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Part 3 â€“ Put it all together. Calculate the concentrations of all species in a 0.280 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4Ã— 10â€“2 and Ka2 = 6.3Ã— 10â€“8. Solution Na2So3(aq) -------------> 2Na^+ (aq) + SO3^2- (aq) 0.28MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2*0.28MÂ Â Â Â Â Â Â Â 0.28M [Na^+]Â Â Â = 2*0.28M = 0.56M [SO3^2-]Â Â = 0.28M SO3^2- (aq) + H2O --------> HSO3^- (aq) + OH^- (aq) IÂ Â Â Â Â Â Â 0.28Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0 CÂ Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x EÂ Â Â Â 0.28-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x kb2Â Â = Kw/Ka2 = 1*10^-14/6.3*10^-8 = 1.58*10^-7 Kb2Â Â Â = [HSO3^-][OH^-]/[SO3^2-] 1.58*10^-7 = x*x/0.28-x 1.58*10^-7*(0.28-x) = x^2 xÂ Â = 0.00021 [OH^-]Â Â = x = 0.00021M [HSO3^-]Â Â = x = 0.00021M HSO3^- + H2O -------------> H2SO3 + OH^- Kb1Â Â = Kw/Ka1 = 1*10^-14/1.4*10^-2Â Â = 7.14*10^-13 Kb1Â Â = [H2So3][OH^-]/[HSO3^-] 7.14*10^-13Â Â = [H2So3]* 0.00021/ 0.00021 7.14*10^-13M = [H2So3] .

Educación

1.58*10^-7*(0.28-x) = x^2
xÂ Â = 0.00021
[OH^-]Â Â = x = 0.00021M
[HSO3^-]Â Â = x = 0.00021M
HSO3^- + H2O -------------> H2SO3 + OH^-
Kb1Â Â = Kw/Ka1
= 1*10^-14/1.4*10^-2Â Â = 7.14*10^-13
Kb1Â Â = [H2So3][OH^-]/[HSO3^-]
7.14*10^-13Â Â = [H2So3]* 0.00021/ 0.00021
7.14*10^-13M = [H2So3]

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