Question

Consider a the titration of 0.531 L of 0.969 M ascorbic acid (H₂C₆H₆O₆) with 1.91 M NaOH. What is the pH at the second equivalence point of the titration?

ka1: 7.9*10^-5

ka2: 1.6*10^-12

Answer 1

in the 2nd equivalence point:

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

its a case of salt hydrolysis, so

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

N1V1 = N2V2   at equivalence point

note im using normality, so N =2 M

2 *0.531*0.9696 = 1 x 1.69 x V_NaOH

V_NaOH = (2*0.513*0.969)/(1.91) = 0.52 L = 520 mL

Total moles of salt produced = (0.969*0.531) = 0.514 moles

Total final volume =   0.531 + 0.52 = 1.05 L

Final Salt concentration =    (0.514 moles) /(1.05 litres) = 0.489 M

substitute

A-2 + H2O <--> HA- + OH-

Kb = [HA-][OH-]/[A-2]

(6.25*10^-3) = x*x/(0.489-x)

x = 0.05224

[OH-] = 0.05224

[OH] = 0.05224

pOH = -log(0.05224) = 1.28199

ph = 14- poH = 14 -1.28199= 12.72