Question

Molar heat of neutralization of \[{\text{NaOH}}\] with \[{\text{HCl}}\] in comparison to that of \[{\text{KOH}}\] with \[{\text{HN}}{{\text{O}}_{\text{3}}}\]is:
A) Less
B) More
C) Equal
D) Depends on pressure

Answer 1

Hint: Heat of neutralization is the amount of heat evolved when one gram equivalent of an acid is completely neutralised by one gram equivalent of the base in dilute solution. Determine the strength of given acid-base pairs and then compare the heat of neutralization.

Complete step-by-step solution:
We have to compare the molar heat of neutralization of \[{\text{NaOH}}\] with \[{\text{HCl}}\] and \[{\text{KOH}}\] with \[{\text{HN}}{{\text{O}}_{\text{3}}}\].
The heat of neutralisation of strong acid against a strong base is always constant. It is because all strong acids and bases ionise completely and thus the heat of neutralization in such cases is actually the heat of formation of water from \[{{\text{H}}^{\text{ + }}}{\text{ and O}}{{\text{H}}^{\text{ - }}}\] ions.
 \[{{\text{H}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq)}} \to {{\text{H}}_{\text{2}}}{{O(l) \Delta H = - 13}}{\text{.7 kcal}}\]
The neutralization reaction of \[{\text{NaOH}}\] with \[{\text{HCl}}\] is as follows:
\[{\text{HCl (aq) + NaOH(aq) }} \to {\text{NaCl(aq) + }}{{\text{H}}_{\text{2}}}{\text{O (l)}}\]
As both \[{\text{HCl}}\]and \[{\text{NaOH}}\] are strong acid and base so will dissociate completely in solution. Hence the ionic reaction of \[{\text{NaOH}}\] with \[{\text{HCl}}\] is as follows:
\[{{\text{H}}^{\text{ + }}}{\text{ (aq) + C}}{{\text{l}}^{\text{ - }}}{\text{(aq) + N}}{{\text{a}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq) }} \to {\text{N}}{{\text{a}}^{\text{ + }}}{\text{(aq)}} + {\text{C}}{{\text{l}}^{\text{ - }}}{\text{(aq) + }}{{\text{H}}_{\text{2}}}{\text{O (l)}}\]
The net ionic reaction after canceling the common ions is as follows:
\[{{\text{H}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq)}} \to {{\text{H}}_{\text{2}}}{{O(l) \Delta H = - 13}}{\text{.7 kcal}}\]
Now, similarly, we will write the neutralisation reaction of \[{\text{KOH}}\] with \[{\text{HN}}{{\text{O}}_{\text{3}}}\].
\[{\text{HN}}{{\text{O}}_{\text{3}}}{\text{(aq) + KOH(aq) }} \to {\text{KN}}{{\text{O}}_{\text{3}}}({\text{aq) + }}{{\text{H}}_{\text{2}}}{\text{O (l)}}\]
As both \[{\text{HN}}{{\text{O}}_{\text{3}}}\]and \[{\text{KOH}}\] are strong acid and base so will dissociate completely in solution. Hence the ionic reaction of \[{\text{KOH}}\] with \[{\text{HN}}{{\text{O}}_{\text{3}}}\] is as follows:
\[{{\text{H}}^{\text{ + }}}{\text{ (aq) + N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{(aq) + }}{{\text{K}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq) }} \to {{\text{K}}^{\text{ + }}}{\text{(aq)}} + {\text{N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{(aq) + }}{{\text{H}}_{\text{2}}}{\text{O (l)}}\]
The net ionic reaction after canceling the common ions is as follows:
\[{{\text{H}}^{\text{ + }}}{\text{(aq) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq)}} \to {{\text{H}}_{\text{2}}}{{O(l) \Delta H = - 13}}{\text{.7 kcal}}\]
As both \[{\text{HCl}}\]and \[{\text{NaOH}}\] and \[{\text{HN}}{{\text{O}}_{\text{3}}}\]and \[{\text{KOH}}\] are strong acid-base pair so the heat of neutralization in both cases is actually the heat of formation of water from \[{{\text{H}}^{\text{ + }}}{\text{ and O}}{{\text{H}}^{\text{ - }}}\] ions. So will have equal Molar heat of neutralization.

Hence, the correct answer is an option (C) Equal.

Note: Heat of neutralisation of all strong and strong base pairs is always equal to the heat of ionisation of water -13.7kcal. The heat of neutralisation of weak acid or weak base is less than the heat of neutralisation of strong acid and strong base.