Question

Calculate the concentrations of all species in a 1.05 M Na₂SO₃ (sodium sulfite) solution. The ionization constants for sulfurous acid are K_a1 = 1.4x10^-2 and K_a2=6.3x10^-8.

Na⁺⁼

SO₃^2-=

HSO₃^-=

H₂SO₃₌₌₌

OH^-=

H⁺⁼

Answer 1

1.05 M Na2SO3

Na2SO3 --------------------> 2 Na+ + SO3-2

[Na+] = 2 x 1.05 = 2.10 M

Ka1 = 1.4 x 10^-2 ,

Ka2 = 6.3 x 10^-8

Kw = 1.0 x 10^-14

Kb1 = Kw / Ka2 = 1.0 x 10^-14 / 6.3 x 10^-8 = 1.59 x10^-7

Kb2 = Kw / Ka1 = 1.0 x 10^-14 / 1.4 x 10^-2 = 7.14 x 10^-13

SO3^-2 + H2O -------------------> HSO3-   + OH-

1.05                                               0               0    ------------------> initial

1.05-x                                            x                x ---------------------> equilibrium

Kb1 = [HSO3-][OH-]/[SO3-2]

Kb1 = x^2 / 1.05-x

1.59 x 10^-7   = x^2 / 1.05-x

x^2 + 1.59 x 10^-7 x - 1.67 x 10^-7 = 0

x = 4.09 x 10^-4

x = [OH-] = [HSO3-] = 4.09 x 10^-4 M

[OH-] = [HSO3-] = 4.09 x 10^-4 M

[SO3^-2] = 1.05 - x

[SO3^-2]      = 1.05 M

[H+] = Kw / [OH-] = 1.0 x 10^-14 / 4.09 x 10^-4

[H+] = 2.44 x 10^-11 M

H2SO3 = Kb2

[H2SO3 ] = 7.14 x10^-13 M