Question

Calculate the concentrations of all species in a 0.300 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

Answer 1

Calculate the concentrations of all species in a 0.300 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

Solution:

Kb1 for SO₃^2- = Kw/Ka2

                        = 1.0 * 10^-14 / 6.3 * 10^-8

                        = 1.59 * 10^-7

Kb2 for SO₃^2- = Kb for HSO3-

                         = Kw/Ka1

                           = 1.0 * 10^-14 / 1.2 * 10^-2

                          = 7.14 * 10^-13

    SO₃^2- + H2O ßà HSO3- + OH-

I 0.300     ------              0             0

C   -x                               x             x

E 0.300-x                         x             x

Kb = x * x/(0.300-x)

1.59 * 10^-7(0.300-x)-x^2 = 0

On solving above equation, we get

X = 0.0002184 = 2.18 * 10^-4 M

[HSO3-]=[OH-]= 2.18 * 10^-4 M

[SO₃^2-] = 0.300 M - 2.18 * 10^-4 M

             = 0.299 M

            HSO3-          + H2O ßà H2SO3 + OH-

I 2.18 * 10^-4 M --------                 0         2.18 * 10^-4 M

C     -x                                                  +x                +x

E       2.18 * 10^-4 – x                      x           2.18 * 10^-4 M + x

Kb = x * 2.18 * 10^-4 M +x / (2.18 * 10^-4 M – x)

7.14 * 10^-13*(2.18 * 10^-4 M – x) – x (2.18 * 10^-4 M +x) = 0

On solving the equation, we get

X = 7.139 * 10^-13 M

[H2SO3] = x

                = 7.139 * 10^-13 M

[HSO3-] = 2.18 * 10^-4 M – x

               = 2.18 * 10^-4 M – 7.139 * 10^-13 M

               = 2.18 * 10^-4 M

[OH-] = 2.18 * 10^-4 M + x

               = 2.18 * 10^-4 M + 7.139 * 10^-13 M

               = 2.18 * 10^-4 M

[SO₃^2-] = 0.299 M

[Na+] = 2 * 0.300 M

            = 0.600 M

[H+] = Kw/[OH-]

         = 1.0 * 10^-14 / 2.18 * 10^-4

         = 4.58 * 10^-19 M