Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,1.0,06=0,006\left(mol\right)\\n_{HCl}=0,4.0,02=0,008\left(mol\right)\end{matrix}\right.\)
PTHH: `Ba(OH)_2 + 2HCl -> BaCl_2 + 2H_2O`
0,006------->0,003
`=>` \(\left[H^+\right]=C_{M\left(HCl\right)}=\dfrac{0,008-0,003}{0,1+0,4}=0,01M=10^{-2}M\)
`=> pH = 2`