chemistry
posted by Johnny on .
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^+] =

HSO3^ comes from the hydrolysis of SO3^2.
..........SO3^2 + H2O ==> HSO3^ + OH^
I.........1.67...............0.......0
C...........x...............x.......x
E........1.67x..............x.......x
Kb for SO3^2 = Kw/Ka2 for H2SO3 = (x)(x)/(SO3^2)
Substitute and solve for x = (OH^) = (HSO3^) 
I found all concentrations of all species except H2SO3. I keep getting wrong. The balanced equation is
H2SO3 > H^+ + HSO3
Is that correct? 
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*6.3E8 = (H^+)^2(SO3^=)/(H2SO3) and solve for (H2SO3). 
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.