Question

Assuming you weighed out a 2.3684 g sample of your unknown, and dissolved and diluted it as in the procedure below, and it took 35.63 mL of a 0.1025 M HCl titrant to reach the endpoint, what are the weight percents of Na₂CO₃ and NaHCO₃ in your unknown sample?

Hints:

-Set g Na₂CO₃ = X

-Eqn A: bicarbonate + carbonate = diluted weighted mass (remember, the mass is not 2.3684, you diluted it before you titrated it.Calculate the diluted g of unknown)

-Convert the g of Na₂CO₃ to mol of HCl

-Convert the g of NaHCO₃ to mol of HCl

-mol HCl = mol HCl from Na₂CO₃ + mol HCl from NaHCO₃

-Before taking the final weight percent, undo the dilution to get back to the original wt % in the unknown

I've never done a question like this before and it is for a pre-lab.

Here's the Procedure:

Boil ~700 mL of distilled water to expel CO₂ and pour the water into a 1000-mL plastic bottle. Screw the cap on tightly and allow the water to cool to room temperature in a water/ice bath. Keep tightly capped when not in use.

Share bottles between 2 people, each person needs ~350 mL of CO₂-free DI water

CO₂ reacts with Na₂CO₃ to form NaHCO₃. If you do not use CO₂-free water, your ratio of carbonate to bicarbonate will be off since some of the carbonate will react with the CO₂ and water and be converted to bicarbonate.

Accurately weigh 2.0–2.5 g of you assigned unknown into a 250-mL volumetric flask by weighing the sample in a weigh boat, delivering some to a funnel in the volumetric flask, and reweighing the boat. Continue this process until the desired mass of reagent has been transferred to the funnel.

This procedure is called “weighing by difference”, and is used to transfer an exact known amount of solid.

Rinse the funnel repeatedly with small portions of room-temperature CO₂-free water to dissolve the sample. Remove the funnel, dilute to the mark with CO₂-free DI H₂O, and mix well.

Pipet a 25.00 mL aliquot of unknown solution into a 250 mL Erlenmeyer flask and titrate with the standardized HCl, using bromocresol green indicator (as in Experiment 7, for standardizing HCl).

This will be your rapid test titration to find the approximate endpoint

Repeat this procedure with at least four more 25.00 mL aliquots of unknown.

You need to end up with at least 3 replicates with good precision

Answer 1

Moles of HCl reacted = 0.1025 * 0.03563 = 3.65*10-3 moles

Assume 'x' g of Na2CO3 is present and 'y' g of NaHCO3 in the 25mL aliquot.

So,

moles of Na2CO3 = x/106

moles of NaHCO3 = y/84

1 mole Na2CO3 requires 2 moles HCl, and 1 mole NaHCO3 requires 1 moles HCl

So,

2*(x/106) + 1*(y/84) = 3.65*10^-3

Also, since original volume was 250 mL out of which this aliquot was taken, so actual masses are 10*x and 10*y.

Thus,

10*x + 10*y = 2.3684

Solve the above two equations to get the values of x and y.