Question

Calculate the concentrations of all species in a 0.860 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8. Find Molarity of Na+, SO3^2-, HSO3-, H2SO3, OH-, and H+. Thank you.

Answer 1

Na2SO3 dissociates completely to form Na+ ions and sulfite ions.
Na2SO3 <==> 2Na⁺ + SO₃^2-

[Na⁺] = 0.86 * 2 = 1.72 M

Initially concentration of SO₃^2- is 0.86 M.

SO₃^2- is a base with K_b = (10^-14)/(6.3× 10^–8) = 1.587*10^-7

SO₃^2-      +H₂O      <--->      HSO₃^-     +             OH^-
0.86M                                 0                            0

0.86 - x                                x                            x

K_b = [OH^-][ HSO₃^-] / [SO₃^2-]
1.587*10^-7 = x² / (0.86 - x ) = x² / 0.86                       [assuming 0.86-x = 0.86 since x is small ]
x = 3.69 x10^-4

[SO₃^2-] = 0.86 - 3.69 x10^-4 = 0.859 M

[HSO₃^-] =[OH-]= 3.69 x10^-4

Negligible amount of the hydrogen sulfite ion (HSO3–) will react with water:

HSO₃^-     +             H₂O        <--->      H₂SO₃    +             OH^-
3.69 x10^-4                                          0                            0

3.69 x10^-4 - x                                     x                            x

K_b = (10^-14)/( 1.4× 10^–2) = 7.14*10^-13 = [OH^-][ H₂SO₃] / [HSO₃^-]

7.14*10^-13 = x² / (3.69 x10^-4 - x ) = x² / 3.69 x10^-4                  [assuming 3.69 x10^-4 -x = 3.69 x10^-4 since x is small ]
x = 1.62 x10^-8

[H₂SO₃] = 1.62 x10^-8

Ans: [Na+] = 1.72 M

[SO₃^2-] = 0.859 M

[HSO₃^-] = 3.69 x10^-4

[H₂SO₃] = 1.62 x10^-8

[OH-]= 3.69 x10^-4

[H+] = (10^-14)/( 3.69 x10^-4) = 2.7 x 10^-11