Question

The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)₂ is

(Assume complete neutralization)

(1) 2.5 mL

(2) 5.0 mL

(3) 10.0 mL

(4) 7.5 mL

Answer 1

Correct option is (3) 10.0 mL

m.eq. of HBr = m.eq. of Ba(OH)₂ 

M₁ × n₁ × V₁(mL) = M₂ × n₂ × V₂(mL) 

0.02 × 1 × V₁(mL) = 0.02 × 2 × 10 

V₁(mL) = 10 mL

Comments

This is very helpful for me

Answer 2

Correct option is (3) 10.0 mL

Comments

Thanks