If $p H = 5.25$ $pH=5.25$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?

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If $p H = 5.25$ $pH=5.25$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?

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Answer 1 · 2016-04-27T07:58:05

$[H_{3} O^{+}]$ $[H_3O^+]$ $=$ $=$ $10^{- 5.25} \cdot m o l \cdot L^{- 1}$ $10^(-5.25)*mol*L^-1$ .

Explanation:

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

It is worthwhile reviewing your grasp of the logarithmic function.

When we write, ${log}_{a} b$ $log_ab$ , we ask to what power we raise the base $a$ $a$ to get $b$ $b$ . Thus ${log}_{10} 100 = 2$ $log_(10)100=2$ , and ${log}_{10} 0.1 = - 1$ $log_(10)0.1=-1$ .

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If $p H = 5.25$ $pH=5.25$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?

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If $p H = 5.25$ $pH=5.25$ , what is $[H_{3} O^{+}]$ $[H_3O^+]$ ?