$pH=1\to [H^+]=0,1\ (M)$
$\to n_{H^+}=0,1.0,3=0,03\ (mol)$
$\to n_{H_2SO_4}=\dfrac{0,03}{2}=0,015\ (mol)$
$n_{Ba(OH)_2}=0,2.0,0625=0,0125\ (mol)$
$H_2SO_4+Ba(OH)_2\to BaSO_4+2H_2O$
Do $0,015>0,0125$ nên $H_2SO_4$ dư
Theo PT: $n_{H_2SO_4\ pứ}=n_{Ba(OH)_2}=0,0125\ (mol)$
$\to n_{H_2SO_4\ dư}=0,015-0,0125=0,0025\ (mol)$
$\to n_{H^+\ dư}=0,0025.2=0,005\ (mol)$
$V_{dd\ spứ}=0,3+0,2=0,5\ (l)$
$\to [H^+]=\dfrac{0,005}{0,5}=0,01\ (M)$
$\to pH=-\log[H^+]=-\log0,01=2$
