# Question #00919

Mar 18, 2017

$\text{0.46 M}$

#### Explanation:

Sodium hydroxide and hydrochloric acid react in a $1 : 1$ mole ratio to produce aqueous sodium chloride and water

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

This means that a complete neutralization will always consume equal numbers of moles of sodium hydroxide and of hydrochloric acid.

Now, a solution's molarity tells you the number of moles of solute present in exactly $\text{1 L}$ of that solution.

You know that the sodium hydroxide solution has a molarity of $\text{0.20 M}$. You can say that if you were to add equal volumes of the two solutions, then the hydrochloric acid solution would also have a molarity equal to $\text{0.20 M}$.

In other words, if you were to mix $\text{54.0 mL}$ of $\text{0.20 M}$ sodium hydroxide solution with $\text{54.0 mL}$ of hydrochloric acid solution and get a complete neutralization as a result, then the molarity of the hydrochloric acid solution would be equal to $\text{0.20 M}$.

However, you know that it takes $\text{23.5 mL}$ of hydrochloric acid solution to get a complete neutralization.

This tells you that the hydrochloric acid solution is more concentrated than the sodium hydroxide solution because it contains the same number of moles of acid in a smaller volume.

In fact, the ratio that exists between the volumes of the sodium hydroxide solution and the volume of the hydrochloric acid solution

$\left(54.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL"))))/(23.5color(red)(cancel(color(black)("mL}}}}\right) = \textcolor{b l u e}{2.298}$

must also exist between the molarity of the hydrochloric acid solution and the molarity of the sodium hydroxide solution.

${c}_{\text{HCl"/c_"NaOH}} = \textcolor{b l u e}{2.298}$

Therefore,

${c}_{\text{HCl" = color(blue)(2.298) * c_"NaOH}}$

In your case, you will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{c}_{\text{HCl" = 2.298 * "0.20 M" = "0.46 M}}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the sodium hydroxide solution.