Homework Help: Chemistry

Posted by Katherine on Friday, April 20, 2012 at 1:40am.

Calculate the concentrations of all species in a 0.810 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4 x 10^-2 and Ka2 = 6.3 x 10^-8

so I'm trying to find Na, SO3 2-, HSO3-, H2SO3, OH-, H+ and heres what I tried to do so far but I'm not even sure if it's right:
Na2SO3 -> 2Na + SO3 2-
0.810 M ...... 0 ...... 0
-x ................+x.....+x
1.4 x 10^-2 = (x^3)/(0.810-x)
x = 0.203959

H2SO3 + OH- = H2O + HSO3-
and then an ICE chart for this one..?

I'm not sure... please help!

Chemistry - DrBob222, Friday, April 20, 2012 at 10:02pm
Na2SO3 is a salt which ionizes 100%; therefore,
Na2SO3 ==> 2Na + SO3^2- and Na^+ is just 2x0.810M = ?

Then SO3^2- hydrolyzes.
........SO3^2- + HOH ==> HSO3^- + OH^-
I.......0.810.............0........-
C.........-x..............x........x
E.....0.810-x..............x.......x

Kb1 for SO3^2- =(Kw/k2 for H2SO3) = (HSO3^-)(OH^-)/(SO3^2-) and you can solve for OH^- and HSO3^-. Note that Kb1 = (about 1.6E-7 which is relatively small so not much hydrolyzes.)

The second hydrolysis equation looks like this.
..........HSO3^- + HOH ==> H2SO3 + OH^-
For this, we look at Kb1 versus Kb2.
Kb2 for SO3^2- = (Kw/k1 for H2SO3) = about (1E-14/1.4E-2) = 7.1E-13. Considering that Kb1 is about 10^-7 and not much is hydrolyzed, Kb2 is even smaller (much smaller by a factor of about 100,000 or so); therefore, the OH^- contributed by this hydrolysis is negligibly small and we can ignore that. If we recognize that OH^- = HSO3^- (from Kb1 equation), then if we write Kb2 expression it is
(H2SO3)(OH^-)/(HSO3^- and (H2SO3) = just Kb2 or about 7E-13.
That gives you Na^+, HSO3^-, OH^-, and SO3^2-.
I think you can take it from here.
Chemistry - Katherine, Saturday, April 21, 2012 at 4:28pm
thanks!!!!
Chemistry - N, Sunday, June 3, 2012 at 1:32am
This is all wrong.

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