Question

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?

A shipment of crude sodium carbonate must be assayed for its `"Na"_2"CO"_3` content. You receive a small jar containing a sample from the shipment and weigh out `9.709` grams into a flask, where it is dissolved in water and diluted to `"1.000 L"` with water. 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 answer is `94.28%`. However using the formula VMn (of acid) = VMn (of base), then I get Molarity of Base = (16.9 * 0.1022 * 1) / (10 * 2). And from there on However, why do I multiply by 2 in the denominator?

Answer 1

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")`

Answer 2

The sample is 94.28 % sodium carbonate.

Explanation:

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 %`