What are the equilibrium concentrations of H+,OH- ,(HSO3)-, (SO3)2- and Na+ in the solution that results from combining 50.0 mL of 0.200 M NaHSO3 with 50.0 mL of 0.600 M Ba(OH)2 ? Ka1 = 1.5 x 10-5 and Ka2 = 1.0 x 10-7 for H2SO3

does it matter that Ba(OH)2 has 2 OH's?When finding the dominant equation i keep getting the kb and ka to be equal which equation do i then use?

Yes, it makes a difference that Ba(OH)2 has two hydroxyl groups. The concn of Ba(OH)2 is 0.600 M which makes the (OH^-) = 1.2 M and when 50 mL is diluted to 100 mL the final (OH^-) will be 0.6 M (and you will have 0.06 mols in 100 mL).

I don't understand your statement about "getting the Kb and Ks to be equal".
I would write the ionic equation something like this for the neutralization of the base + acid.
OH^- + HSO3^- ==> HOH + SO3^=

Then I would calculate mols OH^- from the Ba(OH)2 and mols HSO3^- from the NaHSO3 and determine the number of mols of SO3^= formed. That leaves an excess of Ba(OH)2 and that will basically determine the (OH^-). BaSO3 is largely insoluble (Ksp = 9.5 x 10^-10 in an OLD OLD book I have and the Internet gives variable results from about 10^-7 to 10^-10 BUT that is insoluble enough to know that (SO3^=) will be determined largely by the Ksp for BaSO3. I would use that to determine SO3^= and from there I would go with the hydrolysis of SO3^= to form HSO3^- to get (HSO3^-). (H^+) will follow from (OH^-).
I hope this is helpful.

To find the equilibrium concentrations of H+, OH-, (HSO3)-, (SO3)2-, and Na+ in the solution, we need to follow the steps outlined below:

Step 1: Write out the balanced equation for the reaction.
Ba(OH)2 + NaHSO3 → Ba(HSO3)2 + NaOH

Step 2: Determine the initial moles and concentrations of each species.
Given that the volume is the same for both solutions (50.0 mL each), we can assume that the moles and concentrations of NaHSO3 and Ba(OH)2 are equal to each other:
Moles of NaHSO3 = Moles of Ba(OH)2
Concentration of NaHSO3 = Concentration of Ba(OH)2

Step 3: Write out the equilibrium expression for each species and apply the appropriate equilibrium constants (Ka or Kb).

For (HSO3)-:
HSO3- ⇌ H+ + SO3^2-
Ka1 = [H+][SO3^2-]/[HSO3-] = 1.5 x 10^-5

For H2SO3:
H2SO3 ⇌ H+ + HSO3-
Ka2 = [H+][HSO3-]/[H2SO3] = 1.0 x 10^-7

Step 4: Construct the ICE table and determine the changes in concentration.
We start with equal initial concentrations of NaHSO3 and Ba(OH)2, and from the balanced equation, we can see that they react in a 1:1 ratio. Therefore, the initial concentration of NaOH is also equal to the concentration of NaHSO3 and Ba(OH)2.

ICE table:
NaHSO3 + Ba(OH)2 → NaOH + Ba(HSO3)2

Species | Initial Concentration | Change in Concentration | Equilibrium Concentration
NaHSO3 | 0.200 M | -x (due to reaction) | 0.200 - x M
Ba(OH)2 | 0.600 M | -x (due to reaction) | 0.600 - x M
NaOH | 0 M | +x (due to reaction) | x M
Ba(HSO3)2 | 0 M | +x (due to reaction) | x M

Step 5: Set up the equilibrium expression using the appropriate equilibrium constant for each species, and solve for x.

Using Ka1:
1.5 x 10^-5 = (x)(x)/(0.200 - x)

Using Ka2:
1.0 x 10^-7 = (x)(0.200 - x)/(0.600 - x)

Solve these two equations simultaneously to find x. Once you have the value of x, substitute it back into the ICE table to find the equilibrium concentrations of each species.

Regarding your second question, it does matter that Ba(OH)2 has two OH- ions. Since Ba(OH)2 is a strong base, it will dissociate completely in water to give two OH- ions per formula unit. This affects the stoichiometry and equilibrium calculations for the reaction.

If you are finding the equilibrium constants, Kb and Ka, to be equal for different equations, it is likely due to a calculation error. Check your values and ensure you are using the correct equations and values for Kb and Ka.

I hope this helps! Let me know if you have any further questions.