# 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;

- Strong acid vs strong base
- Strong acid vs weak base
- Weak acid vs strong base
- 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 + H_{2}O

H_{2}SO_{4} + 2KOH à K_{2}SO_{4} + 2H_{2}O

HNO_{3} + KOH à KNO_{3} +H_{2}O

**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 + NH_{4}OH à NH_{4}Cl + H_{2}O

3H_{2}SO_{4} + 2Al(OH)_{3} à Al_{2}(SO_{4})_{3} + 6H_{2}O

**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.

CH_{3}COOH + NaOH à CH_{3}COONa + H_{2}O

H_{3}PO_{4} + KOH à K_{3}PO_{4} + H_{2}O

H_{2}CO_{3 }+ NaOH à Na_{2}CO_{3} + H_{2}O

**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 equation**s 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 + H_{2}O

Extracting ions

H^{+ } Cl^{–} + Na^{+} OH^{–} à Na^{+} Cl^{–} + H_{2}O

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 H_{2}O

Task 2

H_{2}SO_{4 }+ 2KOH à K_{2}SO_{4} + 2H_{2}O

Extracting ions

2H^{+ }SO_{4}^{2-} + 2K^{+} 2OH^{–} à 2K^{+} SO_{4}^{2-} + 2H_{2}O

Cancel out the spectator ions

The spectator ions are SO_{4}^{2- }and K^{+} because the oxidation state of the two ions remained the same.

2H^{+} + 2OH^{–} à 2H_{2}O

**Few Titration problems**

**Example 1**

20cm^{3} of 0.004 mol/dm^{3} HCl reacted with 25cm^{3} of unknown concentration of NaOH. Calculate the concentration of NaOH.

Solution

HCl + NaOH à NaCl + H_{2}O

C_{A}V_{A}/C_{B}V_{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_{A}V_{A}/C_{B}V_{B} = N_{A}/N_{B}

C_{A} = 0.004 mol/dm^{3}

C_{B }= ?

V_{A}= 20cm^{3}

V_{B }= 25cm^{3}

N_{A}= 1

N_{B} = 1

Making C_{B }the subject of formula

C_{B} = C_{A}V_{A}N_{B} / V_{B}N_{A}

C_{B} = 0.004 X 20 X 1 /25 X 1

C_{B} = 0.0032 mol/dm^{3}

Example 2

If 30cm^{3} of 0.025 mol/dm^{3} HCl reacted with 0.035 mol/dm^{3} of KOH. Calculate the volume of KOH.

Solution

HCl + KOH à KCl + H_{2}O

C_{A}V_{A}/C_{B}V_{B} = N_{A}/N_{B}

V_{B } = C_{A}V_{A}N_{B}/C_{B}N_{A}

V_{B} = 0.025 x 30 x 1 /0.035 x 1

V_{B} = 2.14 dm^{3}

Example 3

Calculate the concentration of 30cm^{3} of sulphuric acid that will react with 0.045 mol/dm^{3} 20cm^{3} of KOH.

Solution

H_{2}SO_{4} + 2KOH à K_{2}SO_{4} + H_{2}O

C_{A}V_{A}/C_{B}V_{B} = N_{A}/N_{B}

C_{A }= C_{B}V_{B}N_{A}/V_{A}N_{B}

C_{A} = 0.045 x 20 x 1 /30 x 1

C_{A} = 0.03 mol/dm^{3}

You may be wondering why I didn’t change volume to dm^{3}. 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:

CH_{3}COOH + NaOH à CH_{3}COONa + H_{2}O

The titration was carried out using phenolpthalein indicator. If the concentration of base is 0.0033mol/dm^{3}, calculate the volume of base required to neutralize 30cm^{3} of 0.0035mol/dm^{3}

Solution

C_{A}V_{A}/C_{B}V_{B} = N_{A}/N_{B}

V_{A} = C_{A}V_{A}N_{B}/C_{B}N_{A}

V_{A} =0.0035 x 30 x1 /0.0033 x 1

V_{A} = 3.18 dm^{3}

**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**

**C _{1}V_{1} = C_{2}V_{2} OR M_{1}V_{1} = M_{2}V_{2}**

**Where C _{1} = concentration of solute V_{1} = volume of solute M_{1} = molarity of solute**

** C _{2} = concentration of solution V_{2} = volume of solution M_{2} =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}) = C_{1}/C_{2} or V_{1}/V_{2}

**Few Dilution problems**

Task 1

What volume of distilled water should be added to 400cm^{3} of 2.0 mol/dm^{3} HCl to obtain 0.30 mol/dm^{3} of solution?

Solution

Using C_{1}V_{1} = C_{2}V_{2}

V_{2} = C_{1}V_{1}/C_{2}

V_{2} = 2.0 x 400 /0.30

Task 2

Calculate the volume of a 12.0 mol/dm^{3} HCl that should be diluted with distilled water to obtain 2.5 dm^{3} of 0.050 mol/dm^{3}?

Solution

Using C_{1}V_{1} = C_{2}V_{2}

V_{1} = C_{2}V_{2}/C_{1}

V_{1 }=2.5 x 0.05/12

V_{1} = 0.0104 dm^{3}

Task 3

The label on a stock bottle of hydrochloric acid was prepared and it contains 0.4 dm^{3} of the acid Calculate the concentration of the HCl in the stock solution required to make I dm^{3} of 2.0 mol/dm^{3} HCl.

Solution

Using C_{1}V_{1} = C_{2}V_{2}

C_{1} = C_{2}V_{2}/V_{1}

_{ }C_{1 }= 1 x 2 / 0.04

C_{1} = 5.0 mol/dm^{3}

**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/dm^{3} 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/dm^{3} 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/dm^{3}.

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/dm^{3}. Calculate the pH of the acid.

Solution

Sulphuric acid is dibasic

H_{2}SO_{4} = 2[H^{+}] [SO_{4}^{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**

- Neutralization reactions are employed in determining the unknown centration of acid or base
- It is applied in treatment of acidic soils
- It is applied wastewater effluent treatment before releasing it to the environment
- It is employed in the production of antacid tablets to neutralize excess gastric acid n the stomach
- It is applied in neutralizing bee stings or insect bites
- 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.