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

thats how its supposed to be

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.