How is pOH determined?

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How is pOH determined?

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Answer 1 · 2016-06-04T17:21:12

$p O H$ $pOH$ is detemined equivalently to $p H$ $pH$ , but $p H + p O H = 14$ $pH+pOH=14$ under standard conditions.

Explanation:

$p H$ $pH$ $=$ $=$ $- {log}_{10} [H_{3} O^{+}]$ $-log_10[H_3O^+]$

$p O H$ $pOH$ $=$ $=$ $- {log}_{10} [H O^{-}]$ $-log_10[HO^-]$

Now $K_{w}$ $K_w$ $=$ $=$ $[H_{3} O^{+}] [H O^{-}]$ $[H_3O^+][HO^-]$ $=$ $=$ $10^{- 14} at 298K$ $10^(-14)" at 298K"$ .

Thus $- {log}_{10} K_{w} = - {log}_{10} [H_{3} O^{+}] - {log}_{10} [H O^{-}]$ $-log_10K_w=-log_10[H_3O^+] -log_10[HO^-]$

$= - {log}_{10} (10^{- 14})$ $=-log_10(10^-14)$ $=$ $=$ $+ 14$ $+14$

And thus $p O H + p H = 14$ $pOH+pH=14$ under standard conditions.

You should review the logarithmic function if you have any problem with this treatment. You should review it anyway, because I might have made a mistake.

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How is pOH determined?

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