How do you do acid base titration calculations?

Acid-base titrations are really stoichiometry problems. The central key to these problems is to find the molar ratio of the acid to the base. This comes from the balanced chemical equation. Then you use the volume and molarity to convert to moles and vice versa. Here’s how it works.

If 11.6 mL of 3.00 mol/L sulfuric acid are required to neutralize the sodium hydroxide in 25.0 mL of NaOH solution, what is the molarity of the NaOH?

First, write the balanced chemical equation for the reaction.
H₂SO₄ + 2NaOH → 2H₂O + Na₂SO₄

The steps in your calculation will be
1. Convert the volume of H₂SO₄ to moles of H₂SO₄
2. Convert the moles of H₂SO₄ to moles of NaOH
3. Convert the moles of NaOH to molarity of NaOH

Here we go.
1. 0.0116 L H₂SO₄ × (3.00 mol H₂SO₄/1 L H₂SO₄) = 0.0348 mol H₂SO₄
2. 0.0348 mol H₂SO₄ × (2 mol NaOH/1 mol H₂SO₄) = 0.0696 mol NaOH
3. 0.0696 mol NaOH/0.0250 L NaOH = 2.78 mol/L

Note: Many titration calculations use the formula M₁V₁ = M₂V₂, where M stands for molarity and V stands for volume, but this formula works only if the molar ratio of acid to base is 1:1. You are always safe if you use the molar ratios explicitly in your calculations.

Very detailed and simple explanation here: http://preparatorychemistry.com/Bishop_Titration.htm