# 50.0 ml of 0.10 M NaOH completely neutralizes 30.0 ml of hydrochloric acid. How would you calculate the concentration of HCl?

Oct 31, 2015

$\text{0.17 M}$

#### Explanation:

Start by making sure that you have a clear understanding of what a complete neutralization actually means.

In this case, you're dealing with sodium hydroxide, $\text{NaOH}$, a strong base, and hydrochloric acid, $\text{HCl}$, a strong acid. Both these compounds will completely dissociate in aqueous solution to form

${\text{NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

and

${\text{HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

When these two solutions are mixed, the hydroxide ions, ${\text{OH}}^{-}$, that can be found in the sodium hydroxide solution will react with the hydronium ions, ${\text{H"_3"O}}^{+}$, to form water.

This is what a neutralization reaction is all about.

The balanced chemical equation for the reaction looks like this - I'll show you the net ionic equation

${\text{OH"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) -> 2"H"_2"O}}_{\textrm{\left(l\right]}}$

This tells you that in order for the neutalization to be complete, you need equal numbers of moles of hydroxide and hydronium ions.

Use the volume and molarity of the sodium hydroxide solution to find how many moles of hydroxide you have

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

$n = {\text{0.10 M" * 50.0 * 10^(-3)"L" = "0.0050 moles OH}}^{-}$

This means that the hydrochloic acid solution you used must provide $0.0050$ moles of hydrochloric acid to the mix. This means that its molarity will be

$c = \text{0.0050 moles"/(30.0 * 10^(-3)"L") = "0.1667 M}$

Rounded to two sig figs, the answer will be

c_"HCl" = color(green)("0.17 M")