If $p H = 8$ $pH=8$ , what are $[H_{3} O^{+}]$ $[H_3O^+]$ and $[H O^{-}]$ $[HO^-]$ ?

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If $p H = 8$ $pH=8$ , what are $[H_{3} O^{+}]$ $[H_3O^+]$ and $[H O^{-}]$ $[HO^-]$ ?

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Answer 1 · 2017-11-02T07:43:33

0.00000001

Explanation:

The conversion from pH to [H+] is: $10^{- p} H$ $10^-pH$

Therefore, you can just do the math having that formula and the pH value: $10^{- 8} = 1 E - 8 or 0.00000001$ $10^-8 = 1 E-8 or 0.00000001$

Keep in mind that some professors/textbooks will use [H3O+] instead of [H+], but they are both referring to the same thing.

Hope this helped (c:

Answer 2 · 2017-11-02T08:39:18

$[H_{3} O^{+}] = 10^{- 8} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=10^-8*mol*L^-1$

Explanation:

By definition, $p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$ ....(note that we use $H^{+}$ $H^+$ and $H_{3} O^{+}$ $H_3O^+$ interchangeably). And thus if $p H = 8$ $pH=8$ , $[H_{3} O^{+}] = 10^{- 8} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=10^(-8)*mol*L^-1$ .

Note that in aqueous solution,...

$p H + p O H = 14$ $pH+pOH=14$ ...and so....

....... since $[H O^{-}] = 10^{- 6} \cdot m o l \cdot L^{- 1}$ $[HO^-]=10^(-6)*mol*L^-1$ .....this solution is ALKALINE given that there are more hydroxide ions than hydronium ions....

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If $p H = 8$ $pH=8$ , what are $[H_{3} O^{+}]$ $[H_3O^+]$ and $[H O^{-}]$ $[HO^-]$ ?

2 Answers

Explanation:

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If $p H = 8$ $pH=8$ , what are $[H_{3} O^{+}]$ $[H_3O^+]$ and $[H O^{-}]$ $[HO^-]$ ?