Question

You are titrating 25 mL of 0.25 M Ascorbic Acid solution ith 0.15 M NaOH. Ascorbic Acid is an organic diprotic acid, H2A, that is found in many natural materials. Ka1 = 7.94 x 10^-5 and Ka2 = 1.62 x 10^-12. What is the a) initial pH, b) at the 1st equivalence point, c) the 2nd equivalence point, and d) after 27.5 mL of NaOH have been added?

Answer 1

a)

initially, assume just 1st ionization, since 10^-5 vds 10^-12 is large difference

H2A = H+ + HA-

Ka = [H+][HA-]/[H2A]

7.94*10^-5 = x*x/(0.25-x)

x = 0.00442

pH = -log(x) = -log(0.00442) = 2.354

b)

1st equivalence point... this is speical point for a diprotic acid

since

pH is given by:

pH = 1/2*(pKa1+pKa2)

pKA1 0 -log(ka1) = -log(7.94*10^-5) = 4.100

pKa2 = -log(Ka2) = -log(1.62*10^-12) =11.7904

then

pH = 1/2*(4.10+11.7904) = 7.9452

c)

in the 2nd equivalence, expect hydrolysis:

A-2 + H2O = HA- + OH-

Kb2 = [HA-][OH-]/[A-2]

KB2 = Kw/(Ka2) = (10^-14)/(1.62*10^-12) = 0.0061728

0.0061728 = y*y/(M-y)

find total volume requireD:

mmol of acid = MV = 25*0.25 = 6.25 mmol of acid = 6.25*2 = 12.5 mmol of H+

then

mmol of NaOH = 12.5 mmol requried

V = mmol/M = 12.5/0.15 = 83.33 mL

total V = 25+83.33 = 108.33 mL

M = M1*V1/(Vt = 25*0.25 /(108.33) =0.05769

now

0.0061728 = y*y/(0.05769-y)

y = 0.016

[OH-] = y= 0.016

pOH= -log(0.016)I = 1.795

pH = 14-pOH = 14-1.795 = 12.205

e)

this is abuffer

mmol of base = MV = 0.15*27.5 = 4.125

the first ionization buffer so:

mmol of OH = 4.125

mmol of H2A neutralized/ HA- formed = 4.125

mmol of H2A left = 6.25-4.125 = 2.125

pH = pKa1 + log(HA-/HN2A)

pH = 4.10+ + log(4.125/2.125) = 4.3880