Enthalpy of Neutralization of Strong Acid and Strong Base

The enthalpy of neutralization will always be constant for a strong acid and a strong base. This is because every strong acid and strong base is completely ionized in a dilute solution. Enthalpy changes in neutralization are always negative—when an acid and alkali react, heat is given out. Enthalpy of neutralization is the heat change when an amount of an acid is completely neutralized by a base or vice versa. In the case of strong acids and strong bases, it is usually a simple and strongly exothermic reaction whereby a hydrogen ion, H⁺, from the acid, combines with a hydroxide ion, OH⁻, from the base to form water, H₂O.

Heat of Neutralization

It is an enthalpy change during the neutralization of 1 equivalent of an acid and base.

Case of Strong Acid and Strong Base:

$\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$

The enthalpy change of the above reaction is equal to -13.7 kcal eq^-1 or -57.3 kJ eq^-1.

It is to be noted that the enthalpy of neutralization of 1 equivalent of any strong acid or strong base is equal to -13.7 kcal eq^-1 or -57.3 kJ eq^-1.

$\mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}=-13.7 \mathrm{kcal} \mathrm{mol}^{-}$

Case in which any one of the Acids or Base (or Both) are weak:

In case one of the acids or base is weak, then some part of the heat will be used up for ionization of the weak component and the heat liberated will be lesser than the above values

For example

$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O},|\Delta \mathrm{H}|<13.7 \mathrm{kcal} \mathrm{eq}^{-1}$

Properties of enthalpy of neutralization:

• It is always exothermic $\Delta \mathrm{H}=-\mathrm{ve}$
• The heat of neutralization of strong acid and strong base is always 13.7 Kcal/mole or 57 kJ/mole.
• It is independent of the nature of strong acid or strong base.
• If one electrolyte is weak then $\Delta \mathrm{H}$ will be less than —13.7 Kcal as some amount of heat will be absorbed in the ionization of the weak electrolyte.
• In the case of HF, $\Delta \mathrm{H}$ is more due to more hydration of F^- ions.

Recommended topic video on (Enthalpy of Neutralization of Strong Acid and Strong Base)

Some Solved Examples

Example 1: Equal amounts of 1M HCl and 1 M H₂SO₄ are neutralized by 1 M NaOH solution and A and B are the respective heat liberated after neutralization. What is the relation between A and B?
1) $A=B$
2) $2 A=B$
3) $A=2 B$
4) $A=4 B$

Solution

The number of equivalents of H₂SO₄ is 2

while the number of equivalents of HCl is just 1.

It means that H₂SO₄ gives 2 moles of H⁺ ions while HCl only gives 1 mole of H⁺.

So the energy liberated by H₂SO₄ will be somewhat equal to twice the energy liberated by HCl.

Therefore, the correct relation between A and B is$2 A=B$

Example 2 : The enthalpy of neutralization of which of the following acids and base is nearly -13.7 Kcal

1)HCN and NaOH

2)HCN and NH₄OH

3) HCl and NaOH

4)HCl and NH₄OH

Solution

Enthalpy of Neutralisation - Amount of Enthalpy change during Neutralisation of one gm equivalent acid & one gm equivalent base.

$\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} ; \quad \Delta \mathrm{H}=-13.70 \mathrm{kcal}$

The heat of neutralization of strong acid and strong base is equal to -13.7 kcal. HCl is a strong acid and NaOH is a strong base.

Hence, the answer is the Option (3).

Example 3: While performing a thermodynamics experiment, a student made the following observations.

$\begin{aligned} & \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-55.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

The enthalpy of ionization as calculated by the student is ____________kJmol^-1. (nearest integer)

1) 2

2)4

3)7

4)9

Solution

The ionization of $\mathrm{CH}_3 \mathrm{COOH}$.
$
\mathrm{CH}_3 \mathrm{COOH} \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{+} \Delta \mathrm{H}^{-}=?
$

Given,
(1) $\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}, \Delta \mathrm{H}=-57.3 \mathrm{KJ} / \mathrm{mol}$
(2) $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}=-55.3 \mathrm{KJ} / \mathrm{mol}$

