Respuesta :

superman1987
we will use this two reaction equation:

H2SO3 + H2O ↔ H3O+  +  HSO3-    Ka1 = 1.3 x 10^-2

HSO3-  + H2O ↔ H3O+   + SO3 2-    Ka2= 6.3 x 10^-8

we will use the ICE table for the first equation:

              H2SO3 + H2O ↔ H3O+ +  HSO3- 

initial     0.025                        0            0

change   -X                             +X          +X

Equ       (0.025-X)                     X             X 

 
Ka1 = [H3O+] [HSO3-] / [H2SO3]

1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X

∴ X = 0.0127

when [H3O+] = X
                   
 ∴[H3O+] = 0.0127 M


and when [HSO3-] = X

∴[HSO3-] = 0.0127 M

and when [H2SO3] = 0.025 - X

∴[H2SO3] = 0.025 - 0.0127

                 = 0.0123 M

when Kw = [OH-][H3O+]

and Kw = 1.1 x 10^-14 / 0.0127

∴[OH-] = 1.1 x 10^-14 / 0.0127

            = 8.66 x 10^-13 M

- by using the ICE table for the second equation:

              HSO3- + H2O ↔ H3O+         + SO3 2-

initial    0.0127                      0.0127            0

change    -X                            +X                +X

Equ      (0.0127-X)                (0.0127+X)        X


when Ka2 = [SO32-] [H3O+] / [HSO3-]

by substitution:

6.3 x 10^-8 = X(0.0127+X) / (0.0127-X) 

as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X

6.3 x 10^-8 = 0.0127X /0.0127

∴X = 6.3 x 10^-8

when [SO3 2-] = X 

∴[SO32-] = 6.3 x 10^-8

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