Consider the equilibrium system $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ Given \(\Delta H_{\mathrm{f}}^{\circ} \mathrm{HF}(a q)=-320.1 \mathrm{~kJ} / \mathrm{mol}\), $$\begin{aligned} \Delta H_{\mathrm{f}}^{\circ} \mathrm{F}^{-}(a q) &=-332.6 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ} \mathrm{F}^{-}(a q)=-13.8 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ K_{\mathrm{a}} \mathrm{HF} &=6.9 \times 10^{-4} \text { at } 25^{\circ} \mathrm{C} \end{aligned}$$ calculate \(S^{\circ}\) for \(\mathrm{HF}(a q)\).

At equilibrium, the change in Gibbs free energy (ΔG) is equal to 0. The relationship between ΔG, ΔH, and ΔS is given by the equation: $$\Delta G = \Delta H - T \Delta S$$ Since ΔG=0 at equilibrium, this equation can be rewritten as: $$\Delta H = T \Delta S$$

Using the provided standard enthalpy of formation values for HF(aq) and F⁻(aq), we can calculate the ΔH for the dissociation reaction as follows: $$\Delta H_{\mathrm{rxn}}=\Delta H_{\mathrm{f}}^{\circ} \mathrm{F}^{-}(a q) - \Delta H_{\mathrm{f}}^{\circ} \mathrm{HF}(a q)$$ Substitute the given values: $$\Delta H_{\mathrm{rxn}}=(-332.6 \mathrm{~kJ} / \mathrm{mol}) - (-320.1 \mathrm{~kJ} / \mathrm{mol}) = -12.5 \mathrm{~kJ/mol}$$

We are given the equilibrium constant for the dissociation of HF at 25°C, \(K_\mathrm{a}\). This can be used to calculate the change in entropy (ΔS) for the reaction at this temperature. The relationship between K, ΔG, and ΔS is given by: $$K_{\mathrm{a}} = e^{\frac{-\Delta G}{R T}}= e^{\frac{\Delta S}{R}}$$ Here, R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298.15 K). We can calculate the ΔS for the reaction as follows: $$\Delta S = R \ln K_{\mathrm{a}}$$ Substitute the given value for the equilibrium constant, Ka, and the gas constant, R: $$\Delta S = (8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}) \ln (6.9 \times 10^{-4}) = -80.57 \mathrm{J/mol \cdot K}$$

Now that we have the ΔH and ΔS values for the dissociation reaction, we can calculate the S⁰ for HF(aq). We know that at equilibrium: $$\Delta H = T \Delta S \Rightarrow S^{\circ} \mathrm{HF}(a q)= \frac{\Delta H - T \cdot S^{\circ} \mathrm{F}^{-}(a q)}{T} - \Delta S_{\mathrm{rxn}}$$ Substitute the values for ΔH, T, ΔS, and \(S^{\circ} \mathrm{F}^{-}(a q)\): \begin{align*} S^{\circ} \mathrm{HF}(a q) &= \frac{-12.5 \mathrm{~kJ/mol} - (298.15 \mathrm{K} \times -13.8 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K})}{298.15 \mathrm{K}} - (-80.57 \mathrm{J/mol \cdot K}) \\ &= \frac{-12.5 \times 10^3 \mathrm{J/mol} + 4112.67 \mathrm{J/mol}}{298.15 \mathrm{K}} + 80.57 \mathrm{J/mol \cdot K} \\ &= -28.01 \mathrm{J/mol \cdot K} \end{align*}

The standard entropy (S⁰) of aqueous hydrogen fluoride (HF) is -28.01 J/mol·K.