Calculate the concentration of all species in a 0.530 M solution of [H2SO3].
[H2SO3], [HSO3-], [SO32-], [H3O+], [OH-]
The following two reactions will be there in the process:
H2SO3 + H2O <==> H3O+ + HSO3- . . .Ka1 = 1.3 x 10^-2
HSO3- + H2O <==> H3O+ + SO3 2-...... Ka2 = 6.3 x 10^-8
Because Ka1 >>> Ka2, the amount of H3O+ produced (and the amount of HSO3- reacted) in reaction 2 is negligible compared to the amount from Step 1.
Using ICE chart for the first hydrolysis step, we have:
H2SO3 + H2O <==> H3O+ + HSO3-
MOLARITY
Initial . . . . . . . .0.530 .. . . . . . . . . . . . .0 . . . . .
. .0
Change . . .. . . . .-x . . . . . . . . . . . . . . .x . . . . . .
.x
Equilibrium . . .0.530-x . . . . . . . . . . . . .x . . . . . .
.x
Ka1 = [H3O+][HSO3-] / [H2SO3] = (x)(x) / (0.530-x)
1.3 x 10^-2 = x^2 / (0.530 - x)
x^2 = (1.3 x 10^-2)(0.530-x)
x^2 = -0.013x + 0.0069
x^2 +0.013x - 0.0069 = 0
Solving the equation, we get
x = 0.076 and x= -0.089, -0.089 can be neglected because concentration can't be negative.
[H2SO3] = 0.530 - x
= 0.530 - 0.076
= 0.454 M
[H3O+] = [HSO3-] = x = 0.076 M
Since [H3O+][OH-] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / 0.076
= 1.3 x 10^-13
Now, for second reaction and set up an ICE chart.
Molarity . . . . .HSO3- + H2O <==> H3O+ + SO3 2-
Initial . . . . . . . 0.076 . . . . . . . . . . . . 0.076 . . .
.0
Change . . . . . . .-x . . . . . . . . . . . . . . . .x . . . .
.x
Equilibrium . . .0.076-x . . . . . . . . . . .0.076+x . . x
Ka2 = [H3O+][SO3 2-] / [HSO3-]
6.3 x 10^-8 = (0.076+x)(x) / (0.076-x)
Since Ka2 is so small (10^-8), then the x term will be very small compared to 0.076 and we can neglect it from 0.076-x and 0.076+x.
6.3 x 10^-8 = 0.076x / 0.076
6.3 × 10 ^-8 = x = [SO3 2-]
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