Question #13edc

1 Answer
Jun 7, 2017

Here's how you can do that.


The idea here is that the sodium carbonate will react with the hydrochloric acid to produce sodium chloride and carbonic acid, which decomposes to produce water and carbon dioxide.

#"Na"_ 2"CO"_ (3(aq)) + color(red)(2)"HCl"_ ((aq)) -> 2"NaCl"_ ((aq)) + "CO"_ (2(g)) + "H"_ 2"O"_ ((l))#

On the other hand, the sodium chloride present in the mixture will not react with the hydrochloric acid.

So, start by calculating the number fo moles of hydrochloric acid needed to titrate the solution.

#25 color(red)(cancel(color(black)("cm"^3))) * "0.1 moles HCl"/(10^3color(red)(cancel(color(black)("cm"^3)))) = "0.0025 moles HCl"#

Notice that the reaction consumes #color(red)(2)# moles of hydrochloric acid for every #1# mole of sodium carbonate that takes part in the reaction. This tells you that the solution contained

#0.0025 color(red)(cancel(color(black)("moles HCl"))) * ("1 mole Na"_2"CO"_3)/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = "0.00125 moles Na"_2"CO"_3#

Use the molar mass of sodium carbonate to convert this to moles

#0.00125 color(red)(cancel(color(black)("moles Na"_2"CO"_3))) * "105.99 g"/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3)))) = "0.1325 g"#

Now, you know that you have #"0.1325 g"# of sodium carbonate in #"25 cm"^3# of solution. This implies that #"1000 cm"^3# of solution will contain

#1000 color(red)(cancel(color(black)("cm"^3color(white)(.)"solution"))) * ("0.1325 g Na"_2"CO"_3)/(25 color(red)(cancel(color(black)("cm"^3color(white)(.)"solution")))) = "5.30 g Na"_2"CO"_3#

Since the total mass of the salt mixture is equal to #"20 g"# in #"1000 cm"^3# of solution, the percent purity will be

#"purity" = (5.3 color(red)(cancel(color(black)("g"))))/(20color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)("27% Na"_2"CO"_3)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the concentration of the acid, the mass of the mixture, and the volume of the stock mixture solution.