Question

Sulfurous acid, H2SO3, is a diprotic acid with Ka1= 1.3*10^-2 and Ka2= 6.3*10^-8. What is the pH of 0.200M solution of the sulfurous acid? What is the concentration of the sulfite ion, SO3^2-, in the solution? What is the percent ionization of H2SO3?

Please include a brief explanation of your reasoning if you can; I'd really like to understand this. Thank you!

Answer 1

H2SO3 ---------------------> H+ + HSO3-

0.200                                  0          0   ------------------> initial

0.200 -x                              x           x --------------------> equilibrium

Ka1 = [H+][HSO3-]/[H2SO3]

1.3 x 10^-2 = x^2 / 0.2 -x

x^2 + 1.3 x 10^-2 x - 2.6 x 10^-3 = 0

by solving this  

x = 0.045

x = [H+] = 0.045 M

pH = -log[H+] = -log (0.045)

pH = 1.35

percentage of ionisation = ( x / C ) x 100

                                        = (0.045 / 0.200) x 100

percentage of ionisation = 22.5 %

[SO3^-2] = Ka2

for dibasic acid second ionisation constant value = A^-2 concnetration

[SO3^-2] = 6.3 x 10^-8 M