# Question #93be2

Jun 4, 2015

The exact same volume, 5 mL.

Notice how the balanced chemical equation for this neutralization reaction looks like

$H C {l}_{\left(a q\right)} + N a O {H}_{\left(a q\right)} \to N a C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

The $1 : 1$ mole ratio that exists between hydrochloric acid and sodium hydroxide tells you that, in order for a complete neutralization to take place, you need equal numbers of moles of each compound.

Since your solutions have the same strength, you'll get the same number of moles per unit of volume for both the acid, and the base.

A solution's strength is expressed by its concentration, i.e. the number of moles of the solute you get per unit of volume of solution.

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

$C$ - the molarity of the solution;
$n$ - the numbe of moles of solute;
$V$ - the volume of the solution.

If you have equal concentrations for both solutions, you can write

$C = {n}_{H C l} / {V}_{H C l}$ and $C = {n}_{N a O H} / {V}_{N a O H}$, or

${n}_{H C l} / {V}_{H C l} = {n}_{N a O H} / {V}_{N a O H}$

Moreover, if you need equal numbers of moles of each compound, you'd get

$\frac{\cancel{n}}{V} _ \left(H C l\right) = \frac{\cancel{n}}{V} _ \left(N a O H\right) \implies {V}_{H C l} = {V}_{N a O H}$