Answer
Please see the work below.
Work Step by Step
We know that
$log[H_3O^+]=-PH=-3.85$
$log[H_3O^{+}]=-3.85$
$[H_3O^+]=anti log(-3.85)$
$[H_3O^+]=10^{-3.85}=1.41\times 10^{-4}M$
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