Question

If 63.99 mL of Ba(OH)2 is required to neutralize 56.3 mL of 0.0410 molar HCl, what is the molarity of the Ba(OH)2 ?

Answer 1

0.0181 M

Step-by-step explanation:

Let's consider the following neutralization reaction.

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

56.3 mL of 0.0410 M HCl react. The moles of HCl are:

0.0563 L × 0.0410 mol/L = 2.31 × 10⁻³ mol

The molar ratio of HCl to Ba(OH)₂ is 2:1. The moles of Ba(OH)₂ are:

1/2 × 2.31 × 10⁻³ mol = 1.16 × 10⁻³ mol

There are 1.16 × 10⁻³ mol of Ba(OH)₂ in 63.99 mL. The molarity of Ba(OH)₂ is:

1.16 × 10⁻³ mol/ 63.99 × 10⁻³ L = 0.0181 M