The heats fo neutralization of `HCl` with `NH_(4) OH` and `NaOH` with `CH_(3) COOH` are `- 51.4 kJ eq^(-1)` and `- 50.6 kJ eq^(-1)`, respectively. The heat of neutralization of acetic acid with `NH_(4) OH` will be

Options

A. `-44.6 kJ eq^(-1)`
B. `-50.6 kJ eq^(-1)`
C. `-51.4 kJ eq^(-1)`
D. `-57.4 kJ eq^(-1)`

Answer

A

Explanation

`(i) HCl+NH_(4)OHtoNH_(4)Cl+H_(2)O`, `DeltaH=-51.4 kJ`
`(ii) CH_(3)COOH+NaOHtoCH_(3)COONa+H_(2)O`, `DeltaH=-50.6 kJ`
`(iii) CH_(3)COOH+NH_(4)OHtoCH_(3)COONH_(4)+H_(2)O`, `DeltaH=?`
We know
`(iv) HCl+NaOH to NaCl+H_(2)O`, `DeltaH=-57.4 kJ`
To get `(iii)` add `(i)` and `(ii)` and subtract `(iv)`.
(Please note that salts formed in case `(i)` and `(ii)` are almost completely ionised. Also suppose that the salt formed in case `(iii)` is also ionised).
`DeltaH=(-51.4)+(-50.6)-(-57.4)kJ`
`=-51.4-50.6+57.4=-44.6 kJ`