Problem: A sample of solid calcium hydroxide, Ca(OH)2, is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00 × 10−2 M HCl requires 36.6 mL of the acid to reach the end point.Ca(OH)2(aq) + 2HCl(aq) ⟶ CaCl2(aq) + 2H2O(l)What is the molarity?

The titration is at endpoint which means that all of the base is neutralized by the acid. 

Ca(OH)₂(aq) + 2HCl(aq) ⟶ CaCl₂(aq) + 2H₂O(l)

1 mole Ca(OH)₂ : 2 mole HCl

Recall: 

$\boxed{\mathbf{Molarity}\mathbf{ }\mathbf{=}\frac{\mathbf{ }\mathbf{moles}\mathbf{ }\mathbf{solute}\mathbf{ }}{\mathbf{L}\mathbf{ }\mathbf{solution}}}$

Calculate moles Ca(OH)₂:

$\frac{5.00 \times {10}^{-2} \cancel{mol HCl }}{1 \cancel{\mathbf{L}}}\times 36.6 \cancel{\mathbf{m}\mathbf{L}}\times \frac{{10}^{-3} \cancel{\mathbf{L}}}{1\mathbf{ }\cancel{\mathbf{m}\mathbf{L}}}\times \frac{1 mol Ca{\left(OH\right)}_2}{2 \cancel{mol HCl }}$

= 9.15 x 10^-4 mol Ca(OH)₂