Question

Consider a the titration of 0.685 L of 0.773 M ascorbic acid (H₂C₆H₆O₆) with 1.76 M NaOH. What is the pH at the second equivalence point of the titration?

Answer 1

1)

moles of ascorbic acid = 0.685 x 0.773 = 0.5295

H2C6H6O6 + 2 NaOH   -----------> Na2C6H6O6 + 2 H2O

moles of NaOH = 2 x moles of acid

                         = 2 x 0.5295

                         = 1.06

1.76 x V = 1.06

V = 0.602

volume of NaOH at 2nd equivalence point = 0.6017 L

here C6H6O62- remains.

[C6H6O62-] = 0.5295 / 0.685 + 0.602 = 0.4115 M

C6H6O2-   + H2O    ------------> HC6H6O6- +   OH-

0.411                                                 0                  0

0.411 - x                                             x                 x

Kb1 = x^2 / 0.411 - x

1.26 x 10^-10 = x^2 / 0.411 - x

x = 7.22 x 10^-6

[OH-] = 7.22 x 10^-6 M

pOH = 5.14

pH = 8.86