Consider the process $$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad \Delta H^{\circ}=55.8 \mathrm{~kJ}$$ (a) Will the \(\mathrm{pH}\) of pure water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) be \(7.0 ?\) (b) If not, calculate the \(\mathrm{pH}\) of pure water at \(37^{\circ} \mathrm{C}\).

We will start by writing the expression for the equilibrium constant, \(K_{eq}\), for the given process: $$K_{eq} = \frac{[\mathrm{H}^+][\mathrm{OH}^-]}{[\mathrm{H_2O}]}$$ Using the Van't Hoff equation, we can relate the equilibrium constant at different temperatures: $$\ln \frac{K_2}{K_1}=-\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$ Here, \(K_1\) is the equilibrium constant at the reference temperature (\(T_1\)) and \(K_2\) is the equilibrium constant at the new temperature (\(T_2\)). We are given \(\Delta H^{\circ}\), and we can use the gas constant \(R = 8.314 \mathrm{~J\ K^{-1}\ mol^{-1}}\). The reference temperature is \(T_1 = 25^{\circ}\mathrm{C} = 298\ \mathrm{K}\), and the equilibrium constant at this temperature is the ion product of water, \(K_{w} = 1.0 \times 10^{-14}\). The new temperature is \(T_2 = 37^{\circ}\mathrm{C} = 310\ \mathrm{K}\). Now, let's plug these values into the equation and solve for \(K_2\): $$\ln \frac{K_2}{1.0 \times 10^{-14}} = -\frac{(55.8 \times 10^3 \ \mathrm{J\ mol^{-1}})}{(8.314 \mathrm{~J\ K^{-1}\ mol^{-1}})}\left(\frac{1}{310\ \mathrm{K}} - \frac{1}{298\ \mathrm{K}}\right)$$ Solve for \(K_2\): $$K_2 = 1.0 \times 10^{-14} \cdot e^{\left[-\frac{(55.8 \times 10^3 \ \mathrm{J\ mol^{-1}})}{(8.314 \mathrm{~J\ K^{-1}\ mol^{-1}})}\left(\frac{1}{310\ \mathrm{K}} - \frac{1}{298\ \mathrm{K}}\right)\right]}$$

At equilibrium, the concentrations of H⁺ and OH⁻ are equal, so \(\mathrm{[H^+]}=\mathrm{[OH^-]}\). Let's call this common concentration \(x\). Then, we have: $$K_2 = x^2$$ Now, we can determine whether the pH at 37℃ is 7 by checking if \(x = 1.0 \times 10^{-7}\ \mathrm{M}\). If it is, then the pH of pure water at 37℃ is 7.

If the pH at 37℃ is not 7, then we need to calculate the new pH. The pH is defined as: $$\mathrm{pH} = -\log_{10} [\mathrm{H}^+]$$ Using the value of \(x\) obtained from the new equilibrium constant, \(K_2\), substitute \(x\) for \([\mathrm{H}^+]\) and calculate the pH: $$\mathrm{pH} = -\log_{10} x$$