An aqueous solution of calcium hydroxide is standardized by titration with a 0.183 M solution of hydrochloric acid. If 16.6 mL of base are required to neutralize 25.8 mL of the acid, what is the molarity of the calcium hydroxide solution?

0.142 M

Explanation:

Let's consider the neutralization reaction between calcium hydroxide and hydrochloric acid.

Ca (OH) ₂ + 2 HCl → CaCl₂ + 2 H₂O

25.8 mL of 0.183 M HCl are used. The reacting moles are:

0.0258 L * 0.183 mol/L = 4.72 * 10⁻³ mol

The molar ratio of Ca (OH) ₂ to HCl is 1:2. The reacting moles of Ca (OH) ₂ are 1/2 * 4.72 * 10⁻³ mol = 2.36 * 10⁻³ mol

2.36 * 10⁻³ moles of Ca (OH) ₂ are contained in 16.6 mL. The molarity of Ca (OH) ₂ is:

M = 2.36 * 10⁻³ mol/0.0166 L = 0.142 M

Molarity of calcium hydroxide (C1) = ?

Molarity of hydrochloric acid (C2) = 0.183 M

Volume of hydrochloric acid (V2) = 25.8 ml

Volume of calcium hydroxide (V1) = 16.6 ml

Using the equation of

C1*V1 = C2*V2

C2 = 0.183 * 25.8/16.6

C2 = 0.284 M

The molarity of calcium hydroxide is 0.284 M