A shipment of crude sodium carbonate must be assayed for its #"Na"_2"CO"_3# content. [ . . . ] A #"10.00-mL"# sample is titrated to the equivalence point with #"16.90 mL"# of a #"0.1022 M HCl"# solution. What is the #%# #"Na"_2"CO"_3# in the sample?

The acid-base equivalence point is when equal mols of #"H"^(+)# and #"OH"^(-)# have reacted. Here, the idea is:

1. The #"mols"# of #"HCl"# that reacted in the #"16.90 mL"# titrant volume are related to the #"mols"# of #"OH"^(-)# that came from the #"Na"_2"CO"_3# in the #"10.00 mL"# aliquot OF the crude sample that actually reacted.
2. The #"10.00 mL"# aliquot of #"Na"_2"CO"_3(aq)# contains a fraction of the original mass contained in the starting #"1.000 L"# total volume.
3. The #"9.709 g"# was in that total volume of #"1.000 L"#.

So first, we find the #"mols"# of #"HCl"#:

#"0.1022 mol HCl"/cancel"L" xx 0.01690 cancel"L" = "0.001727 mols HCl"#

The #"Na"_2"CO"_3# in water reacts as follows:

#"CO"_3^(2-)(aq) + cancel(2)"H"_2"O"(l) -> cancel("H"_2"CO"_3(aq)) + 2"OH"^(-)(aq)#

#ul(cancel("H"_2"CO"_3(aq)) -> cancel("H"_2"O"(l)) + "CO"_2(g))#

#"CO"_3^(2-)(aq) + "H"_2"O"(l) -> "CO"_2(g) + 2"OH"^(-)(aq)#

As a result, we have #2"OH"^(-)# for every #1"Na"_2"CO"_3(aq)# and #1"H"^(+)# for every #1"HCl"#. The #"mols"# of #"Na"_2"CO"_3# reacted is then found like so:

#0.001727 cancel"mols HCl" xx "1 mol H"^(+)/cancel"1 mol HCl" = "0.001727 mols H"^(+)#

At the equivalence point, this reacts exactly, with #"0.001727 mols OH"^(-)#, so it reacts with

#0.001727 cancel("mols OH"^(-)) xx ("1 mol Na"_2"CO"_3)/(2 cancel("mols OH"^(-))) = "0.0008636 mols Na"_2"CO"_3#

found in #"10.00 mL"# of the solution aliquot.

Mass, mols, and volume are extensive, so:

#("0.0008636 mols Na"_2"CO"_3)/("10.00 mL") = ("0.08636 mols Na"_2"CO"_3)/("1000. mL")#

Therefore, in the original flask, there was

#("0.08636 mols Na"_2"CO"_3)/"1.000 L"#.

As a result, the actual mass of #"Na"_2"CO"_3(s)# contained within the crude #"9.709 g"# sample that was in the #"1.000 L"# flask is:

#0.08636 cancel("mols Na"_2"CO"_3) xx ("105.986 g Na"_2"CO"_3)/cancel("1 mol Na"_2"CO"_3)#

#=# #color(green)("9.153 g Na"_2"CO"_3(s))#

Finally, the percent by mass is:

#("9.153 g Na"_2"CO"_3(s))/("9.709 g sample") xx 100% = color(blue)(94.27% "w/w")#

Step 1. Calculate the amount of #"HCl"# used

#"Moles of HCl" = 16.90 color(red)(cancel(color(black)("mL HCl"))) × "0.1022 mmol HCl"/(1 color(red)(cancel(color(black)("mL HCl")))) = "1.727 mmol HCl"#

Step 2. Calculate the amount of #"Na"_2"CO"_3# in the aliquot

The equation for the reaction is

#"Na"_2"CO"_3 + "2HCl" → "2NaCl + CO"_2 + "H"_2"O"#

#"Moles of Na"_2"CO"_3 = 1.727 color(red)(cancel(color(black)("mmol HCl"))) × ("1 mmol Na"_2"CO"_3)/(2 color(red)(cancel(color(black)("mmol HCl")))) = "0.8636 mmol Na"_2"CO"_3#

Step 3. Calculate the mass of #"Na"_2"CO"_3# in the aliquot

#"Mass of Na"_2"CO"_3 = 0.8636 color(red)(cancel(color(black)("mmol Na"_2"CO"_3))) ×("105.99 mg Na"_2"CO"_3)/(1 color(red)(cancel(color(black)("mmol Na"_2"CO"_3)))) = "91.53 mg Na"_2"CO"_3#

Step 4. Calculate the mass of #"Na"_2"CO"_3# in the original solution

#"Mass of Na"_2"CO"_3 = 1000 color(red)(cancel(color(black)("mL solution"))) × ("91.53 mg Na"_2"CO"_3)/(10.00 color(red)(cancel(color(black)("mL solution")))) = "9153 mg Na"_2"CO"_3 = "9.153 g Na"_2"CO"_3#

Step 5. Calculate the mass percent of #"Na"_2"CO"_3#

#"Mass %" = "Mass of component"/"Mass of sample" × 100 % = (9.153 color(red)(cancel(color(black)("g"))))/(9.709 color(red)(cancel(color(black)("g")))) × 100 % = 94.28 %#