# Question 239b8

Aug 20, 2015

You need to know the mass of sodium carbonate used to make the 250-mL sample.

#### Explanation:

The key to this problem is the ratio between the number of moles of sodium carbonate and the number of moles of hydrochloric acid your titration requires.

Now, the problem with your question is that I think it's incomplete. You need to know the mass of solid sodium carbonate dissolved to make the 250-mL sample in order to be able tocalculate the number of moles of sodium carbonate.

I'll assume that you dissolve $x$ grams of sodium carbonate to make the 250-mL solution.

The number of moles of sodium carbonate present in this sample would be

xcolor(red)(cancel(color(black)("g"))) * ("1 mole Na"""_2"CO"_3)/(106.0color(red)(cancel(color(black)("g")))) = x/106"moles"

You get this many moles of sodium carbonate in the 250-mL sample, which means that the aliquot will contain

20color(red)(cancel(color(black)("mL"))) * (x/106"moles")/(250color(red)(cancel(color(black)("mL")))) = ((0.08 * x)/106)"moles Na"""_2"CO"_3

The balanced chemical equation for the reation between hydrochloric acid and sodium carbonate looks like this

$\textcolor{red}{2} {\text{HCl"_text((aq]) + "Na"_2"CO"_text(3(aq]) -> 2"NaCl"_text((aq]) + "CO"_text(2(g]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between hydrochloric acid and sodium carbonate. This means that you need to use twice as many moles of the former than you have moles of the latter.

This means that your titration required

((0.08x)/106)color(red)(cancel(color(black)("moles Na"""_2"CO"_3))) * (color(red)(2)" moles HCl")/(1color(red)(cancel(color(black)("mole Na"""_2"CO"_3)))) = ((0.16 * x)/106)"moles HCl"

Now use the average titration volume you found for the hydrochloric acid solution to determine its concentration

$C = \frac{n}{V}$

["HCl"] = (((0.16 * x)/106)"moles")/(20.24 * 10^(-3)"L") = (0.0746 * x)" M"#

To get the actual value, replace $x$ with the mass of sodium carbonate you used to make the 250-mL sample.