Well I recently did a titration practical and now I\'m meant to dothe calculatio

Question

Well I recently did a titration practical and now I'm meant to dothe calculations now for the original concentration but I don'tknow how.

1. Diluted 13.75mL of vinegar with 87.5mL of water to make a 100mLsample

2. Put that diluted sample into burette with the 0.108M NaOHunderneath (yes, I know that was meant to be the other way roundbut it's easier for me)

3. It took about 25.8mL of the dilute to reach equilibrium with the20mL 0.108M NaOH sample.

With these pieces of information, how would I find out the originalconcentration of the vinegar?

Thanks in advance.

Explanation / Answer

So, the amount of dilute you used contained the following number ofmoles of acetic acid: 20 mL x 1 L/1000ml x .108 mol/L = .00216 mol So you know that 25.8 mL of the dilute had .00216 mol .00216mol/.0258L = .0837 mol/L Now you have to figure out how much was in the original vinegar,based on how much you diluted it. Since your sample was 13.75% vinegar, that means that the .0837mol/L all comes from that 13.75% So the amount of acetic acid in the vinegar would be ..0837/.1375 =.608. Seems low. Next time, you should probably follow theinstruction ;-)

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