$n_{HCl}=n_{H^+}=0,1(mol)$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$2Na+2H^+\to 2Na^++H_2$
$2Na+2H_2O\to 2Na^+ + OH^-+H_2$
Đặt $x$ là số mol $OH^-$
$\to n_{H_2O}=2x(mol)$
Bảo toàn $H$: $0,1+2x=0,1.2+x$
$\to x=0,1$
$V_{dd}=0,1l$
$\to [OH^-]=\dfrac{0,1}{0,1}=1$
$\to pH=14+\log[OH^-]=14+\log 1=14$