\(nH^+\)= nHCl=0,0002(mol)
\(nOH^-\)=2.0,02.0,03=0,0012(mol)
\(H^+\) + \(OH^-\)\(\rightarrow\)H2O
0,0002(mol) \(\rightarrow\) 0,0002(mol)
=> n\(OH^-\)dư = 0,001(mol)
=> CM OH- =\(\dfrac{0,001}{0,5}\)=\(2.10^{-3}\)(CM)
=> CM H+=\(5.10^{-12}\)
=> ph= -lg(\(5.10^{-12}\))=11,3