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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

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

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

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

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

answer is C.

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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 = $- 10.5 - (- 12.5)$

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

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