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.
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!
6 answers
thanks!!!!
This is all wrong.
It is not wrong, correct way to calculate answers thanks DrBob222
It's definitely wrong... Otherwise my Chemistry program for homework would be accepting my inputs for the answers, which it's not.
DrBob222 is correct, you have different numbers for your homework or you calculated incorrectly.
1 answer