# What volume of 2.50 M "HCl" in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M "Na"_2"CO"_3?

## The balanced equation is: $2 {\text{HCl"_((aq)) + "Na"_2"CO"_(3(aq)) -> 2"NaCl"_((aq)) + "H"_2"O"_((l)) + "CO}}_{2 \left(g\right)}$ I'm not sure I'm doing the calculations right.

Apr 12, 2016

$\text{0.0600 L}$

#### Explanation:

The balanced chemical equation tells you that you need $\textcolor{red}{2}$ moles of hydrochloric acid for every mole of sodium carbonate, ${\text{Na"_2"CO}}_{3}$, in order to have a complete neutralization.

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

If you take into account the fact that sodium carbonate dissociates completely in aqueous solution to form sodium cations, ${\text{Na}}^{+}$, and carbonate anions, ${\text{CO}}_{3}^{2 -}$, and that hydrochloric acid, $\text{HCl}$, is a strong acid that ionizes completely in a $1 : 1$ mole ratio to form hydronium cations, ${\text{H"_3"O}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$, you can rewrite the balanced chemical equation as

${\text{CO"_ (3(aq))^(2-) + color(red)(2)"H"_ 3"O"_ ((aq))^(+) -> overbrace(["H"_ 2"O"_ ((l)) + "CO"_ (2(aq))])^(color(blue)("H"_ 2"CO"_ (3(aq)))) + 2"H"_ 2"O}}_{\left(l\right)}$

This is the net ionic equation for this reaction. The sodium cations and the chloride anions are spectator ions, which is why I didn't include them here.

It's worth noting that the reaction produces carbonic acid, ${\text{H"_2"CO}}_{3}$, which exists in equilibrium with water and aqueous carbon dioxide.

So, use the molarity and volume of the sodium carbonate solution to find how many moles of carbonate anions are present

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you'll have - keep in mind that sodium carbonate dissociates in a $1 : 1$ mole ratio to produce carbonate anions

${n}_{C {O}_{3}^{2 -}} = {\text{0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 0.750color(red)(cancel(color(black)("L"))) = "0.0750 moles CO}}_{3}^{2 -}$

According to the aforementioned $1 : \textcolor{red}{2}$ mole ratio, a complete neutralization would require

0.0750 color(red)(cancel(color(black)("moles CO"_3^(2-)))) * (color(red)(2)color(white)(a)"moles H"_3"O"^(+))/(1color(red)(cancel(color(black)("mole CO"_3^(2-))))) = "0.150 moles H"_3"O"^(+)

Since you know the molarity of the hydrochloric acid solution, you can calculate what volume would contain this many moles by

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies V_"solution" = n_"solute}} / c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will thus have

V_(H_3O^(+)) = (0.150 color(red)(cancel(color(black)("moles"))))/(2.50 color(red)(cancel(color(black)("mol"))) "L"^(-1)) = color(green)(|bar(ul(color(white)(a/a)"0.0600 L"color(white)(a/a)|)))

The answer is rounded to three sig figs.