$n_{HCl}=0,1.0,01=0,001 (mol)$
$n_{H_2SO_4}=0,1.0,005=0,0005 (mol)$
$HCl→H^++Cl^-$
$H_2SO_4→2H^++SO_4^{2-}$
$⇒n_{H^+}=n_{HCl}+2.n_{H_2SO_4}=0,001+0,0005.2=0,002 (mol)$
$-V_{dd}=0,1+0,1=0,2 (l)$
$⇒[H^+]=\frac{0,002}{0,2}=0,01 (M)$
$⇒pH=-log[H^+]=-log(0,01)=2$
---------------------Nguyễn Hoạt---------------------