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

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

K_{w} = autoionization constant of water

**K _{w} = 0.12×10**

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

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

**x = [H _{3}O^{+}] = 3.46×10^{-8}**

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

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