# How many hydroxide ions are needed to completely neutralize 1.0 liter of 0.50 M HCl?

Jun 30, 2017

$3.0 \cdot {10}^{23}$ $\text{OH"^(-) "anions}$

#### Explanation:

The idea here is that the hydronium cations, ${\text{H"_3"O}}^{+}$, and the hydroxide anions, ${\text{OH}}^{-}$, react in a $1 : 1$ mole ratio to reform water.

In other words, it takes $1$ mole of hydroxide anions to completely neutralize $1$ mole of hydronium cations. Consequently, you can say that it takes $1$ hydroxide anion to completely neutralize $1$ hydronium cation.

Now, you know that you're dealing with $\text{1.0 L}$ of a $\text{0.50-M}$ hydrochloric acid solution, so right from the start, you know that the solution contains $0.50$ moles of hydrochloric acid $\to$ think about what molarity represents.

Since hydrochloric acid is a strong acid, it will ionize completely in aqueous solution to produce hydronium cations

${\text{HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

This means that your solution contains $0.50$ moles of hydronium cations.

As a result, you can say that you will need $0.50$ moles of hydroxide anions to completely neutralize the number of moles of hydronium cations present in your sample.

Now, to convert this to the number of anions, you need to use Avogadro's constant, which tells you that in order to have $1$ mole of hydroxide anions you need to have $6.022 \cdot {10}^{23}$ individual anions.

Therefore, you will need

0.50 color(red)(cancel(color(black)("moles OH"^(-)))) * (6.022 * 10^(23)color(white)(.)"OH"^(-)"anions")/(1color(red)(cancel(color(black)("mole OH"^(-))))) = color(darkgreen)(ul(color(black)(3.0 * 10^(23)color(white)(.)"OH"^(-)"anions"

to completely neutralize $\text{1.0 L}$ of $\text{0.50-M}$ hydrochloric acid solution.

The answer is rounded to two sig figs.