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"HCl"_((aq)) + "Na"_2"CO"_(3(aq)) -> 2"NaCl"_((aq)) + "H"_2"O"_((l)) + "CO"_(2(g))#
I'm not sure I'm doing the calculations right.
The balanced equation is:
I'm not sure I'm doing the calculations right.
1 Answer
Explanation:
The balanced chemical equation tells you that you need
#"Na"_ 2"CO"_ (3(aq)) + color(red)(2)"HCl"_ ((aq)) -> 2"NaCl"_ ((aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(aq))#
If you take into account the fact that sodium carbonate dissociates completely in aqueous solution to form sodium cations,
#"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"_((l))#
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,
So, use the molarity and volume of the sodium carbonate solution to find how many moles of carbonate anions are present
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#
In your case, you'll have - keep in mind that sodium carbonate dissociates in a
#n_(CO_3^(2-)) = "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
#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
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(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.