# If 11.08 mL of 0.1111 M NaOH were required to neutralize 25.00 mL of a monoprotic acid, what was the concentration of the acid?

Mar 18, 2018

${\text{0.04924 mol L}}^{- 1}$

#### Explanation:

You know that you're dealing with a monoprotic acid here, so you can represent this unknown acid as $\text{HA}$ and write the balanced chemical equation that describes the neutralization reaction like this

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

So $1$ mole of sodium hydroxide will neutralize $1$ mole of the acid.

Now, you know that you can express the number of moles of solute present in a solution in terms of the volume of the solution and of its molar concentration.

$n = c \cdot V$

Here

• $n$ is the number of moles of solute
• $c$ is the molarity of the solution
• $V$ is the volume of the solution, expressed in liters

So, you know that this neutralization reaction required

$\text{11.08 mL" = 11.08 * 10^(-3) quad "L}$

of a

${\text{0.1111 M" = "0.1111 mol L}}^{- 1}$

sodium hydroxide solution, which means that you have

n_ ("NaOH") = "0.1111 mol" color(red)(cancel(color(black)("L"^(-1)))) * 11.08 * 10^(-3) color(red)(cancel(color(black)("L")))

n_ ("NaOH") = 1.231 * 10^(-3) quad "moles"

You also now that this number of moles of sodium hydroxide neutralized $\text{25.00 mL}$ of a solution of the unknown acid.

Moreover, you know that the number of moles of sodium hydroxide consumed in the reaction must be equal to the number of moles of acid consumed in the reaction.

This means that you have

n_ ("HA") = 1.231 * 10^(-3) quad "moles"

and--don't forget that the volume must be expressed in liters!

$1.231 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{10}^{- 3}}}} \quad \text{moles" = c_ ("HA") * 25.00 * color(red)(cancel(color(black)(10^(-3)))) quad "L}$

The concentration of the acid solution is equal to

c_ "HA" = "1.231 moles"/"25.00 L" = color(darkgreen)(ul(color(black)("0.04924 mol L"^(-1))))

The answer is rounded to four sig figs.