Calculate the concentrations of all species in a 1.01 M Na2SO3 (sodium sulfite)

Question

Calculate the concentrations of all species in a 1.01 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4

Explanation / Answer

Na2SO3 is a salt which ionizes 100%; therefore, Na2SO3 ==> 2Na + SO3^2- and Na^+ is just 2x1.01M = 2.02 Then SO3^2- hydrolyzes. ........SO3^2- + HOH ==> HSO3^- + OH^- I.......1.01.............0........- C.........-x..............x........x E.....1.01-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-.

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