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...

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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**

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.