Question

20.0mL of 0.127M diprotic acid (H2A) was titrated with 0.1018M KOH. The acid ionization constants for the acid are Ka1=5.2

Answer 1

For the diprotic cid following reactions will occur:

H2A + H2O -----> H3O⁺ + HA^- here Ka1 = 5.2 X 10^-5

HA- + H2O ------> A2- + H3O+ here Ka2 = 3.4 X 10^-10


a) What volume of KOH is required to reach the first equivalence point?

At given in question 20ml of 0.127 M H2A , with 0.1018 M of KOH , Ka1 = 5.2 x 10^-5, Ka2 = 3.4 X 10^-10

The first equivalence point occurs when we react all of H2A with KOH producing HA- as in following rxn;

H2A + KOH (aq) --> KHA- + H2O

This is the point at which moles KOH added will be equals moles H2A initially present. As this happens we have the following reaction :

HA- + H2O(l) <=> A2- + H3O +   ( this happens at Ka2.)

Ka2 = 3.4 X 10^-10 as given in the question.

moles H2A = V x M = (0.127mol/L)(20/1000mL) = 0.00254moles
therefore the moles of the base is also 0.00254 moleT which requires to reach the 1st equivalence point.

To calculate the volume of KOH at first equivalence point:

volume KOH = 0.00254 moles/0.1018 M = 0.02495 = 24.95 mL

this gives a total new volume of 20 mL+ 24.95ml = 44.95 mL

The volume of KOH required to go from HA- to A2- will be the same volume which is required to go from H2A to HA- because the same amount of KOH is required to neutralize each of the two protons. That means that an additional 24.95ml of KOH will need to be added to reach the second equivalence point..

Total volume to reach the 2nd equivalence point = 44.95 mL + 24.95mL = 69.9ml