What is the molarity of H+ in a solution formed by mixing 27.00 mL of 0.2445 M Ca(OH)2 with 110.00 mL of 0.1662M HCl?

Ca(OH)₂ is a strong alkali. 1 mole of Ca(OH)₂ completely dissociates to give 2 mole of OH⁻ ions.
[OH⁻] = [Ca(OH)₂]ₒ × 2 = (0.2445 M) × 2 = 0.4890 M

HCl is a strong acid. 1 mole of HCl completely dissociates to give 1 mole of H⁺ ions.
[H⁺] = [HCl]ₒ = 0.1662 M

Equation for the reaction: H⁺(aq) + OH⁻(aq) → H₂O(l)
OR: Mole ratio H⁺ : OH⁻ = 1 : 1

Initial no. of moles of H⁺ ions = (0.1662 mol/L) × (110.00/1000 L) = 0.018282 mol
Initial no. of moles of OH⁻ ions = (0.4890 mol/L) × (27.00/1000 L) = 0.013203 mol < 0.018282 mol
Hence, OH⁻ ions completely react.

No. of moles of H⁺ ions = (0.018282 - 0.013203) mol = 0.005079 mol
Final volume of solution = (110.00 + 27.00) mL = 137.00 mL = 0.13700 L

Molarity of H⁺ ions in the final solution = (0.005079 mol) / (0.13700 L) = 0.03707 M

We need to get the moles of each reactant:

Ca(OH)2 ---> (0.2445 mol/L) (0.02700 L) = 0.0066015 mol
HCl ---> (0.1662 mol/L) (0.1100 L) = 0.018282 mol

The chemical reaction is this:

2HCl + Ca(OH)2 ---> CaCl2 + 2H2O

The key point is the 2:1 molar ratio between HCl and Ca(OH)2. We use this ratio to determine the amount of HCl used to neutralize all of the Ca(OH)2.

0.0066015 mol times 2 = 0.013203 mol

Next, we determine the amount of HCl left over:

0.018282 mol - 0.013203 mol = 0.005079 mol

The final step:

0.005079 mol / 0.137 L = 0.03707 M

The 0.137 L is the sum of the two volumes mixed. Often, the problem goes one more step and asks for the pH:

pH = -log 0.03707 = 1.4310