Question

Given that \(\Delta H_{\mathrm{t}}^{\circ}\) 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 \(25^{\circ} \mathrm{C}\)

Short answer

Answer: The acid dissociation constant (Ka) for hydrofluoric acid (HF) at 25°C is approximately 1.2 x 10^6.

Step-by-step solution

Convert the temperature to Kelvin

In this exercise, the given temperature is 25°C, which needs to be converted to Kelvin. To do this, add 273.15 to the given temperature in Celsius: \(T(K) = 25 + 273.15 = 298.15~K\)

Calculate the Gibbs free energy change (ΔG°)

Now we can plug the given values of \(\Delta H_{\mathrm{t}}^{\circ}\), \(S^{\circ}\) and the temperature (T) into the equation for Gibbs free energy change: \(\Delta G^{\circ} = \Delta H_{\mathrm{t}}^{\circ} - T\Delta S^{\circ}\) \(\Delta G^{\circ} = (-320.1~kJ/mol) - (298.15~K)(88.7~J/mol\cdot K)\) First, note that we need to convert the entropy change from J/mol·K to kJ/mol·K to match the units of enthalpy change: \(S^{\circ} = 88.7~J / (mol\cdot K) \times \frac{1~kJ}{1000~J} = 0.0887~kJ / (mol\cdot K)\) Now, substitute the values and calculate \(\Delta G^{\circ}\): \(\Delta G^{\circ} = -320.1 - (298.15)(0.0887) = -320.1 - 26.4 = -346.5~kJ/mol\)

Calculate the acid dissociation constant (Ka)

With the obtained value of \(\Delta G^{\circ}\), we can now find the acid dissociation constant (\(K_{a}\)) using the equation: \(\Delta G^{\circ} = -RT\ln K_{\mathrm{a}}\) First, let's rearrange the equation to isolate \(K_{\mathrm{a}}\): \(\ln K_{\mathrm{a}} = -\frac{\Delta G^{\circ}}{RT}\) Next, we can plug in the values for \(\Delta G^{\circ}\), R (which is 8.314 J/(mol·K) or 0.008314 kJ/(mol·K)) and T: \(\ln K_{\mathrm{a}} = -\frac{-346.5~kJ/mol}{(0.008314~kJ/(mol\cdot K))(298.15~K)}\) Now calculate the natural logarithm of the \(K_{\mathrm{a}}\): \(\ln K_{\mathrm{a}} = 14.0\) Finally, we need to take the exponent of the natural logarithm to obtain the \(K_{\mathrm{a}}\): \(K_{\mathrm{a}} = e^{14.0} \approx 1.2 \times 10^6\) Therefore, the acid dissociation constant (\(K_{\mathrm{a}}\)) for HF at 25°C is approximately \(1.2 \times 10^6\).