Question

Adipic acid is a diprotic weak acid (H₂A). A polymer chemist wishes to analyze a sample thought to contain 0.1000 M adipic acid. Assume she titrates 10.00 mL of the sample with 0.1000 M NaOH (pka1=4.43 and pka2 = 5.41)

Calculate the pH of the original sample, pH at both equivalence points, pH at both half neutralization points, and when the volume of the titrant used is three times as much as the first equivalent point. Use this information to sketch a rough plot of pH versus volume for this titration.

Answer 1

H2A molarity = 0.100M

Ka1 = 10^-pH = 10^-4.43 = 3.72 x 10^-5

H2A .------------------------> HA- + H+

0.1-x                                  x          x

Ka1 = x^2 / 0.1-x

3.72 x 10^-5 = x^2 / 0.1-x

x = 1.91 x 10^-3

x = [H+] = 1.91 x 10^-3 M

pH = -log [H+] = 2.72

pH of the original sample = 2.72

pH at first equivalence point :

here millimoles of acid = millimoles of base

0.1 x 10 = 0.1 x V1

V1 = 10 ml

here volume of NaOH = 10 ml

here

pH = (pKa1 + pKa2 )/2

pH = (4.43 + 5.41)/2

pH = 4.92

pH at second equivalence point :

here volume of NaOH = 20 mL

salt only remains here salt concentration= 0.1 x 10 / (10 + 20)

                                                                   = 0.033 M

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

0.033-x                                x             x

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

Kw / Ka2 = 2.57 x 10^-9 = x^2 / 0.033 -x

by solving this

x = 9.21 x 10^-6

[OH-] = 9.21 x 10^-6 M

pOH = -log[OH-] = 5.04

pH + pOH = 14

pH = 8.96

pH at first half equivalence point

at this point pH = pKa1

pH = 4.43

pH at second half equivalence point

at this point

pH = pKa2

pH = 5.41