Question

What is the volume in milliliters of 0.350 M KOH needed to completely neutralize 15.0 mL of a 0.250M H2S04 solution?

Accepted Answer

Sulfuric acid, H2SO4, produces two hydrogen ions per molecule. The second dissociation isn't complete since HSO4- is a weak acid, but OH- ions that are present in solution will remove the H+ from the HSO4- ions. 
Neutralization occurs when the moles of H+ and OH- are equal. Moles of H+ ion that will be neutralized is calculated as follows:
(15.0 ml H2SO4)(1 L/1000ml)(0.250 moles/L)(2 moles H+/mole H2SO4) = 0.0075 moles H+
The volume of 0.350 M KOH needed to provide 0.0075 moles of OH- is:
(0.0075 moles)/(0.350 moles/L) x (1000 ml/1 L) = 21.4 ml KOH
Here's a shortcut of the above calculation:
Since moles acid = moles base, and moles = molarity x volume,
(molarity of acid)(volume of acid)(n) = (molarity of base)(volume of base)(n)
(Molarity of the solution must be multiplied by the number of H+ or OH- produced per molecule, which is represented by "n" in the equation.)
(0.250 M)(15.0 ml)(2) = (0.350)(x)(1) where x = volume of OH- = 21.4 ml
Volume doesn't need to be converted to liters, because you'll end up with milliliters if you start with volume in milliliters. The steps of converting from ml to L and then from L back to ml cancel each other out.

Answer

The neutralization reaction of potassium hydroxide and sulfuric acid is:
2KOH + H_2SO_4 -> K_2SO_4 + 2H_2O
One mole of sulfuric acid is neutralized by two moles of potassium hydroxide to give one mole of potassium sulfate and two moles of water.
Each liter of a one M solution of sulfuric acid has one mole of sulfuric acid. 15 ml of a 0.250 M solution has (15/1000)*0.25 = .00375 moles of sulfuric acid. This is neutralized by 0.00375*2 = 0.0075 moles of potassium hydroxide.
Each liter of a 0.350 M solution of potassium hydroxide solution has 0.35 moles. To obtain 0.0075 moles, the volume of 0.035 M solution required is 0.0075*1000/0.35 = 21.4285 mL.
The volume in milliliters of 0.350 M potassium hydroxide required to completely neutralize 15 ml of a 0.25 M solution of sulfuric acid is 21.4285 mL

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What is the volume in milliliters of 0.350 M KOH needed to completely neutralize 15.0 mL of a 0.250M H2S04 solution?

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