Question

50.0 mL of 0.100 M H₂SO₃ is titrated with 0.150 M KOH soluion.

a) What is the initial pH of the sulfurous acid?

b) How much titrant is needed to reach the equivalence points? What is the pH at each equivalence point?

c) What is the pH of the solution after the following points of the titration (with 0.150 M KOH)? (20.0 mL, 33.3 mL, 60.0 mL, 70.0 mL, 100.0 mL, 120.0 mL, 130.0 mL)?

Answer 1

for H2SO3

K_a1 = 1.4×10^–2
K_a2 = 6.3×10^–8

50.0 mL of 0.100 M H₂SO₃ is titrated with 0.150 M KOH soluion.

a) What is the initial pH of the sulfurous acid?

H2SO3 ---------------------> H+ + HSO3-

0.1                                       0           0 ----------------> initial

0.1-x                                    x             x ----------------> equilibrium

Ka1 = [H+][HSO3-]/[H2SO3]

Ka1 = x^2 / 0.1-x

1.4 x 10^-2 = x^2 / 0.1-x

x^2 + 1.4 x 10^-2 x - 1.4 x 10^-3 = 0

x = 0.0311

x = [H+] = 0.0311 M

pH = -log [H+] = -log ( 0.0311)

pH = 1.51 -------------------------> initial pH

b) How much titrant is needed to reach the equivalence points? What is the pH at each equivalence point?

H2SO3 + KOH -------------------> KHSO3 + H2O

it is first equivalence point .

here millimoles of acid = millimoles of base

50 x 0.1 = V x 0.150

V = 33.33 mL

at first equivalence point the volume of KOH needed = 33.33 mL

here pH = (pKa1 + pKa2) / 2

pH = (1.85 + 7.20) /2

pH = 4.52 ---------------------> first equivalence pH

second equivaelce point

H2SO3 + 2KOH -----------------> K2SO4 + 2HO

1                2

1 millimoles of H2SO3 = 2 millimoles KOH

5   = 2 x 0.150 x V

16.67 = V

total volume needed = 33.33 + 16.67 mL = 50 mL needed second equivalece point

here    pH = 7.20