What is Neutralization Reaction?

Neutralization Reaction is a reaction in which an acid and a base react together to form salt. A neutralization reaction is an example of spontaneous reactions. The heat released when neutralization reaction occur is called heat of neutralization.

You can read everything about heats of reactions and how to calculate the problems involved.

It is important to remember that heat of neutralization and heat of combustion are all exothermic.

Neutralization reactions are easily demonstrated using titration experiments.

Neutralization reactions are of different categories;

  1. Strong acid vs strong base
  2. Strong acid vs weak base
  3. Weak acid vs strong base
  4. Weak acid vs weak base (not easily demonstrated)

Strong acid vs strong base titration

For a strong acid against a strong base, any indicator can be used to indicate the endpoint.

The experiment can easily be demonstrated using methyl orange, methyl red, phenolphthalein, etc


HCl + NaOH à NaCl + H2O

H2SO4 + 2KOH à K2SO4 + 2H2O

HNO3 + KOH à KNO3 +H2O


A strong acid is an acid that ionizes completely in the water while a weak acid is an acid that ionizes partially in the water.

A strong base equally ionizes completely in the water while a weak base ionizes partially in the water.

Consequently, since a strong acid and a strong base dissociate completely in water, they are termed strong electrolytes.

Strong acid vs weak base titration

In the titration involving a strong acid and a weak base titration, methyl orange or methyl red could be used to demonstrate the experiment.


HCl + NH4OH à NH4Cl + H2O

3H2SO4 + 2Al(OH)3 à Al2(SO4)3 + 6H2O

Weak acid vs Strong Base

In this category of titration, a weak acid can easily react with a base and the end point shown using Phenolphthalein.


H3PO4 + KOH à K3PO4 + H2O

H2CO3 + NaOH à Na2CO3 + H2O

Why can’t we talk about weak acid vs weak base?

This is primarily because of the stoichiometric amount of solutions that may be needed before a feasible reaction occurs.

Ionic equation of neutralization reaction

Now I just want to show you how to derive the ionic equations from neutralization reactions.

Although, I have written a full post on how to write and balance ionic equations in chemistry.

Task 1

HCl + NaOH à NaCl + H2O

Extracting ions

H+  Cl  + Na+ OH à Na+ Cl + H2O

Cancelling out the spectator ions

Spectator ions are ions that appear the same on both sides.

The spectator ions here are Na+ and Cl ions because the oxidation state of the two ions did not change in both sides of equation.

H+ + OH à H2O

Task 2

H2SO4 + 2KOH à K2SO4 + 2H2O

Extracting ions

2H+ SO42- + 2K+ 2OH à 2K+ SO42- + 2H2O

Cancel out the spectator ions

The spectator ions are SO42- and K+ because the oxidation state of the two ions remained the same.

2H+ + 2OH à 2H2O

Few Titration problems

Example 1

20cm3 of 0.004 mol/dm3 HCl reacted with 25cm3 of unknown concentration of NaOH. Calculate the concentration of NaOH.


HCl + NaOH à NaCl + H2O

CAVA/CBVB =NA/NB  This is called the titration formula

Explanation of terms

CA = concentration of acid

CB = concentration of base

VA= volume of acid

VB = volume of base

NA= number of moles of acid

NB =number of moles of base



CA = 0.004 mol/dm3

CB = ?

VA= 20cm3

VB = 25cm3

NA= 1

NB = 1

Making CB the subject of formula


CB = 0.004 X 20 X 1 /25 X 1

CB = 0.0032 mol/dm3

Example 2

If 30cm3 of 0.025 mol/dm3 HCl reacted with 0.035 mol/dm3 of KOH. Calculate the volume of KOH.


HCl + KOH à KCl + H2O



VB = 0.025 x 30  x 1 /0.035 x 1

VB = 2.14 dm3

Example 3

Calculate the concentration of 30cm3 of sulphuric acid that will react with 0.045 mol/dm3 20cm3 of KOH.


