# 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

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###### 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?