Question

How many milliliters of a 0.221 M HCl solution are needed to neutralize 299 mL of a 0.0459 M Ba(OH)2 solution?

Answer 1

ANSWER:

Given,

Molarity of HCl , M₁ = 0.221 M

Normality of HNO₃, N₁ = 0.221 M {because it releases 1 H⁺ only}

Volume of HCl, V₁ = ?

Molarity of Ba(OH)₂, M₂ = 0.0459 M

Normality of Ba(OH)₂, N₂ = 2 x M₂ {because it releases 2 OH^-}

= 2 x 0.0459 = 0.0918 N

Volume of Ba(OH)₂, V₂ = 299 mL

Now, at neutralization point:

N₁ x V₁ = N₂ x V₂

0.221 N x V₁ = 0.0918 N x 299 mL

V₁ = 124.2 mL

Hence, 124.2 mL of a 0.221 M HCl solution are needed to neutralize 299 mL of a 0.0459 M Ba(OH)2 solution.