\(n_{OH^-}=6.10^{-3}\left(mol\right)\)
\(n_{H^+}=V.10^{-4}\left(mol\right)\)
\(n_{OH^-dư}=0,02.\left(0,0001.V+0,06\right)\left(mol\right)\)
Ta có:
\(n_{OH^-dư}+n_{H^+}=n_{OH^-\text{}}\)
\(\Leftrightarrow0,02.\left(0,0001.V+0,06\right)+V.10^{^{-4}}=6.10^{-3}\)
\(\Leftrightarrow V=47,06\left(ml\right)\)