Question

A 50 mL solution of 0.02500 M maleic acid (ka1=1.3e-2, ka2=5.9e-7) was titrated with 0.100 M NaOH. Calculate the pH after the adiition of a) 0.00 mL, b) 9.00 mL, c) 12.50 mL, d) 20.00 mL, e) 25.00 mL, and f) 26.00 mL of NaOH

Answer 1

mmol of acid = MV = 50*0.025 = 1.25 mmol of acid, so 1.25*2 = 2.50 mmol of H+

a)

initially:

Ka1 = [H+][HA-]/[H2A]

1.3*10^-2 = x*x/(0.025-x)

x = 0.01266

for 2nd ionization, youmay assume this is mainly H+1

so

pH = -log(0.01266) = 1.89

b)

mmol of base = MV = 0.1*9 = 0.9 mmol

this is a buffer

pH = pKa1 + log(A-/HA)

pH = -log(1.3*10^-2) + log(0.9 / (1.25-0.9)) = 2.29623

c)

mmol of base = 1.25

so this is first equivalence point

so

pH = 1/2*(pKa1 + pKa2)

pH = 1/2*(-log(1.3*10^-2) - log(5.9*10^-7)) = 4.0576

d)

mmol of base = MV = 20*0.1 = 2 mmol

so.. 1.25 mmol --> 2-1.25 = 0.75

this is again a buffer, in the second ionization

pH = pKa2 + log(2/0.75 )

pH= -log(5.9*10^-7) + log(2/0.75 )

pH = 6.655

e)

this is second equivalence point:

Vtotal = 50+25 = 75 mL

0.025 *50/(50+25) = 0.016666 of A-2

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

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

Kb2 =Kw/Ka2 = (10^-14)/(5.9*10^-7) = 1.6949*10^-8

1.6949*10^-8 = y*y/(0.016666 -y)

y = 1.677*10^-5

pOH = -log( 1.677*10^-5) = 4.7754

pH = 14-4.7754) = 9.2246

f)

mmol of base = 26*0.1 = 2.6 mmol

0.1 mmol of OH- in exces

Vtotal = 50+26 = 76 mL

[OH-] = 0.1/(76) = 0.00131

pOH= -log(0.00131) = 2.88272

pH = 14-2.88272 = 11.11728