If $p H = 8.19$ $pH=8.19$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?

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If $p H = 8.19$ $pH=8.19$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?

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Answer 1 · 2017-03-07T17:42:23

$p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$ , so $[H_{3} O^{+}] = 6.00 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=6.00xx10^-9*mol*L^-1$ .

Explanation:

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

So that if $p H = 8.19$ $pH=8.19$ , $[H_{3} O^{+}] = 10^{- 8.19}$ $[H_3O^+]=10^(-8.19)$

$6.00 \times 10^{- 9} \cdot m o l \cdot L^{- 1}$ $6.00xx10^-9*mol*L^-1$

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If $p H = 8.19$ $pH=8.19$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?

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If $p H = 8.19$ $pH=8.19$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?