Question

Consider a the titration of 0.643 L of 0.889 M sulfurous acid (H2SO3) with 1.83 M NaOH. What is the pH at the second equivalence point of the titration?

Answer 1

moles of sulfurous acid = 0.643 x 0.889

= 0.572

at first equivalence point

number of moles of acid = number of moles of base

0.572 = 1.83 x V

V = 0.312 L

volume of NaOH = 0.312 L

molarity of H2SO3 = 0.572 / (0.643 + 0.312)

= 0.60 M

Molarity of NaOH = 0.572 / (0.643 + 0.312) = 0.60 M

H2SO3 + NaOH ------------------> HSO3 - + H2O

0.60

HSO3- + H2O ----------------> H2SO3 + OH-

0.60 0 0

0.60 - x x x

Kb1 = x^2 / 0.60 - x

1.59 x 10^-7 = x^2 / 0.60 - x

x = 3.08 x 10^-4

[OH-] =  3.08 x 10^-4 M

pOH = -log [OH-] = 3.51

pH = 10.49