Question

What is the enthalpy change when 100ml of 1M ${{H}_{2}}S{{O}_{4}}$ is completely neutralized by 100ml of 1M $Ca{{(OH)}_{2}}$?
a.) -13.7 kcal
b.) -27.4 kcal
c.) +1.37 kcal
d.) -2.74kcal

Answer 1

Hint: Enthalpy of neutralization is the change in the enthalpy that occurs when one equivalent of both acid and base undergoes a neutralization reaction. The product formed after the neutralization reaction is water and salt.

Complete Solution :
Now we know that ${{H}_{2}}S{{O}_{4}}$ is a diprotic acid which means that ${{H}_{2}}S{{O}_{4}}$ gives two hydrogen on dissociation.
So, 1 mole of ${{H}_{2}}S{{O}_{4}}$ gives 2 moles of ${{H}^{+}}$ ions
Given in the question:
100ml of 1M ${{H}_{2}}S{{O}_{4}}$ is completely neutralized by 100ml of 1M$Ca{{(OH)}_{2}}$
100ml of 1M ${{H}_{2}}S{{O}_{4}}$ will give = $1\times 2$ moles of ${{H}^{+}}$ ions
So, 100ml of 1M ${{H}_{2}}S{{O}_{4}}$ will give $\dfrac{100\times 1\times 2}{1000}$ moles of ${{H}^{+}}$ ions
= 0.2 moles of ${{H}^{+}}$ ions

- Similarly, 1 mole of $Ca{{(OH)}_{2}}$ gives 2 moles of $O{{H}^{-}}$ ions
100ml of 1M $Ca{{(OH)}_{2}}$ will give = $1\times 2$ moles of $O{{H}^{-}}$ ions
So, 100ml of 1M \[Ca{{(OH)}_{2}}\] will give $\dfrac{100\times 1\times 2}{1000}$ moles of $O{{H}^{-}}$ ions = 0.2 moles of $O{{H}^{-}}$ ions
And, the heat of neutralization is given by the number of hydrogen ions and number of hydroxide ions reacts 1 mole of ${{H}^{+}}$ and 1 mole of $O{{H}^{-}}$ reacts to give the heat of neutralization = - 13.7 Kcal
Here we have 0.2 moles of each hydrogen ion and hydroxide ion
- 0.2 mole of ${{H}^{+}}$ and 0.2 mole of $O{{H}^{-}}$ reacts to give the heat of neutralization = $0.2\times (-13.7)$ kcal
= -2.74 Kcal
So, the correct answer is “Option B”.

Note: An acid which only has one acidic proton in its molecule is known as monobasic acid. An acid which has two acidic protons in its molecule is known as dibasic acid and an acid which has three acidic protons in its molecule is known as tribasic acid.