/40 = 0.2moldm3

  1. Concentration of A in mol dm3

By using CA VA/CB VB = nA/nB

CA   x 20.0/0.2 x 25   = 1/2

CA = 0.2 x 25 x 1/ 2 x 20 = 0.125moldm3

  1. A is 0.100 mol dm3 of Hcl

B is a solution containing impure 15gdm3 of KHCO3 if 25cm3 of B required 26cm3 of A for complete neutralization, calculate the:

  1. Concentration of KHCO3 in mol dm3
  2. Mass of KHCO3 in gdm3
  3. Percentage purity of KHCO3 in the mixture

(H=1, O=16, CL=35.5, K=39, C=12)

Solution

Equation of reaction

HCl + KHCO3 -----à KCl + CO2 + H2O

1 mole 1 mole

  1. Concentration of KHCO3 in mol dm3

By using CAVA/CBVB = nA/nB

0.100 x 26/CB x 25 = 1/1

=2.6/25 = 0.104 mol dm3

  1. Mass concentration of KHCO3 in gdm3

Mass concentration= molar concentration x molar mass of KHCO3

Molar mass of KHCO3 = 100g/mol

Mass concentration= 0.104 x 100 = 10.4 gdm3

  1. Percentage purity = mass of pure / mass of impure x 100

= 10.4/15 x 100/1 =69.3%

Balancing redox equations

Balancing redox reactions is slightly more complex than balancing standard reactions, but still follows a relatively simple set of rules. One major difference is the necessity to know the half-reactions of the involved reactants; a half-reaction table is very useful for this. Half-reactions are often useful in that two half reactions can be added to get a total net equation. Although the half-reactions must be known to complete a redox reaction, it is often possible to figure them out without having to use a half-reaction table. This is demonstrated in the acidic and basic solution examples. Besides the general rules for neutral conditions, additional rules must be applied for aqueous reactions in acidic or basic conditions.

The method used to balance redox reactions is called the Half Equation Method. In this method, the equation is separated into two half-equations; one for oxidation and one for reduction.

Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:

  1. Balance elements in the equation other than O and H.
  2. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation.
  3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.
  4. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.)
  5. The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.
  6. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.
  • (If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH- must be added to turn the remaining H+ into water molecules.)
  • The equation can now be checked to make sure that it is balanced.

Example 1: Balancing a neutral solution

Balance the following reaction

Cu+(aq) + Fe(s) → Fe3+(aq) + Cu(s)

Solution

Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:

Cu+(aq) + e−→ Cu(s)

Fe3+(aq)+3e−→ Fe(s)

The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. These yields:

Cu+(aq) + e−→ Cu(s)

 Fe(s) → Fe3+(aq)+3e‑

Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire 

Cu+(aq) + e−→Cu(s) half-reaction by 3 and leaving the other half reaction as it is. This gives:

3Cu+(aq) + 3e−→3Cu(s)

Fe(s) → Fe3+ (aq) + 3e−

Step 3: Adding the equations give:

3Cu+(aq)+3e−+Fe(s)→3Cu(s)+Fe3+(aq)+3e−

The electrons cancel out and the balanced equation is left.

3Cu+(aq) + Fe(s)→3Cu(s) + Fe3+(aq)

Example 2: Balancing in an Acid Solution

Balance the following redox reaction in acidic conditions.

Cr2O72(aq)+HNO2(aq)→Cr3+(aq)+NO3−(aq)

Solution

Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.

Cr2O72-(aq) → Cr3+ (aq)

 HNO2(aq)→NO3−(aq)

Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:

Cr2O72−(aq)→2Cr3+ (aq)

 HNO2(aq)→NO3− (aq)

Step 3: Add H2O to balance oxygen. The chromium reaction needs to be balanced by adding 7H2O molecules. The other reaction also needs to be balanced by adding one water molecule. This yield:

Cr2O72− (aq) → 2Cr3+ (aq) + 7H2O(l)

HNO2 (aq) + H2O (l) → NO3−(aq)

Step 4: Balance hydrogen by adding protons (H+). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.

14H+(aq)+Cr2O72−(aq)→2Cr3+(aq)+7H2O(l)

HNO2(aq)+H2O(l)→3H+(aq)+NO3−(aq)

Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:

6e−+ 14H+ (aq)+Cr2O72−(aq)→2Cr3+(aq)+7H2O(l)

For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side:

HNO2(aq)+H2O(l)→3H+(aq)+NO3−(aq)+2e−

Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives:

3⋅[HNO2(aq)+H2O(l)→3H+(aq)+NO3−(aq)+2e−]

3HNO2 (aq)+3H2O(l)→9H+(aq)+3NO3−(aq)+6e−

6e−+14H+ (aq) +Cr2O72− (aq)→2Cr3+(aq)+7H2O(l)⋅

Step 7: Add the reactions and cancel out common terms.

