Question

Consider a 0.12 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given here. Calculate the contributions to [H3O+] from each ionization step. At what point can the contribution of the second step be neglected?

A. Ka1=1.0×10−4 and Ka2=5.0×10−5

Answer 1

The  [H3O+] in step 1 is 0.0034 M while the  [H3O+] in step 2 is 0.00039 M

What is the contribution of each  step?

Let us set up the ICE table in each case, for K1;

         H2A(aq) + H2O(l)-------->   H3O^+(aq) + HA^-(aq)

I        0.12                                     0                    0

C       -x                                        +x                   +x

E      0.12 - x                                x                     x

Ka1= [H3O^+] [HA^-]/[ H2A]

Ka1= x^2/  0.12 - x  

1.0×10^−4 = x^2/  0.12 - x  

1.0×10^−4(0.12 - x ) = x^2

1.2 * 10^-5 - 1.0×10^−4x =  x^2

x^2 +  1.0×10^−4x - 1.2 * 10^-5  = 0

x =0.0034 M

[H3O+] = 0.0034 M

Again;  [H3O+] = [HA^-] = 0.0034 M

          HA^-(aq) + H20(l)    -------> A^-(aq)   + H3O^+

I       0.0034                                  0               0

C       -x                                          + x            +x

E    0.0034 - x                               x                x

Ka2= [A^-] [H3O^+]/[HA^-]

5.0×10^−5 = x^2/ 0.0034 - x  

5.0×10^−5 (0.0034 - x ) = x^2

1.7 * 10^-7 - 5.0×10^−5x =  x^2

x^2 + 5.0×10^−5x - 1.7 * 10^-7 = 0

x=0.00039 M

Learn more about the dissociation of a polyprotic acid:https://brainly.com/question/14481763

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