Question

Heat of neutralization of any strong acid and strong base is always constant and $\Delta H= -57.3 \, kJ$. This is because

• A. both the acid and base undergo complete ionization
• B. during neutralization salt and water are formed
• C. 1 mole of water is formed from $\ce{H^+}$ and $\ce{OH^-}$ ions
• D. the reaction is exothermic

Answer 1

Answer: (C)

1 mole of water is formed from $\ce{H^+}$ and $\ce{OH^-}$ ions

Answer 2

The enthalpy of neutralization of any strong acid with a strong base is always the same, i.e., 57.1 kJ. This is because the strong acids, strong bases and the salts they form, are all completely ionized in dilute aqueous solution.
${ Na^+ = OH^- + H^+ + Cl^- -> Na^+ + Cl^- + H_2 O}$ or ${ H^+_{(aq)} + OH^-_{(aq)} -> H_2O_{(l)} }$
i.e.; 1 mole of water is formed from ${H^+}$ and ${OH^-}$ ions.