# A 10.0 mL sample of a monoprotic acid is titrated with 45.5 mL of 0.200 M NaOH. What is the concentration of the acid?

Aug 6, 2016

$\text{0.910 M}$

#### Explanation:

A monoprotic acid can contribute one proton to the neutralization reaction that takes place when sodium hydroxide, $\text{NaOH}$, is added to the solution.

If you take $\text{HA}$ to be the general formula of a monoprotic acid, you can say that the balanced chemical equation that describes this neutralization reaction looks like this

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

Iin order to have a complete neutralization, you need to add equal numbers of acid and of base.

As you know, molarity is defined as the number of moles of solute present in one liter of solution. You already know the molarity of the sodium hydroxide solution.

Now, let's assume that the monoprotic acid solution has a molarity equal to $\text{0.200 M}$. In this case, the titration would require $\text{45.5 mL}$ of acid solution, since solutions of equal volumes and equal molarities contain the same number of moles of solute.

However, you know that the acid solution has a volume of $\text{10.0 mL}$. This volume is

$\left(45.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL"))))/(10.0 color(red)(cancel(color(black)("mL}}}}\right) = \textcolor{b l u e}{4.55}$

smaller than what you'd need if the acid solution had the same molarity as the sodium hydroxide solution. This can only mean that the acid solution is $\textcolor{b l u e}{4.55}$ times more concentrated than the sodium hydroxide solution.

You thus have

c_("HA") = color(blue)(4.55) * "0.200 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.910 M")color(white)(a/a)|)))

Therefore, you can say for a fact that $\text{10.0 mL}$ of $\text{0.910 M}$ monoprotic acid solution contains the same number of moles of solute as $\text{45.5 mL}$ of $\text{0.200 M}$ sodium hydroxide solution.