H2SO4 + 2KOH à K2SO4 + H2O



CA = 0.045 x 20 x 1 /30 x 1

CA = 0.03 mol/dm3

You may be wondering why I didn’t change volume to dm3. It is because I only need a ratio, the two u its of volume will cancel out and you’ll get same result.

Example 4

Consider the equation below:


The titration was carried out using phenolpthalein indicator. If the concentration of base is 0.0033mol/dm3, calculate the volume of base required to neutralize 30cm3 of 0.0035mol/dm3




VA =0.0035 x 30 x1 /0.0033 x 1

VA = 3.18 dm3

Dilution and Dilution factor

What is dilution? Dilution is the process of adding water to a solute or adding a solute (e.g acids) to water in order to decrease the concentration or increase the volume.

The formula of dilution is given as

C1V1 = C2V2      OR  M1V1 = M2V2

Where C1 = concentration of solute         V1 = volume of solute        M1 = molarity of solute

            C2 = concentration of solution      V2 = volume of solution     M2 =molarity of solution

Please it is important to note that concentration (molar concentration) is the same as Molarity.

The number of times a solute (acid or base) can be diluted is called dilution factor.

The dilution factor formula is formulated using the dilution formula.

Dilution factor

Dilution factor (Df) = C1/C2    or V1/V2

Few Dilution problems

Task 1

What volume of distilled water should be added to 400cm3 of 2.0 mol/dm3 HCl to obtain 0.30 mol/dm3 of solution?


Using C1V1 = C2V2

V2 = C1V1/C2

V2 = 2.0 x 400 /0.30

Task 2

Calculate the volume of a 12.0 mol/dm3 HCl that should be diluted with distilled water to obtain 2.5 dm3 of 0.050 mol/dm3?


Using C1V1 = C2V2

          V1 = C2V2/C1

           V=2.5 x 0.05/12

           V1 = 0.0104 dm3

Task 3

The label on a stock bottle of hydrochloric acid was prepared and it contains 0.4 dm3 of the acid Calculate the concentration of the HCl in the stock solution required to make I dm3 of 2.0 mol/dm3 HCl.


Using C1V1 = C2V2

          C1 = C2V2/V1

               C1 = 1 x 2 / 0.04

         C1 = 5.0 mol/dm3


pH is the degree of acidity or alkalinity of a solution.

     pH = -log [H+]

    pOH= -log [OH]

    pH + pOH =14

5 pH problems in chemistry

Example 1

What is the pH of 0.0001 mol/dm3 Hydrochloric acid?


HCl à [H+] + [Cl]

pH = -log [H+]

pH =-log [0.0001]

pH = 3

Example 2

What is the pH of 0.0001 mol/dm3 Hydrochloric acid?


HCl à [H+] + [Cl]

pH =-log [H+]

pH = -log [0.0001]

pH = 4

Example 3

Determine the pH  and pOH  of an orange juice solution containing hydroxide ion concentration of  1 x 10 -8 mol/dm3.


Since it is an orange juice, it is citric acid.

Citric acid is a weak organic acid, so it is monobasic.

pOH = -log [OH]

pOh = -log [1 x 10 -8]

pOH= 8

pH + pOH  =14

pOH= 14-8 =6

Example 4

A dibasic acid (conc. H2SO4) has a hydrogen ion concentration of 0.00123 mol/dm3. Calculate the pH of the acid.


Sulphuric acid is dibasic

H2SO4 = 2[H+] [SO42-]

You need to multiply the concentration by 2 before applying the pH formula.

2[H+] = 2 X 0.00123

          = 0.00246

pH = -log [H+]

pH= -log [0.00246]

pH=  2.61

Applications of Neutralization reactions

  1. Neutralization reactions are employed in determining the unknown centration of acid or base
  2. It is applied in treatment of acidic soils
  3. It is applied wastewater effluent treatment before releasing it to the environment
  4. It  is employed in the production of antacid tablets to neutralize excess gastric acid n the stomach
  5. It is applied in neutralizing bee stings or insect bites
  6. Also employed in the manufacture of scrubber to prevent acidic rain formation

The neutralization reaction is a very important concept in chemistry that is widely applied in real life in so many different fields.

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