# Question ce208

##### 1 Answer
May 29, 2016

Here's why that is the case.

#### Explanation:

You're dealing with two neutralization reactions that feature sodium hydroxide, $\text{NaOH}$, a strong base.

The first neutralization reaction has sodium hydroxide reacting with nitric acid, ${\text{HNO}}_{3}$, a strong acid, while the second reaction has sodium hydroxide reacting with hydrofluoric acid, $\text{HF}$, a weak acid.

It's important to remember here that for a strong base, it doesn't matter if the acid is strong or weak, the reaction proceeds the same way.

When sodium hydroxide reacts with nitric acid, the balanced molecular equation looks like this

${\text{NaOH"_ ((aq)) + "HNO"_ (3(aq)) -> "NaNO"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}$

Notice that $1$ mole of sodium hydroxide reacts with $1$ mole of nitric acid.

This means that if the two solutions have the same molarity, you will need equal volumes in order to make sure that you deliver equal numbers of moles to the reaction.

The same is true for the reaction between sodium hydroxide and hydrofluoric acid, for which the balanced molecular equation looks like this

${\text{NaOH"_ ((aq)) + "HF"_ ((aq)) -> "NaF"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

Once again, the $1 : 1$ mole ratio that exists between the two reactants tells you that if you have two solutions of the same molarity, you will once again need equal volumes to ensure that you have a complete neutralization.

Now, it doesn't matter if the acid is strong or weak because the hydroxide anion, ${\text{OH}}^{-}$, coming from the complete dissociation of the base will react with the hydronium cations, ${\text{H"_3"O}}^{+}$, coming from the dissociation of the acid.

If the acid is strong, is dissociates completely and you have

overbrace("OH"_ ((aq))^(-))^(color(purple)("coming from NaOH")) + overbrace("H"_ 3"O"_ ((aq))^(+))^(color(blue)("coming from HNO"_ 3)) -> 2"H"_ 2"O"_((l))#

If the acid is weak, it only partially dissociates

${\text{HF"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "F}}_{\left(a q\right)}^{-}$

However, the hydroxide anions will still react with the hydronium cations, which will cause the equilibrium to shift to the right -- think Le Chatelier's Principle here.

More of the weak acid will dissociate, the resulting hydronium cations being once again consumed by the reaction with the hydroxide anions.

This will go on until all the moles of the weak have dissociated, i.e. until the neutralization is complete.