1 A 20.0 mL sample of an unknown acid was diluted to 250 mL in a volumetric flas

Question

1 A 20.0 mL sample of an unknown acid was diluted to 250 mL in a volumetric flask. A 25.0 mL aliquot of the diluted ascid was then titrated with 0.100 M NaOH and the following titration curve resulted. 14 points 15 (a) The acid is strong , weak (b) By inspection of the titration curve estimate 10 1. The pKa 2. The pH at the equivalence point 3. Volume of NaOH required to reach the equivalence point. 0 100 50 Volume NaOH(mL) (c) Determine the original concentration of the unknown acid. 3 points (d) What technique did we use in the lab to be able to accurately determine the equivalence point/3 points) e) Explain how you could have performed a second experiment to determine the pKa of this acid without performing a titration. 14 points

Explanation / Answer

1

a)

this acid is weak, since the pH wont change from V 0 to 50

b)

1.

pKa --> half point is V = 25mL

pH at this poiunt --> 5 approx, so pKa = 5

2.

pH in equivalence point is approx 9

3

volume ofNaOh for equivalence point --> drastic change in pH = equivalence

so

Vequiv. = 50 mL

c)

original concentration:

mmol of base = MV = 25*0.1 = 2.5

[acid] =mmol/V = 2.5/20 = 0.125 M

d)

this must have been titration with an indicator

e)

use back titration, which improves the error due to excess addition of titrant

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