How many moles of of KOH are needed to exactly neutralize 500. mL of 1.0 M HCl?

Jun 10, 2017

$\text{0.50 moles KOH}$

Explanation:

Potassium hydroxide, $\text{KOH}$, will react with hydrochloric acid, $\text{HCl}$, to produce aqueous potassium chloride and water.

The balanced chemical equation that describes this neutralization reaction looks like this

${\text{KOH"_ ((aq)) + "HCl"_ ((aq)) -> "KOH"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

So, the reaction consumes potassium hydroxide and hydrochloric acid in a $1 : 1$ mole ratio, which implies that the number of moles of potassium hydroxide needed to completely neutralize the hydrochloric acid solution will be equal to the number of moles of hydrochloric acid present in said solution.

As you know, molarity tells you the number of moles of solute present in $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

This implies that the hydrochloric acid solution contains

500. color(red)(cancel(color(black)("mL"))) * "1.0 moles HCl"/(10^3color(red)(cancel(color(black)("mL")))) = "0.50 moles HCl"

Therefore, you can say that a complete neutralization requires $0.50$ moles of potassium hydroxide. The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the solution.