Question

13.0 mL of hydrochloric acid (HCl) required to neutralize 36.11 ml of 0.045 M sodium hydroxide (NaOH). What is the molarity of hydrochloric acid?

Answer 1

Molarity of HCl = 0.12 M

Step-by-step explanation:

Given data:

Volume of HCl = 13.0 mL

Volume of NaOH = 36.11 mL

Molarity of NaOH = 0.045 M

Molarity of HCl = ?

Solution:

Formula:

M₁V₁ = M₂V₂

M₁= Molarity of HCl

V₁ = Volume of HCl

M₂ = Molarity of NaOH

V₂ = Volume of NaOH

M₁ × 13.0 mL = 0.045 M × 36.11 mL

M₁ = 1.62 M.mL / 13.0 mL

M₁ =0.12 M