# What volume of 0.1292 M NaOH is needed to neutralize 25.00 mL of HCl of unknown concentration?

## This is a titration curve question. I know that the initial pH of the hydrochloric acid is 1.65. I know that you can find the concentration of HCl by plugging it into the pH formula. So, $- \log \left[H +\right] = 1.65$ ${10}^{-} 1.65 = 0.02239 = \left[H +\right]$. However, that's as far as I've gotten. I know the volume of the 0.1292 M NaOH is between 22 and 23 mL, but I can't seem to figure out how to find the exact value. Any help is greatly appreciated!

Mar 31, 2018

Here's what I got.

#### Explanation:

Your first step is correct because the first thing that you need to do here is to use the $\text{pH}$ of the hydrochloric acid solution to find the concentration of the acid.

As you know, hydrochloric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations, ${\text{H"_3"O}}^{+}$.

This means that the hydrochloric acid solution has

$\left[{\text{HCl"] = ["H"_3"O}}^{+}\right]$

and since

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

you can say that

["HCl"] = 10^(-1.65) quad "M"

Now, sodium hydroxide and hydrochloric acid neutralize each other in a $1 : 1$ mole ratio.

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

Since sodium hydroxide is a strong base, i.e. it dissociates completely in aqueous solution to produce hydroxide anions, the neutralization reaction can be described using the net ionic equation

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

This tells you that in order to neutralize $1$ mole of hydrochloric acid, you need $1$ mole of sodium hydroxide. To find the number of moles of hydrochloric acid present in the hydrochloric acid solution, use the molarity and the volume of the sample.

25.00 color(red)(cancel(color(black)("mL solution"))) * (10^(-1.65) quad "moles HCl")/(10^3color(red)(cancel(color(black)("mL solution")))) = 2.500 * 10^(-3.65) quad "moles HCl"

This means that the sodium hydroxide solution must contain $2.500 \cdot {10}^{- 3.65}$ moles of sodium hydroxide in order for the neutralization to be complete. Once again, use the molarity of the sodium hydroxide solution to figure out the volume of the sample.

2.500 * 10^(-3.65) color(red)(cancel(color(black)("moles NaOH"))) * (10^3 quad "mL solution")/(0.1292 color(red)(cancel(color(black)("moles NaOH")))) = 1.935 * 10^(0.35) quad "mL"

Rounded to four sig figs, the answer would be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{volume NaOH solution = 4.332 mL}}}}$

Based on the values that you have provided here, the volume of the sodium hydroxide cannot be between $22$ and $\text{23 mL}$ because the sodium hydroxide solution is

color(white)( (color(black)("the more concentrated solution " ->)/(color(black)("the less concentrated solution " ->)))) (0.1292 color(red)(cancel(color(black)("M"))))/(0.022387color(red)(cancel(color(black)("M")))) = color(blue)(5.771)

times more concentrated than the hydrochloric acid solution. Since the two reactants neutralize each other in a $1 : 1$ mole ratio, it follows that the volume of the less concentrated solution must be $\textcolor{b l u e}{5.771}$ times higher than the volume of the more concentrated solution in order for the two solutions to contain the same number of moles of each reactant.

In this case, you have

color(white)( (color(black)("the less concentrated solution " ->)/(color(black)("the more concentrated solution " ->)))) (25.00 color(red)(cancel(color(black)("mL"))))/(4.332color(red)(cancel(color(black)("mL")))) = color(blue)(5.771)

which confirms that the volume of the sodium hydroxide solution is indeed equal to $\text{4.332 mL}$.