Answer
$CH_4$ has a negligible acidity:
Conjugate base: ${CH_3}^-$, which is a strong base.
Work Step by Step
$CH_4(aq) + H_2O(l) --> C{H_3}^-(aq) + H_3O^+(aq)$
As we can see in this reaction, $C{H_3}^-$ is the conjugate base of $CH_4$.
Since $CH_4$ is a very weak acid, $C{H_3}^{-}$ will be a strong base.
