Comprehension Enthalpy of neutralization is defined as the enthalpy change when \( 1 \mathrm{~mo...
0Comprehension
Enthalpy of neutralization is defined as the enthalpy change when \( 1 \mathrm{~mole} \) of an acid/a base is completely neutralized by a base/an acid in dilute solution.
For strong acid and strong base, neutralizations net chemical change is
\[
\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell) ; \Delta_{\mathrm{r}} \mathrm{H}^{\mathrm{o}}=-55.84 \mathrm{~kJ} / \mathrm{mol}
\]
\( \Delta \mathrm{H}_{\text {ionization }}^{\mathrm{o}} \) of aqueous solution of strong acid and strong base is zero.
When a dilute solution of a weak acid or base is neutralized, the enthalpy of neutralization is some what less because of the absorption of heat in the ionization of the weak acid or weak base, for a weak acid/ weak base
\[
\Delta \mathrm{H}_{\text {neutralization }}^{\mathrm{o}}=\Delta \mathrm{H}_{\text {ionization }}^{\mathrm{o}}+\Delta_{\mathrm{r}} \mathrm{H}^{\circ}\left(\mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}\right)
\]
Under the same condition, how many \( \mathrm{mL} \) of \( 0.1 \mathrm{M} \mathrm{NaOH} \) and \( 0.05 \mathrm{M} \mathrm{H}_{2} \mathrm{~A} \) (strong diprotic acid) solution should be mixed for a total volume of \( 100 \mathrm{~mL} \) to produce the highest rise in temperature :
(a) \( 25: 75 \)
(b) \( 50: 50 \)
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(d) \( 66.66: 33.33 \)
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