Question

Consider a the titration of 0.729 L of 0.519 M ascorbic acid (H₂C₆H₆O₆) with 1.69 M NaOH. What is the pH at the second equivalence point of the titration?

Answer 1

Ascorbic acid (Vitamin C)

H2C6H6O6

7.9 x 10-5

1.6 x 10-12

The ka values of ascorbic acid is not given....its needed so i am taking my from end...

At second equilvence point we will have salt of ascorbic acid.

its a case of salt hydrolysis, so

K_h = Kw/K_a2 = (10^-14 )/1.6x10^-12 = 6.25 x10^-3

N1V1 = N2V2   at equivalence point

2 x 0.729x 0.519 = 1 x 1.69 x V_NaOH

V_NaOH = 0.4477 L

Total moles of salt = (0.729 x 0.519 ) = 0.3783 moles

Total final volume =   0.729 + 0.4477 Litre   = 1.1767 litres

Concentration of salt =    (0.3783 moles) /(1.1767 litres) = 0.3214 M

Kh = ch2/(1-h) = 6.25 x10^-3 solving for h we get Its a quadratic equation...

51.4h² + h -1 = 0

h = 0.13

[OH] = ch = 0.3214 x 0.13 = 0.0417

pOH = -log(0.041) = 1.38

ph = 14- poH = 14 -1.38 = 12.62