Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}\) for \(\mathrm{HF}(a q)\) is \(88.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\), find \(K_{\mathrm{a}}\) for \(\mathrm{HF}\) at \(25^{\circ} \mathrm{C}\).

The balanced chemical equation for the dissociation of aqueous hydrogen fluoride (HF) is: HF(aq) \(\rightleftharpoons\) H\({^{+}}\)(aq) + F\({^{-}}\)(aq)

We can use the following equation to calculate the Gibbs free energy, \(\Delta G^{\circ}\), for the reaction: \(\Delta G^{\circ} = \Delta H_{\mathrm{f}}^{\circ} - T\Delta S^{\circ}\) where \(T\) is the temperature in Kelvin. Since the given temperature is \(25^{\circ} \mathrm{C}\), we need to convert it into Kelvin: \(T = 25 + 273.15 = 298.15\) K Now, plug in the given values to calculate \(\Delta G^{\circ}\): \(\Delta G^{\circ} = -320.1\mathrm{~kJ} / \mathrm{mol} - (298.15 \mathrm{K})(88.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})\) \(\Delta G^{\circ} = -320.1\mathrm{~kJ} / \mathrm{mol} - (298.15 \mathrm{K})(0.0887 \mathrm{~kJ} / \mathrm{mol} \cdot \mathrm{K})\) \(\Delta G^{\circ} = -320.1 - 26.44 = -346.54\) kJ/mol

We can use the relationship between Gibbs free energy and the equilibrium constant, \(K\), as follows: \(\Delta G^{\circ} = -RT\ln{K}\) where \(R\) is the gas constant (8.314 J/mol⋅K). To find the acid dissociation constant, \(K_{\mathrm{a}}\), we can rearrange the equation and solve for \(K_{\mathrm{a}}\): \(K_{\mathrm{a}} = e^{(-\Delta G^{\circ}/RT)}\) First, convert the Gibbs free energy into J/mol: \(\Delta G^{\circ} = -346.54\mathrm{~kJ/mol} \times 1000\mathrm{~J/kJ} = -346540\mathrm{~J/mol}\) Now, plug in the values and calculate \(K_{\mathrm{a}}\): \(K_{\mathrm{a}} = e^{(-(-346540\mathrm{~J/mol})/(8.314 \mathrm{J/mol⋅K} \times 298.15\mathrm{~K}))}\) \(K_{\mathrm{a}} = e^{146.96}\) \(K_{\mathrm{a}} \approx 3.70 \times 10^{63}\) So, the acid dissociation constant, \(K_{\mathrm{a}}\), for HF at \(25^{\circ} \mathrm{C}\), is approximately \(3.70 \times 10^{63}\).