# Question a034f

##### 1 Answer
May 4, 2017

Here's what I got.

#### Explanation:

The $\text{pH}$ of the nitric acid solution will tell you its molarity. As you know, nitric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations and nitrate anions.

${\text{HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

This means that a nitric acid solution has

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

The $\text{pH}$ of the solution is defined as

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

This means that you have

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

which, in your case, will get you

["H"_3"O"^(+)] = 10^(-0.29) = "0.513 M"

Now, notice that you have

${\text{Na"_ 2"CO"_ (3(aq)) + color(red)(2)"HNO"_ (3(aq)) -> "CO"_ (2(g)) uarr + "H"_ 2"O"_ ((l)) + 2"NaNO}}_{3 \left(a q\right)}$

The $1 : \textcolor{red}{2}$ mole ratio that exists between the sodium carbonate and the nitric acid tells you that the reaction consumes $\textcolor{red}{2}$ moles of nitric acid for every $1$ mole of sodium carbonate that takes part in the reaction.

In your case, the reaction consumed

25.4 color(red)(cancel(color(black)("cm"^3))) * ("0.10 moles Na"_2"CO"_3)/(10^3color(red)(cancel(color(black)("cm"^3)))) = "0.00254 moles Na"_2"CO"_3

This implies that it also consumed

0.00254 color(red)(cancel(color(black)("moles Na"_2"CO"_3))) * (color(red)(2)color(white)(.)"moles HNO"_3)/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3)))) = "0.00508 moles HNO"_3

If you know the concentration of the nitric acid solution

["HNO"_3] = "0.513 M"

you can find the volume used in the titration

0.00508 color(red)(cancel(color(black)("moles HNO"_3))) * (10^3color(white)(.)"cm"^3)/(0.513 color(red)(cancel(color(black)("moles HNO"_3)))) = "9.9 cm"^3#

Therefore, you can say that you titrated ${\text{25.4 cm}}^{3}$ of ${\text{0.10 mol dm}}^{- 3}$ sodium carbonate solution to the equivalence point by using ${\text{9.9 cm}}^{3}$ of $\text{0.513 M}$ nitric acid solution.