Problem: The Kw for water at 0 C is 0.12 x 10^-14 . Calculate the pH of a neutral aqueous solution at 0°C?

The relationship between [H⁺] and [OH^-] is connected by the following equation:

$\boxed{{\mathbf{K}}_{\mathbf{w}}\mathbf{=}\left[{\mathbf{H}}^{\mathbf{+}}\right]\left[{\mathbf{OH}}^{\mathbf{-}}\right]}$

K_w = autoionization constant of water

K_w = 0.12×10^-14 at T = 0°C

A neutral solution has [H⁺] = [OH^-]. Denoting these concentrations as x, we can determine the concentrations of these ions:

$$\begin{gathered} {\mathbf{K}}_{\mathbf{w}}\mathbf{=}\left[H^{+}\right]\left[{\mathrm{OH}}^{-}\right] \\ \mathbf{0}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\mathbf{=}\left(x\right)\left(x\right) \\ \mathbf{0}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}} \\ {\mathbf{x}}^{\mathbf{2}}\mathbf{=}\sqrt{\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}} \end{gathered}$$

x = [H₃O⁺] = 3.46×10^-8