Asked by Chris
                Calculate the concentrations of all species in a 1.41 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.
Na^+ = 2.82M
SO3^2- = 1.40M
HSO3^- = 0.00047M
H2SO3 =
OH^- =
H^+ =
I have been able to calculate the first three, but do not know where to go from there to find the last three. Any help would be very much appreciated.
            
        Na^+ = 2.82M
SO3^2- = 1.40M
HSO3^- = 0.00047M
H2SO3 =
OH^- =
H^+ =
I have been able to calculate the first three, but do not know where to go from there to find the last three. Any help would be very much appreciated.
Answers
                    Answered by
            DrBob222
            
    I agree with Na^+ and HSO3^-. I would have used 1.41 M for SO3^2- since 1.41-0.00047 = essentially 1.41 M.
You must have written this equation to obtain the 0.00047.
SO3^2- + HOH ==> HSO3^- + OH^-
So (HSO3^-) = 0.00047 which makes OH^- = 0.00047 M.
For H^+, you know OH and
(H^+)(OH^-) = Kw which gives you H^+.
Finally, for H2SO3, you have the second hydrolysis of HSO3^- as
HSO3^- + HOH ==> H2SO3 + OH^-
Kb2 = (Kw/k2) = 1E-14/1.4E-2 and
1.4E-2 = (H2SO3)(OH^-)/(HSO3^-).
However, in the first hydrolysis (kb1) you said (HSO3^-) = (OH^-) so Kb2 = (H2SO3) = about 7.1E-13 M.
    
You must have written this equation to obtain the 0.00047.
SO3^2- + HOH ==> HSO3^- + OH^-
So (HSO3^-) = 0.00047 which makes OH^- = 0.00047 M.
For H^+, you know OH and
(H^+)(OH^-) = Kw which gives you H^+.
Finally, for H2SO3, you have the second hydrolysis of HSO3^- as
HSO3^- + HOH ==> H2SO3 + OH^-
Kb2 = (Kw/k2) = 1E-14/1.4E-2 and
1.4E-2 = (H2SO3)(OH^-)/(HSO3^-).
However, in the first hydrolysis (kb1) you said (HSO3^-) = (OH^-) so Kb2 = (H2SO3) = about 7.1E-13 M.
                    Answered by
            ]swag
            
    YOLO
    
                    Answered by
            Anonymous
            
    How did you get 0.00047?
    
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