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.
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?
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