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$\mathrm{pH} + \mathrm{pOH} = 14$

For water at 60 degrees Celsius,

$K_\mathrm{w} = 1 \times 10^{-13} = [\ce{H+}] \times [\ce{OH-}]$

Hence, $[\ce{H+}] = 1 \times 10^{-6.5} = [\ce{OH-}]$

So $\mathrm{pH} = -\log(1\times 10^{-6.5}) = 6.5 = \mathrm{pOH}$

which gives $\mathrm{pH} + \mathrm{pOH} = 6.5+6.5 = 13$, not $14$

So is my math or chemistry wrong, or are there limitations on $\mathrm{pH} + \mathrm{pOH} = 14$?

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$\mathrm{pH + pOH = 14}$

This equation only holds true at around $25~^\circ\mathrm{C}$, where the water autodissociation constant $K_\mathrm{w} = 10^{-14}$.

Mathematically, you can write

$$\begin{align} K_\mathrm{w} = [\ce{H+}][\ce{OH-}] &= 10^{-14} & &\\ \log([\ce{H+}][\ce{OH-}]) &= \log(10^{-14}) & &\text{(logarithms of both sides)} \\ \log[\ce{H+}] + \log[\ce{OH-}] &= -14 & &(\log(ab) = \log a + \log b) \\ -\log[\ce{H+}] - \log[\ce{OH-}] &= 14 & & \\ \mathrm{pH + pOH} &= 14 & & \text{(definitions of pH and pOH)} \end{align}$$

It's fairly clear from this that if $K_\mathrm{w}$ is $10^{-x}$ at some other temperature, then you will obtain $\mathrm{pH + pOH} = x$.

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  • 1
    $\begingroup$ IOW, pH+pOH=-log(Kw). $\endgroup$ – MSalters Dec 1 '16 at 11:28

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