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Q.

The heats of neutralization of $HCl with {NH}_{4} OH$ and that of NaOH with CH₃ COOH are respectively -51.4 and 1 50.6kJ eq^-1 then the heat of neutralisation of acetic acid with NH₄OH will be

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a

$- 44 .6 {kJeq}^{- 1}$

b

$- 50 .6 kJ {eq}^{- 1}$

c

$- 57 .4 kJ {eq}^{- 1}$

d

$- 51 .4 {kJeq}^{- 1}$

answer is A.

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Detailed Solution

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$\begin{array}{l} HCl + {NH}_{4} OH; ΔH = - 51 .4 \\ \Rightarrow △ H_{ionisation} = 57 .4 - 51 .4 = 6 kJ \\ NaOH + {CH}_{3} COOH ΔH = - 50 .6 \\ \Rightarrow △ H_{ionization} = 57 .4 - 50 .6 = 6 .8 kJ \\ {CH}_{3} COOH + {NH}_{4} OH ΔH = - 57 .4 + 6 + 6 .8 = - 44 .6 kJ \end{array}$

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