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

Examples

HCl + NaOH à NaCl + H₂O

H₂SO₄ + 2KOH à K₂SO₄ + 2H₂O

HNO₃ + KOH à KNO₃ +H₂O

Hint

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.

Examples

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.

CH₃COOH + NaOH à CH₃COONa + H₂O

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⁺ SO₄^2- + 2K⁺ 2OH^– à 2K⁺ SO₄^2- + 2H₂O

Cancel out the spectator ions

The spectator ions are SO₄^2- 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.

Solution

HCl + NaOH à NaCl + H₂O

C_AV_A/C_BV_B =N_A/N_B  This is called the titration formula

Explanation of terms

C_A = concentration of acid

C_B = concentration of base

V_A= volume of acid

V_B = volume of base

N_A= number of moles of acid

N_B =number of moles of base

Method

C_AV_A/C_BV_B = N_A/N_B

C_A = 0.004 mol/dm³

C_B = ?

V_A= 20cm³

V_B = 25cm³

N_A= 1

N_B = 1

Making C_B the subject of formula

C_B = C_AV_AN_B / V_BN_A

C_B = 0.004 X 20 X 1 /25 X 1

C_B = 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.

Solution

HCl + KOH à KCl + H₂O

C_AV_A/C_BV_B = N_A/N_B

V_B  = C_AV_AN_B/C_BN_A

V_B = 0.025 x 30  x 1 /0.035 x 1

V_B = 2.14 dm³

Example 3

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

Solution

H₂SO₄ + 2KOH à K₂SO₄ + H₂O

C_AV_A/C_BV_B = N_A/N_B

C_A = C_BV_BN_A/V_AN_B

C_A = 0.045 x 20 x 1 /30 x 1

C_A = 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:

CH₃COOH + NaOH à CH₃COONa + H₂O

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

Solution

C_AV_A/C_BV_B = N_A/N_B

V_A = C_AV_AN_B/C_BN_A

V_A =0.0035 x 30 x1 /0.0033 x 1

V_A = 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 (D_f) = 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?

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?

Solution

Using C₁V₁ = C₂V₂

          V₁ = C₂V₂/C₁

           V₁  =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.

Solution

Using C₁V₁ = C₂V₂

          C₁ = C₂V₂/V₁

               C₁ = 1 x 2 / 0.04

         C₁ = 5.0 mol/dm³

pH

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?

Solution

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?

Solution

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.

Solution

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.

Solution

Sulphuric acid is dibasic

H₂SO₄ = 2[H⁺] [SO₄^2-]

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.