Question

Calculate the concentrations of all species in a 0.600 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

[HSO3^-]=?
[H2SO3]=?
[OH^-]=?
[H^+]

Answer 1

Well, if we see all species present in the sodium sulfite solution , there is absolutely no molecular H2SO3 will exist in aqueous solution in that form but rather as SO2 and water. The concentrations of the rest of the species listed here can be calculated as below -

The sodium sulfite will dissociate completely in aq. solution to give Na⁺ and SO₃^2- ions.

Na₂SO₃ ----> 2Na⁺ + SO₃^2-

It can be seen here, that 1 mole of sodium sulfite will produce 2 moles of Na⁺ cations and 1 mole of SO3^2- anions. That means ; [Na⁺] = 2 . [Na2SO3] = 2 . 0.600 = 1.2 M

                     [SO3^2-] = [Na2SO3]   = 0,600 M

The sulfite anion will act as a base and react with water to form bisulfate ion or HSO3^- ion . The base dissociation constant , K_b, for the sulfate ion will be equal to

K_b = K_w / Ka₂   

    = 10^-14 / 6.3 x 10^-8     = 1.59 x 10^-7

see the raection below:

                               SO₃^2-_(aq) + H₂O _(l) -----> HSO₃^-_(aq) + OH^-_(aq)

                  I            0.600                               0                 0

                  C            -x                                    +x              +x

                  E      (0.600-x)                                x                  x

The base dissociation constant will be equal to

K_b = [HSO₃^-] . [OH^-] / [SO₃^2-]      = x .x /( 0.600 -x)         = x² / (0.600 - x)

Because the value of K_b is so small, you can approximate (0.600 - x ) as 0.600 . This means ,

1.59 x 10^-7 = x² / 0.600

x = 3.08 x 10^-4

As a result yo will get,

[OH^-] = 3.08 x 10^-4 M

[HSO₃^-] = 3.08 x 10^-4 M

Now you can calculate the concentration of H⁺ ion as well,

[H⁺] = 10^-14 / [OH^-]   =   10^-14 / 3.08 x 10^-4    = 3.25 x 10^-11 M