What is the logarithmic concentration of the hydroxide ion in aqueous solution?

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What is the logarithmic concentration of the hydroxide ion in aqueous solution?

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Answer 1 · 2016-01-26T16:15:54

I presume you mean $p O H$ $pOH$ ?

Explanation:

We would calculate $p O H$ $pOH$ the same way we calculate $p H$ $pH$ . That is $p O H$ $pOH$ $=$ $=$ $- {log}_{10} [O H^{-}]$ $-log_(10)[OH^-]$ , where $[O H^{-}]$ $[OH^-]$ is the concentration of hydroxide ion.

Given that $10^{- 14}$ $10^-14$ $=$ $=$ $[H_{3} O^{+}] [O H^{-}]$ $[H_3O^+][OH^-]$ ,

then ${log}_{10} 10^{- 14}$ $log_(10)10^(-14)$ $=$ $=$ ${log}_{10} [H_{3} O^{+}]$ $log_(10)[H_3O^+]$ $+$ $+$ ${log}_{10} [H O^{-}]$ $log_(10)[HO^-]$

And thus $14$ $14$ $=$ $=$ $p H + p O H$ $pH + pOH$ (because by definition ${log}_{10} 10^{- 14}$ $log_(10)10^(-14)$ $=$ $=$ $- 14$ $-14$ .

So you need to find hydroxide (or acid) concentration, and take its negative logarithm to the base 10.

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What is the logarithmic concentration of the hydroxide ion in aqueous solution?

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