Answer:
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
