Question

A monobasic acid is dissociated in 0.1 M solution. When 200ml of the acid solution is neutralized by 0.1 M NaOH heat evolved is 244 cal. If the heat of neutralization of a strong acid with a strong base is –13.7 kcal, the molar heat of dissociation of the acid is

• A. $2.20kcalmo{l}^{-1}$
• B. $1.50kcalmo{l}^{-1}$
• C. $-1.50kcalmo{l}^{-1}$
• D. $2.0kcalmo{l}^{-1}$

Answer 1

The correct option is D

$2.0kcalmo{l}^{-1}$

Moles of the acid neutralized $=\frac{200}{1000}R\times 0.1=0.02$
Moles of acid remaining undissociated =$0.02(1-\frac{25}{100})=0.015$
Heat evolved during neutralization for 100% ionization = $13700\times 0.02=274$ cal
Heat used up for dissociation of 0.015 mole = 274 – 244 = 30cal
Heat of dissociation per mole = $\frac{30}{0.015}=2000cal=2.0k$ cal