\(n_{OH^-}=0,04a\left(mol\right)\)
\(n_{H^+}=6\cdot10^{-4}\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
\(PH=13\) là môi trường bazo. vậy axit hết.
\(\Rightarrow n_{OH^-dư}=0.04a-6.10^{-4}\)
\(1=-log\left(\dfrac{0.04a-6.10^{-4}}{0.1}\right)\Rightarrow a=0.265M\)