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Topic: URGENT: Chemistry Back-titration lab Help Please!  (Read 3809 times)

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Offline PopindaChopz98

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URGENT: Chemistry Back-titration lab Help Please!
« on: May 31, 2017, 11:30:17 PM »
Hey there! I need some help ASAP on a lab assignment on a back-titration of an unknown carbonate. I will provide the data obtained:

Mass of unknown carbonate samples:

Sample                               1                2                3
Mass of vial + unknown   14.7803 g   14.6020 g   14.4273 g
Mass of vial                   14.6020 g   14.4273 g   14.2418g
Mass of sample transferred   0.1783 g   0.1747 g   0.1855 g

Back-titrations with NaOH (50.00 mL standardized HCl added)

Sample                                     1        2             3
Final burette reading (mL)           24.61     44.71   43.72
Initial burette reading (mL)           4.91     24.61   28.04
Net volume of NaOH added (mL)   19.70    20.10   15.68

I calculated the concentration of NaOH to be 0.1212 mol/L and the mol of HCl in 50.00 mL to be 0.004908 mol. Also, I calculated the moles of NaOH needed to titrate the excess HCl in the flask as 0.002388 mol. I'm stuck with the following, please calculate (just for sample 1 only):

1. The moles of HCl remaining in flask 1 after the reaction with the unknown carbonate
2. The moles of HCl used up by the reaction with the unknown carbonate
3. The moles of carbonate ion that reacted.
4. The mass of carbonate ion that reacted.
5. The % Carbonate (by mass) in the unknown sample.

« Last Edit: June 01, 2017, 09:28:02 AM by sjb »

Offline Borek

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Re: URGENT: Chemistry Back-titration lab Help Please!
« Reply #1 on: June 01, 2017, 03:42:40 AM »
I calculated the moles of NaOH needed to titrate the excess HCl in the flask as 0.002388 mol

Quote
1. The moles of HCl remaining in flask 1 after the reaction with the unknown carbonate

Come on. How many moles of HCl will react with 2.388 mmoles of NaOH?
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Offline PopindaChopz98

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Re: URGENT: Chemistry Back-titration lab Help Please!
« Reply #2 on: June 01, 2017, 08:55:43 AM »
I calculated the moles of NaOH needed to titrate the excess HCl in the flask as 0.002388 mol

Quote
1. The moles of HCl remaining in flask 1 after the reaction with the unknown carbonate

Come on. How many moles of HCl will react with 2.388 mmoles of NaOH?

Yes I figured this one out now I know it's dumb  ;D still the others though?

Offline sjb

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Re: URGENT: Chemistry Back-titration lab Help Please!
« Reply #3 on: June 01, 2017, 09:20:01 AM »
2) How many moles of acid did you add (note, not the answer to the question)

3) From your answer to 2) - how many moles of carbonate react with that number of moles of acid.

4) What is the mass of that amount of carbonate? (note, not mass of salt)

5) Compare the mass of carbonate to that of the salt

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