Question

Calculate the concentrations of all species in a 1.38 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

These are the following concentrations that I need to find.

[Na+]

[SO3^2-]

[HSO3-]

[H2SO3]

[OH-]

[H+]

Please show me how to solve this problem. Anything help is greatly appreciated. Thank you.

Answer 1

[Na+] = 2 * [Na2SO3] = 2 * 1.38 = 2.76 M

[HSO3-] = 1.38 M

[OH-] = 1.38 M

[H+] =

Na2SO3 + H2O <-------> 2Na+ + HSO3- + OH-

HSO3- <-------> H+ + SO3^-        Ka = 6.3 x 10^-8
1.38       0        0            ( Initial)

-y                     +y       +y         ( change)

1.38- y                y         y          ( final)

Ka = [H+][SO3^2-] / [HSO3-]

6.3 x 10^-8 = y^2 / 1.38    { y is very small so ( 1.38 - y) = 1.38 }

y = 2.95 x 10^-4

[SO3^2-] = 2.95 x 10^-4 M

HSO3^- <--------> SO2(aq) + OH-         Kb = 7.1x10^-13

1.38        0            1.38M

-y        +y             +y

1.38 - y                   y            1.38 + y

Kb = [SO2] [OH-] / [HSO3-]

7.1 x 10^-13 = y (1.38 + y) / (1.38 - y)

y = 7.1x10^-13

[SO2] = 7.1 x 10^-13 M

[H2SO3] = 7.1 x 10^-13 M

[H+] = 10^-14 / 0.138 = 7.25 x 10^-14 M