3. A student measured 20 drops of vinegar and one drop of phenolphthalein soluti

Question

3. A student measured 20 drops of vinegar and one drop of phenolphthalein solution into a well of her H solution to the same well, the well was full, but she transferred some of the well plate. After she had added 35 drops of NaO had still not reached the titration end point. She then took a clean pipet and titration mixture into a second well. She rinsed the pipet with a few drops of distilled water and transferred the rinse water into the titration mixture in the second well. Then she continued adding NaOH solution to both wells until she reached the end point in both wells. She then recorded the total number of drops of NaOH solution added to both wells and used it to calculate the data for he vinegar titration. Briefly comment on her procedure and the accuracy of her results. custom page 160

Explanation / Answer

Ans. At titration endpoint, the number of moles of acid is equal to that of the base.

Note that it is the number of moles of acid and base which is accounted, but NOT the concentration of these solution.

So, presence of water (say, traces of water after rinsing the cell plat wells with distilled water) does NOT affect the number of moles of acid or bases. However, it surely affect the concentration, but dilution is NOT our concern because it does not affect the number of moles of the solutes in solute.

# Theoretically, titration vinegar in one single well and in midway-partitioned two wells shall have no effect on the actual effect.

However, the accuracy MAY BE affected due to following reasons-

I. Loss of sample during transfer of solution: Loss of solution during transfer of some volume from original well to the second well may occur due to spillage, residual solution volume retained in the pipette, etc. would cause loss of moles of acetic acid. So, there is reduction in the overall number of moles of acetic acid.

So, endpoint may be reached a bit earlier than the actual. The calculated molarity of acetic acid in vinegar solution thus is also lower than the actual value.

II. Larger NaOH volume consumed at endpoint:

At the theoretical endpoint, there is JUST ONE EXTRA DROP of NaOH added. However, in this case, two extra drops – one to each well – would be required to reach the overall cumulative endpoint.

That is, the amount of NaOH consumed to reach the endpoint would be greater than the actual value.

Consumption of greater volume of NaOH would mean greater moles of acetic acid in vinegar solution. So, the calculated molarity of acetic acid in vinegar solution would be higher than the actual value.

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