Question

What volume of 0.67 mol/L sulphuric acid can be neutralized by 22.7 mL of 0.385 mol/L barium hydroxide

Answer 1

H₂SO₄+Ba(OH)₂=BaSO₄+H₂O

1.Find the amount of barium hydroxide substance:

$M=\frac{n}{V}$

$0.385=\frac{n}{0.0227}$

n=0.0087395

2.Find the volume of sulfuric acid:

$M=\frac{n}{V}$

$0.67=\frac{0.0087395}{V}$

V=0.013 L