# If you mix NaOH and HCl, both having 45.0mL of volume, which of the two is the limiting reactant?

Oct 23, 2015

You can't tell without knowing the molarities of the two solutions.

#### Explanation:

Sodium hydroxide, $\text{NaOH}$, a strong base, will react with hydrochloric acid, $\text{HCl}$, a strong acid, to form sodium chloride, $\text{NaCl}$, and water.

The balanced chemical equation for this neutralization reaction looks like this

${\text{NaOH"_text((aq]) + "HCl"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you have a $1 : 1$ mole ratio between sodium hydroxide and hydrochloric acid. This means that the reaction consumes equal numbers of moles of each reactant.

In order to be able to determine which of the two reactants is a limiting reagent, you need to know how many moles of each you have.

When mixing two solutions of equal volumes, like you have here, the mole ratio becomes equivalent to a concentration ratio.

You know that a solution's molarity is defined as

$C = \frac{n}{V} \text{ }$, where

$n$ - the number of moles of solute;
$V$ - the volume of the solution.

In your case, you would have

${C}_{\text{HCl" = n_"HCl"/V " }}$ and $\frac{\text{ "C_"NaOH" = n_"NaOH}}{V}$

This is equivalent to

$V = {n}_{\text{HCl"/C_"HCl" implies n_"NaOH"/C_"NaOH" = n_"HCl"/C_"HCl}}$

The mole ratio between sodium hydroxide nd hydrochloric acid is

${n}_{\text{NaOH"/n_"HCl" = C_"NaOH"/C_"HCl}}$

Since equal volumes of equal concentration solutions will have equal numbers of moles of each reactant, it follows that the solution that has the lower concentration will act as a limiting reagent.

So remember, when you add equal volumes of each reactant, their respective molarities will determine which one acts as a limiting reagent and which one is in excess.