By doing (2) - (1), will become ionization of $\mathrm{CH}_3 \mathrm{COOH}$
$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{\ominus} \mathrm{Na}^{\oplus}+\mathrm{HCl} \\
& \text { So, } \Delta \mathrm{H}=-55.3-(-57.3) \\
& \qquad \mathrm{H}=2 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$

Example 4: When 600 mL of $0.2 \mathrm{M} \mathrm{HNO}_3$ is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is $\qquad$ $\times 10^{-2 \circ} \mathrm{C}$. (Enthalpy of neutralisation $=57 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and Specific heat of water $=4.2 \mathrm{JK}^{-1} \mathrm{~g}^{-1}$ ) (Neglect heat capacity of the flask)

1) 54

2)44

3)34

4)24

Solution

The reaction will be
$
\begin{aligned}
& \mathrm{HNO}_3+\mathrm{NaOH} \longrightarrow \mathrm{H}_2 \mathrm{O} \\
& 0.2 \mathrm{M} \quad 0.1 \mathrm{M} \\
& \text { initial } 120 \mathrm{~m} \mathrm{~mol} \quad 40 \mathrm{~m} \mathrm{~mol} 0 \\
& \text { final } 80 \mathrm{~m} \mathrm{~mol} \quad 0 \quad 40 \mathrm{~m} \mathrm{~mol} \\
&
\end{aligned}
$

So, a total of 40 m mol is neutralizing.
So total enthalpy of neutralisation for 40 m mol will be
$
\begin{aligned}
& \Delta \mathrm{H}=40 \mathrm{~m} \mathrm{~mol} \times 57 \mathrm{KJ} \mathrm{mol} \\
& \Delta \mathrm{H}=40 \times 10^{-3} \times 57 \times 10^3 \mathrm{~J} \\
& \Delta \mathrm{H}=2280 \mathrm{~J}
\end{aligned}
$
we know the formula,
$
\begin{aligned}
& \mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}=\Delta \mathrm{H} \\
& \mathrm{m}=\text { mass of water } \\
& \mathrm{s}=\text { specific heat of water }=4.2 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~g}^{-1} \\
& \Delta \mathrm{T}=? \\
& \text { Now, } \mathrm{m}=\text { volume } \times \text { density }=1000 \mathrm{~mL} \times 1 \mathrm{~g} \mid \mathrm{ml} \\
& \mathrm{m}=1000 \mathrm{~g} \\
& \text { then, } 2280=1000 \times 4.2 \times \Delta \mathrm{T} \\
& \qquad \mathrm{T}=0.54 \\
& \Delta \mathrm{T}=54 \times 10^{-2} \mathrm{~K}
\end{aligned}
$

Hence, the answer is (54).
Example 5: 200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is -57.1 kJ. The increase in temperature in the system on mixing is $x \times 10^{-2}$. The value of $x$ is $\qquad$ (est integer)
[Given: Specific heat of water $=4.18 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$;Density of water $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$ ]

(Assume no volume change on mixing)

1) 82

2)32

3)46

4)52

Solution
meq. of Acid = 200 *0.2=40
meq. of Base = 300 *0.1=30
Total volume = 500 ml
Now, the heat liberated in the neutralization reaction will be used to heat up the solution. Heat balance gives us:
$
\begin{aligned}
& \text { Heat }=\mathrm{m} \cdot \mathrm{s} \cdot \Delta \mathrm{T} \\
& \Rightarrow 171.3=500 \times 4.18 \times \Delta \mathrm{T} \\
& \Rightarrow \Delta \mathrm{T}=0.82^{\circ} \mathrm{C}=82 \times 10^{-2}{ }^{\circ} \mathrm{C}
\end{aligned}
$

Hence, the answer is (82).

Summary

The enthalpy of neutralization is the heat developed during the reaction of one mole of H⁺ ions from an acid with one mole of OH⁻ ions from the base to form water. It is an exothermic reaction, which means that it releases a considerable amount of heat. Because of this reason, the enthalpy change for the neutralization of strong acids and bases is going to be approximately constant, usually -57 kJ/mol. This value is constant because strong acids and bases are completely dissociated in water, and the main reaction is indeed the formation of water from H⁺ and OH⁻ ions.