# 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?

###### FREE Expert Solution

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

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

Kw = autoionization constant of water

Kw = 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:

${\mathbf{K}}_{{\mathbf{w}}}{\mathbf{=}}\left[{H}^{+}\right]\left[{\mathrm{OH}}^{-}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathbf{0}}{\mathbf{.}}{\mathbf{12}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{14}}{\mathbf{=}}\left(x\right)\left(x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathbf{0}}{\mathbf{.}}{\mathbf{12}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{14}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{=}}\sqrt{\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}$

x = [H3O+] = 3.46×10-8

###### Problem Details

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

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the pH and pOH concept. You can view video lessons to learn pH and pOH. Or if you need more pH and pOH practice, you can also practice pH and pOH practice problems.