CM(NaOH) = 0,01 (mol)=> nNaOH = 0,1.0,01=0,001(mol)
V = 200 ml = 0,2 (l)
2NaOH + H2SO4 -> Na2SO4 +H2O
0,001............0,0005 (mol)
nH+ dư = 0,01.0,2=0,002 (mol)
\(\Sigma n_{H^+ban.dau}=0,0005.2+0,001=0,002\left(mol\right)\)
CM (H+ bđ) = 0,002/0,2=0,01 => pH = 2
