Calculate the concentrations of h2so3, hso−3, so2−3, h3o+ and oh− in 0.025 m h2so3.

1 answer:

3 0

We will use this two reaction equation:

H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2

HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8

we will use the ICE table for the first equation:

H2SO3 + H2O ↔ H3O+ + HSO3-

initial 0.025 0 0

change -X +X +X

Equ (0.025-X) X X

Ka1 = [H3O+] [HSO3-] / [H2SO3]

1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X

∴ X = 0.0127

when [H3O+] = X

∴[H3O+] = 0.0127 M

and when [HSO3-] = X

∴[HSO3-] = 0.0127 M

and when [H2SO3] = 0.025 - X

∴[H2SO3] = 0.025 - 0.0127

= 0.0123 M

when Kw = [OH-][H3O+]

and Kw = 1.1 x 10^-14 / 0.0127

∴[OH-] = 1.1 x 10^-14 / 0.0127

= 8.66 x 10^-13 M

- by using the ICE table for the second equation:

HSO3- + H2O ↔ H3O+ + SO3 2-

initial 0.0127 0.0127 0

change -X +X +X

Equ (0.0127-X) (0.0127+X) X

when Ka2 = [SO32-] [H3O+] / [HSO3-]

by substitution:

6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)

as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X

6.3 x 10^-8 = 0.0127X /0.0127

∴X = 6.3 x 10^-8

when [SO3 2-] = X

∴[SO32-] = 6.3 x 10^-8

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