\(n_{OH^-}=2n_{H_2}=2\cdot\dfrac{0.448}{22.4}=0.04\left(mol\right)\)
\(n_{HCl}=0.02\left(mol\right)\)
\(OH^-+H^+\rightarrow H_2O\)
\(n_{OH^-\left(dư\right)}=0.04-0.02=0.02\left(mol\right)\)
\(\left[OH^-\right]_{dư}=\dfrac{0.02}{0.2}=0.1\left(M\right)\)
\(pH=14+\log\left[OH^-\right]=14+\log\left[0.1\right]=13.\)