Question

Calculate the concentrations of all species in a 1.54 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8

Answer 1

so is this AP Chem?

Answer 2

Nope it is Chem IC the last class of general chemistry in my University.

Answer 3

hmm i see

Answer 4

Na2SO3 + H2O <==> 2Na+ + HSO3- + OH-

Answer 5

you got that right?

Answer 6

I have the first concentrations Na+ , SO3^2-, HSO^3-,and OH- but I can not figure out the concentration of H2SO3 and H+

Answer 7

ok so Ka = [H+][SO3^2-] / [HSO3-] right?

Answer 8

yeah that is correct

Answer 9

then 6.3x10^-8 = x² / 1.54 you can also assume that 1.54 -x = 1.33 since x is small
x = 2.9x10^-4

Answer 10

sorry it took long just working it out on the paper

Answer 11

What I did was I started out with Na3SO3<---> 2Na+ + SO3^2- and I knew that the concentration dissociated completely, so the first concentration were 2Na+=3.08 and SO3^2- =1.54. Then I did the next reaction here is where I think I got confused I did
SO3- + H2O <---> HSO3-+OH- converted my Ka2 to Kb using the Kw=Ka x Kw equation and found I got HSO3-=0.000494 and OH-=0.000494 . The problem is I can not figure out the last reaction HSO3- + H2o <---> H2SO3 +0H but they are asking my to find H+ but the reaction is a basic reaction.

Answer 12

Sorry I think that is wrong I typed it in that is the concentration of H+ and SO3- right?

Answer 13

hmmm
HSO3- <--> H+ + SO3- so you know that Ka = 6.3x10^-8
1.54 M 0 0
-x +x +x
1.54 M - x x x
and yes

Answer 14

and then the eqaution that i wrote out before the one that i wrote comes after that reaction

Answer 15

this then 6.3x10^-8 = x² / 1.54 you can also assume that 1.54 -x = 1.33 since x is small
x = 2.9x10^-4

Answer 16

This is what I have so far I am attaching the online home work problem.

Answer 17

2.9x10^-4*

Answer 18

yeah that what I typed in

Answer 19

hmmmm

Answer 20

ok im not sure but i believe that this might help. well you have the balanced equation H2SO3 --> 2H+ + SO3^2- which isn't the way it ionizes BUT this can be used when one knows H^+ and SO3^2- as in this case.
k1k2 = 0.014 * 6.3E - 8 = (H+)^2(SO3+)/(H2SO3) and solve for (H2SO3).

Answer 21

that would solve for H2SO4

Answer 22

SO3*

Answer 23

hmm I am a little confused been staring at this problem for way to long lol

Answer 24

i can see why...

Answer 25

im having trouble also

Answer 26

How would it ionize originally do you know what would be the steps because maybe I solved it in the wrong order?

Answer 27

How would the NaSO3 break down to find the different concentrations?

Answer 28

Sodium sulfite (NaSO3) can dissociate completely to form Na+ ions with a concentration of 2.66 M, and the bisulfite ion with a concentration of very nearly half (1.33M).

Answer 29

I do not know I think I am going to give up for tonight and try the problem again tomorrow thanks for the help.

Answer 30

no prob i wish helped alot

Answer 31

ok thank you