nH+=nHCl=0,04.0,75=0,03(mol)
nBa(OH)2=0,16.0,08=0,0128(mol)
nKOH=0,16.0,04=6,4.10^-3
=>nOH-=2.0,0128+6,4.10^-3=0,032 (mol)
PT ion: H++OH-→H2O
0,03 0,032
=>nOH- dư=0,032-0,03=2.10^-3 (mol)
=>[OH-] dư=$\frac{2.10^-3}{0,2}$=0,01 (mol/l)
=>[H+]=$\frac{10^-14}{0,01}$=10^-12
=>pH=-log([H+])=12