Question #33da5

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Question #33da5

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Answer 1 · 2016-08-18T16:04:54

An aqueous solution for which $[H O^{-}] = 4.2 \times 10^{- 8} \cdot m o l \cdot L^{- 1}$ $[HO^-]=4.2xx10^-8*mol*L^-1$ is slightly acidic.

Explanation:

For neutrality under standard conditions $[H_{3} O^{+}] = [H O^{-}] = 10^{- 7} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=[HO^-]=10^-7*mol*L^-1$ .
Alternatively since, $p H + p O H = 14$ $pH+pOH =14$ , where $p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$ , if $p H < 7$ $pH<7$ , then the solution is acidic.

Here $p O H = 7.38$ $pOH=7.38$ , thus $p H = 6.62$ $pH=6.62$ .

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