[3HNO2(aq)+3H2O(l)→9H+(aq)+3NO3−(aq)+6e−]

[6e−+14H+(aq)+Cr2O72-(aq)→2Cr3+(aq)+7H2O(l)] 3HNO2(aq)+3H2O(l)+6e−+14H+(aq)+Cr2O72→9H+(aq)+3NO3−(aq)+6e−+2Cr3+(aq)+7H2O(l)

The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:

3HNO2(aq) + 5H+(aq)+ Cr2O72−(aq)→3NO3−(aq)+2Cr3+(aq)+4H2O(l)

Example 3: Balancing in Basic Solution

Balance the following redox reaction in basic conditions.

Ag(s) + Zn2+(aq)→Ag2O(aq) + Zn(s)

Solution

Go through all the same steps as if it was in acidic conditions.

Step 1: Separate the half-reactions.

Ag(s) →Ag2O(aq)

 Zn2+ (aq)→Zn(s)

Step 2: Balance elements other than O and H.

2Ag(s) → Ag2O(aq)

 Zn2+ (aq)→ Zn(s)

Step 3: Add H2O to balance oxygen.

H2O(l) + 2Ag(s)→Ag2O(aq)

 Zn2+(aq)→Zn(s)

Step 4: Balance hydrogen with protons.

H2O(l) + 2Ag(s)→Ag2O(aq)+2H+(aq)

 Zn2+(aq)→Zn(s)

Step 5: Balance the charge with e-.

H2O (l) + 2Ag (s) → Ag2O (aq) + 2H+ (aq) + 2e−

Zn2+ (aq) +2e−→Zn(s)

Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.

Step 7: Add the reactions and cancel the electrons.

H2O(l)+2Ag(s)+Zn2+(aq)→Zn(s)+Ag2O(aq)+2H+(aq)

Step 8: Add OH- to balance H+. There are 2 net protons in this equation, so add 2 OH- ions to each side.

H2O(l)+2Ag(s)+Zn2+(aq)+2OH−(aq)→Zn(s)+Ag2O(aq)+2H+(aq)+2OH−(aq)⋅

Step 9: Combine OH- ions and H+ ions that are present on the same side to form water.

H2O(l)+2Ag(s)+Zn2+(aq)+2OH−(aq)→Zn(s)+Ag2O(aq)+2H2O(l)

Step 10: Cancel common terms.

2Ag(s) + Zn2+(aq) + 2OH− (aq)→ Zn(s) +Ag2O(aq) + H2O(l)

Test for oxidants and reductants: The table below shows the common and important oxidizing agents:

Oxidizing Agent

Half Equation

Colour change when added to reducing agent

Application

Acidified potassium manganate (VII)

 

KMnO4

 MnO4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)

Purple to pale pink (or colorless)

 Used to test for reducing agent

 Acidified potassium dichromate(VI)

 

K2Cr2O7

Cr2O72− (aq)+14H+(aq)+6e−→2Cr3+ +7H2O

 Orange to green

 Oxidizes alcohol to acids; used to test for SO2SO2 gas

 Chlorine

 

Cl2

 Cl2(g)+2e−→2Cl−(g)

 Greenish yellow to colorless

 Oxidizes bromide to bromine and iodide to iodine

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Testing for the presence of reducing agent:

Add an oxidizing agent, e.g. aqueous potassium manganate (VII) to the reducing agent

  • Shake the mixture
  • The aqueous potassium manganate (VII) is decolorized

The table below shows the common and important reducing agents.

 

Reducing Agent

Half Equation

Colour change when added to an oxidizing agent

Application

Aqueous potassium iodide

 

KI

2I−(aq)→I2(aq)+2e−

Colourless to brown

Used to test for oxidizing agent

Aqueous iron(III) sulphate

 

FeSO4

Fe2+(aq)→Fe3+(aq)+e−

Green to brown

–

Carbon Monoxide

 

CO

–

–

Used to reduce iron oxides to iron in blast furnace

Hydrogen

H2(g)→2H+(aq)+2e−

–

Reduces copper(II) oxide to copper

Metals

 

E.g. Na

–

–

Displacement of less reactive metals

 

Testing for the presence of oxidizing agent:

  • Add a reducing agent, e.g. aqueous potassium iodide to the oxidizing agent.
  • Shake the mixture.
  • A brown solution of iodine is produced.
  • The presence of iodine can be confirmed by adding starch solution.
  • A dark blue coloration is obtained.