The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to \(\mathrm{H}^{+}\) ions; that is, \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}^{+}(a q)\right]=0\). (a) For the following reaction $$ \begin{aligned} \mathrm{HCl}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{H}^{+}(a q) &+\mathrm{Cl}^{-}(a q) \\ \Delta H^{\circ} &=-74.9 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for the \(\mathrm{Cl}^{-}\) ions. (b) Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{OH}^{-}\) ions is \(-229.6 \mathrm{~kJ} / \mathrm{mol}\), calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as \(\mathrm{HCl}\) ) is titrated by 1 mole of a strong base (such as \(\mathrm{KOH}\) ) at \(25^{\circ} \mathrm{C}\)

The enthalpy change for the given reaction, which is the formation of H+ and Cl- from HCl, is given as \(\Delta H^{\circ} = -74.9 \mathrm{~kJ/mol}\). Since the standard enthalpy of formation of H+ ions is set to zero, the whole enthalpy change for this reaction corresponds to the enthalpy of formation of Cl- ions. Therefore, \(\Delta H_{\mathrm{f}}^{\circ}\) for the \(\mathrm{Cl}^{-}\) ions is equal to the given \(\Delta H^{\circ}\), which is \(-74.9 \mathrm{~kJ/mol}\).

The enthalpy of neutralization is the heat evolved when one mole of an acid reacts with one mole of a base to form water and a salt. For the reaction of HCl with KOH, the reaction can be written as: \(\mathrm{HCl}(a q) + \mathrm{KOH}(a q) \longrightarrow \mathrm{KCl}(a q) + \mathrm{H}_{2} \mathrm{0}(l)\) The enthalpy change for this reaction can be calculated from the enthalpies of formation of the reactants and products using the formula: \(\Delta H = \Sigma \Delta H_{\mathrm{f}}^{\circ}(\mathrm{products}) - \Sigma \Delta H_{\mathrm{f}}^{\circ}(\mathrm{reactants})\) For this reaction, the products are \(KCl(aq)\) and \(H2O(l)\), and the reactants are \(HCl(aq)\) and \(KOH(aq)\). For \(H2O(l)\), the standard enthalpy of formation is \(-285.8 kJ/mol\); for \(KCl(aq)\) and \(HCl(aq)\), we can use the standard enthalpy of formation of Cl- as a proxy, so \(\Delta H_{\mathrm{f}}^{\circ}(KCl) = \Delta H_{\mathrm{f}}^{\circ}(HCl) = -74.9 kJ/mol\); for \(KOH(aq)\), we need to consider the enthalpy of formation for \(OH^-\) and \(K^+\), when K+ is taken to be zero, the standard enthalpy of formation of \(KOH(aq)\) is \(-229.6 kJ/mol\). Substituting all the above values into the formula, we get: \(\Delta H = [1(-285.8)+1(-74.9)] - [1(-229.6)+1(-74.9)] = -56.3 kJ/mol\) Therefore, the enthalpy of neutralization for the reaction of HCl with KOH is \(-56.3 kJ/mol\).