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Workshop 5: Titration Curves A pH or titration curve can be produced by plotting

ID: 490267 • Letter: W

Question

Workshop 5: Titration Curves A pH or titration curve can be produced by plotting the pH of a solution (containing an analyte) versus the volume of added titrant. The equivalence or end point of a titration occurs when enough titrant has been added to consume all the acid or base initially present in the analyte. You know you have reached the endpoint when you see a significant change in the pH ofthe solution. Part1 Let's begin by analyzing a titration curve When your analyte is a weak acid or base the titration curve shows the effects of buffering before the equivalence point. The midpoint of the titration occurs in the buffer zone and is the point at which the pH is equal to the pK ofthe analyte. In the case of a weak acid-strong base titration the pH at the equivalence point will always be higher than 7. This is because of the basicproperties of the anion (the conjugate base ofthe weak acid) that is present in the solution at this point. To the right is a titration curve for the Titration of unknown acid titration of 25.0 mL of unknown compound X with 0.200 M NaoH. Use this 14 titration curve to answer the following 12 questions. (a) What is the identity of compound X? pH HINT: Use the Chempendix on Sapling to look at the Ka values for possible weak acids. 10 20 30 40 50 volume of 0.200 M NaOH added, mL tration of compound X? (b) What s the starting concen

Explanation / Answer

At half equivalence point pH = pka

Volume to reach equivalence point ~32 mL

volume to reach half neutralization point = 16 mL

pH at this point is 4.2

This is most probably Benzoic acid.

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Assum that the acid is represented by HA.

when V= 0.00 mL

HA <==> H+ + A-

Major species present = HA i.e undissocoated acid

pH active species = A-

pH = 2

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when V = 20mL

Major species present = HA + A-

pH = 4.5

[ph active species] = 10^-pH = 3.16*10^-5 M

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when V= 30.75mL

major species : A-

pH = 6

[pH active species] = 10^-6 M

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when V = 40mL

major species : A- and OH-

pH = 13

[pH active species] = 10^-13 M

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