The [H3O+] in step 1 is 0.0034 M while the [H3O+] in step 2 is 0.00039 M
<h3>What is the contribution of each step?</h3>
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:brainly.com/question/14481763
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