The enthalpies of neutralization of NaOH and ${NH}_{4} OH$ with HCl are 57.30 and 51.43 kJ respectively. The enthalpy of ionization of ${NH}_{4} OH$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2.93 kJ ${mol}^{- 1}$

5.87 kJ ${mol}^{- 1}$

31.58 ${mol}^{- 1}$

63.17 ${mol}^{- 1}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The basic reaction that occurs when a strong acid is neutralized by a strong base is

$H^{+} + {OH}^{-} \to H_{2} O + x$

Hence, the amount of heat liberated x is the same when any strong acid is neutralized by any strong base. However if either the acid or base is weak this amount is less since the weak acid/base is to be ionized by absorbing heat. Thus the difference in the two quantities is equal to the heat of ionization of weak acid / base.

Heat of ionization of ${NH}_{4} OH$

= 57.30 – 51.43 = +5.87 kJ ${mol}^{- 1}$ .

Watch 3-min video & get full concept clarity