Question

Heat of neutralization is equal to heat of formation of water from $H^{+}$ and $O{H}^{-}$ ions.
If true enter 1, else enter 0.

Answer 1

When the acid and the salt are strong electrotytes and solutions are dilute, the process of neutralization can be represented as:

$HA+BOH⟶AB+H_{2}O+57.3kJ$

or $H^{+}+A^{-}+B^{+}+O{H}^{-}⟶A^{+}+B^{-}+H_{2}O+57.3kJ$

Canceling the ions present on both sides, we have

$H^{+}+O{H}^{-}⟶H_{2}O+57.36kJ$

It is thus seen that the net reaction, which takes place in the process of neutralization is the combination of hydrogen and hydroxyl ions to form feebly ionised water. It follows, therefore, that the heat of neutralization of a strong acid against a strong base is merely the heat of formation of water from hydrogen and hydroxyl ions, which is equal to $57.3kJ$. Thus the heat of neutralization of all strong acids by strong alkalies in dilute solution should practically be the same.