Posted by **Johnny** on Sunday, June 17, 2012 at 2:41pm.

Calculate the concentrations of all species in a 1.67 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

I know how to find the Na^+ and SO3^-2 but I don't know how to find molarity of rest of it. Can you show me how to find molarity please? Show me how to find molarity for HSO3^- and I will do the rest of them when I get it.

[Na^+] = 3.34M

[SO3^-2] = 1.67M

[HSO3^-] =

[H2SO3] =

[OH^-] =

[H^+] =

- chemistry -
**DrBob222**, Sunday, June 17, 2012 at 2:57pm
HSO3^- comes from the hydrolysis of SO3^2-.

..........SO3^2- + H2O ==> HSO3^- + OH^-

I.........1.67...............0.......0

C...........-x...............x.......x

E........1.67-x..............x.......x

Kb for SO3^2- = Kw/Ka2 for H2SO3 = (x)(x)/(SO3^2-)

Substitute and solve for x = (OH^-) = (HSO3^-)

- chemistry -
**Johnny**, Sunday, June 17, 2012 at 3:45pm
I found all concentrations of all species except H2SO3. I keep getting wrong. The balanced equation is

H2SO3 ---> H^+ + HSO3

Is that correct?

- chemistry -
**DrBob222**, Sunday, June 17, 2012 at 4:12pm
I would look at this. Yes, your equation is right and I think if you substitute the values for H^+ and HSO3^- you will get H2SO3. An easier way may be this way. Try it and let me know

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).

- chemistry -
**Johnny**, Sunday, June 17, 2012 at 4:40pm
I finally got H2SO3 right.

H2SO3 ---> 2H^+ + SO3^-2 is work this way finding the H2S03. Thank you so much for your help. I appreciated your help.

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