# Calculate the concentration of H2SO3 present in 0.198 M Na2SO3(aq). Ka1 = 0.015 Ka2 = 1.2 × 10−7 Kw = 1 × 10−14 I calculated this using Kb1 and Kb2 in ICE tables, and I ended up finding...

Calculate the concentration of H2SO3 present in 0.198 M Na2SO3(aq).

Ka1 = 0.015

Ka2 = 1.2 × 10−7

Kw = 1 × 10−14

I calculated this using Kb1 and Kb2 in ICE tables, and I ended up finding the concentration to be the equivalent of Kb2. However, that is not the correct answer and I don't know what I'm doing wrong. Can someone please explain how to do this problem? Thank you!

My answer, 8.333E-8, was incorrect.

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### 1 Answer

To solve this problem you need to calculate the Kb's from the Ka's, using the relationship Kb = 10^(-14)/Ka, and keeping in mind that Ka1 gives Kb2, and vice versa.

The first hydrolysis is:

`CO_3^2- + H_2O -> HCO_3^- + OH^-`

Kb = 1.00 x10^(-14)/1.2 x10^(-7) = 8.3 x10^(-8)

Using an ICE table, you will see that at equilibrium:

if [HCO3-] = x, then [OH-] = x and [CO3 2-] = 0.198-x

8.3x10^(-8) = (x^2)/(.198-x)

neglecting x in the denominator, because it's small, we get

x= 1.28 x10^(-4)

This is less than 5% of 0.192 s it is okay to neglect x.

For the second hydrolysis:

`HCO_3^- + H_2O -> H_2CO_3 + OH^-`

Kb = 1.00 x 10^(-14)/.015 = 6.7 x 10^(-13)

[H2CO3] = x, [OH-] = x, [HCO3] = 1.28 x 10^(-4)-x

6.7 x 10^(-13) = (x^2)/1.28 x 10^(-4) -x

Again, we neglect x in the denominator so that x = **9.2 x 10^(-9) **as the final answer. This again less than 5% of the initial concentration of the base so the assumption that it is negligible in the calculation is valid.