if you calculate your first and second derivative and find your endpoint to be a

Question

if you calculate your first and second derivative and find your endpoint to be at 16.16mL of 0.1052M NaOH, what is the molecular weight of your unknown acid if you weighed out 0.5102g of an unknown to start?

o Assume you have a diprotic acid where 16.16mL is the 2nd equivalence point for this example and take this into account for your molar ratio.
o don't forget the dilution ratio for the original volume you dissolved your solid in.

o what is th pka2 value for this unknown, given the information above and the following pH data?

vol, pH
4.04, 3.5
8.08, 4.3
12.112, 5.2
15.12, 5.5
16.16, 8.2
17.53, 10.1

Explanation / Answer

If the end point is the same of the second equivalence point, then, the volume for the first equivalence point would be:

Ve1 = Ve/2

Ve = 16.16 / 2 = 8.08 mL

This volume has a pH of 4.3 so:

[H+] = 10-4.3 = 5.01x10-5 M

The Pka2 is the mid point between the first equivalence point and the second one. This has a volume of 12.112 mL and a pH = 5.2

At this point, pH = pKa, so the pKa2 = 5.2

Now, with the H+ concentration we know the following:

H2A + OH- -------> HA- + H2O

And we know that [H+] = [H2A] - [OH]. this is because at firts equivalence point, we still have excess of acid:

[H2A] = 5.01x10-5 + 0.1052 M = 0.1053 M

and at the first equivalce point:

MaVa = MbVb

But it's a diprotic acid so:

Va = 2MbVb / Ma

Va = 2 * 0.1052 * 8.08 / 0.1053

Va = 16.12 mL

moles of A = 0.1053*0.01612 = 1.7x10-3 moles

MM = 0.5102 / 1.7x10-3 = 300 g/mol

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