Question

Calculate the pH in 0.020 M H2SO3(Ka1=1.5×10−2, Ka2=6.3×10−8)

Answer 1

H2SO3 <----> H+ + HSO3-
0.020              0          0 ( Initial)

0.020-x x           x ( Equilibrium)

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

1.5 x 10^-2 = x^2 / 0.02-x

x^2 + 0.015 x - 0.0003 = 0

x = [H+] = [HSO3-] = 0.0114 M

HSO3- <---->   H+   + SO3^2-
0.0114     0.0114       0 (Initial)
0.0114-x       0.0114+x     x (Equilibrium)

Ka2 = [H+][SO3^2-] / [HSO3-]

6.3 x 10^-8 = (0.0114+x)x / (0.0114-x)

Since x is very small (0.0114 + x) and (0.0114 - x) = 0.0114

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

[H+] = 0.015 + 6.3 x 10^-8 = 0.0114 M

pH = -log[H+] = - log(0.0114) = 1.943