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?

0.0181 M

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