Question

150 mL of 0.5N nitric acid solution at 25.35℃ was mixed with 150 mL of 0.5N sodium hydroxide solution at the same temperature. The final temperature was recorded to be 28.77 ℃. Calculate the heat of neutralization of nitric acid with sodium hydroxide.

Answer 1

Reactions between organic materials and nitric can result in violent or explosive reactions due to the oxidation power of nitric acid. All neutralization reactions are exothermic in nature and heat will be released.

Complete step by step solution:

First, let's try to understand the basic double replacement reaction between nitric acid ($HN{{O}_{3}}$) and sodium hydroxide (NaOH). Here, Sodium hydroxide is a base and nitric acid is an acid. The general equation for an acid-base reaction is
base + acid → salt + water.
Therefore, the reaction between nitric acid and sodium hydroxide can be written as:
$HN{{O}_{3}}+NaOH\to NaN{{O}_{3}}+{{H}_{2}}O$
It is also an example of neutralization reaction. A neutralization reaction is one in which a strong acid and strong base react with each other to yield salt and water molecules.
In a neutralization reaction, there is a combination of ${{H}^{+}}$ ions and $O{{H}^{-}}$ ions to form water.
The heat of neutralization is the change in enthalpy (heat) that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt. It is also defined as the energy released with the formation of 1 mole of water.
The heat of neutralization can be calculated using the formula,
Q = mcΔT
where, Q = the heat of neutralization,
m = the mass of solution
c = the specific heat capacity for aqueous solutions, 4.1814 Joules
ΔT = the change in temperature (measured using your calorimeter)
For the above reaction,
Mass of solution (m) = 150 + 150 = 300g.
Since, density of water = 1g/mL. Therefore, mass = volume.
Therefore, total heat produced (Q) = 300g × 1cal.× (28.77-25.35)
=1026 calories.
The heat of neutralization= $\dfrac{Q}{MV}$
where, M = molarity,
V= volume
The heat of neutralization = $\dfrac{Q}{MV}$
= $\dfrac{1026\times 1000}{150\times 0.5}$
=13.68 kcal.
The heat is released, so a negative sign will come.

Therefore, the neutralization point of acid, for the above reaction is -13.68 kcal.

Note: In reality, the nitrate and sodium ions do not actively participate in the reaction. The hydroxide ions ($O{{H}^{-}}$) from the base react with hydronium ions (${{H}^{+}}$) from the acid to give water as a result.