The enthalpy of neutralization of $N H_{4} O H$ and $C H_{3} C O O H$ is $- 10.5 K c a l m o l^{- 1}$ and enthalpy of neutralization of $C H_{3} C O O H$ with strong base is $- 12.5 K c a l m o l^{- 1}$ The enthalpy of ionization of $N H_{4} O H$ will be

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$4 K c a l m o l^{- 1}$

$3 K c a l m o l^{- 1}$

$2 K c a l m o l^{- 1}$

$3.2 K c a l m o l^{- 1}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The enthalpy of ionization of $N H_{4} O H$ (weak base) = enthalpy of neutralization of $N H_{4} O H$ and $C H_{3} C O O H$ - enthalpy of neutralization of $C H_{3} C O O H$ and strong base = $- 10.5 - (- 12.5)$

$= - 10.5 + 12.5 = 2 K c a l m o l^{- 1}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Get Expert Academic Guidance – Connect with a Counselor Today!