Question

titration of 100.0 mL of 0.6 M H3A by 0.6 M KOH for the next three questions. The triprotic acid has Ka1 = 1.0 x 10-3, Ka2 = 1.0 x 10-6, and an unknown value for Ka3. 1) Calculate the pH after 100.0 mL of KOH has been added. Calculate the pH after 150.0 mL of KOH has been added.The pH of the solution after 200.0 mL of KOH has been added is 8.00. Determine the value of Ka3 for this triprotic acid.

Answer 1

a)

mmol of base = MV = 100*0.6 = 60 mmol

mmol of H3A = MV = 100*0.6 = 60 mmol of H3A

H3A + OH- <-> H2A- + H2O

this is first equivalence point

pH is given via

pH = 1/2*(pKa1 + pKa2)

pKa1 = -log(Ka) = -log(10^-#) = 3

pKa2 = -log(Ka2) = -log(10^-6) = 6

pH = 1/2*(3+6) = 4.5

b)

find pH when V = 150 mL of KOH is added

mmol of OH = MV = 150*0.6 = 90 mmol

this is the SECOND HALF EQUIVALENCE POINT

this is

pH = pKa2 + log(HA-2 / H2A-)

since HA-2 = H2A-

then

pH = pKa2 = 6

c)

V = 200 mL of KOH so

200*0.6 = 120 mmol of base

mmol ofH3A = 60

mmol of H2A = 60

therefore, only HA-2 is left

this is the 3rd half equivalence point

then

pH = 1/*(pKa2 + pKa3)

8 = 1/2*(6 + pKa3)

pKa3 = 8*2 - 6 = 10

pKa3 = 10

Ka3 = 10^-10