a,
Gọi x, y là nồng độ mol $HCl$ và $H_2SO_4$
- TN1:
$n_{HCl}=0,05x$
$AgNO_3+HCl\to AgCl+HNO_3$
$\Rightarrow 143,5.0,05x=2,87$ (1)
- TN2:
$n_{H_2SO_4}=0,1y$
$BaCl_2+H_2SO_4\to BaSO_4+2HCl$
$\Rightarrow 233.0,1y=4,66$ (2)
(1)(2)$\Rightarrow x=0,4; y=0,2$
b,
10ml X có:
$n_{HCl}=0,01.0,4=0,004 mol$
$n_{H_2SO_4}=0,01.0,2=0,002 mol$
$NaOH+HCl\to NaCl+H_2O$
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
$\Rightarrow n_{NaOH}=0,004+0,002.2=0,008 mol$
$\to V_{NaOH}=\dfrac{0,008}{2}=0,004l=4(ml)$