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!

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.

DrBob222 is correct, you have different numbers for your homework or you calculated incorrectly.

thanks!!!!

It is not wrong, correct way to calculate answers thanks DrBob222

To calculate the concentrations of all the species in the solution, we can use the ionization constants for sulfurous acid (H2SO3) to determine the concentration of H+, HSO3-, and SO3 2-.

Let's start by writing the ionization equations for sulfurous acid:

H2SO3 ⇌ H+ + HSO3- (Ka1)
HSO3- ⇌ H+ + SO3 2- (Ka2)

We are also given that the concentration of Na2SO3 is 0.810 M, so we know the concentration of Na+ is also 0.810 M.

Now, let's consider the ionization of H2SO3:

H2SO3 ⇌ H+ + HSO3- (Ka1)

Using an ICE (Initial, Change, Equilibrium) table, we start with an initial concentration of 0 for H+ and HSO3-, and an initial concentration of 0.810 M for H2SO3.

H2SO3 ⇌ H+ + HSO3-
I: 0.810 M 0 0
C: -x +x +x
E: 0.810-x x x

From the equation H2SO3 ⇌ H+ + HSO3-, we can say that the concentration of H+ and HSO3- at equilibrium is equal to x.

Now, we can set up the equilibrium expression using the given Ka1:

Ka1 = [H+][HSO3-] / [H2SO3]
1.4 x 10^-2 = (x^2) / (0.810 - x)

The value of x will give us the concentration of both H+ and HSO3-.

Now, let's consider the ionization of HSO3-:

HSO3- ⇌ H+ + SO3 2- (Ka2)

Using a similar ICE table, we start with an initial concentration of 0 for H+ and SO3 2-, and an initial concentration of x for HSO3-.

HSO3- ⇌ H+ + SO3 2-
I: x 0 0
C: -x +x +x
E: x-x x x

Again, the concentration of H+ and SO3 2- at equilibrium is equal to x.

Now, we can set up the equilibrium expression using the given Ka2:

Ka2 = [H+][SO3 2-] / [HSO3-]
6.3 x 10^-8 = x * x / (x)

Solving this equation will give us the concentration of H+ and SO3 2- at equilibrium, which is also equal to x.

Now, with the concentrations of H+, HSO3-, and SO3 2- known, we can calculate the concentration of OH- by using the fact that water (H2O) is a neutral solution, so the concentration of H+ multiplied by the concentration of OH- must equal the value of the ion product of water (1.0 x 10^-14).

[H+][OH-] = 1.0 x 10^-14
x * [OH-] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / x

Lastly, we can calculate the concentration of Na+ because it does not undergo ionization and remains at the initial concentration of 0.810 M.

In summary, to find all the concentrations, you need to solve the following equations:

1.4 x 10^-2 = (x^2) / (0.810 - x) (Ka1)
6.3 x 10^-8 = x * x / (x) (Ka2)
[OH-] = 1.0 x 10^-14 / x
Na+ = 0.810 M (from Na2SO3)

Using these equations, you can calculate the concentrations of all the species.

It's definitely wrong... Otherwise my Chemistry program for homework would be accepting my inputs for the answers, which it's not.

This is all wrong.