# chemistry

posted by 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^+] =

• chemistry - ,

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 - ,

I found all concentrations of all species except H2SO3. I keep getting wrong. The balanced equation is

H2SO3 ---> H^+ + HSO3

Is that correct?

• chemistry - ,

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 - ,

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.