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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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` =...

(The entire section contains 2 answers and 324 words.)

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