Homework Answers
H2A is diprotic acid
H2A -------------------> HA- + H+
I 0.0700 0 0
C -x +x +x
E ( 0.0700-x) x x
Ka1 = [H+] [ HA- ] / [ H2A ]
=> 4.0*10-6 = [ x2 ] / [0.07-x]
=> 4.0*10-6 * [0.07-x] = [ x2 ]
=> x = 0.00053 M
[H+ ] = 0.00053 M
pH = -log [H+]
pH = -log [0.00053 ]
pH = 3.27
HA- -------------------> A2- + H+
I 0.00053 0 0.00053
C -x +x +x
E ( 0.00053-x) x (0.00053+x)
Ka2 = [H+] [ A2- ] / [ HA- ]
=> 8.2*10-9 = (0.00053+x)(x) / (0.00053-x)
=> x = 8.2*10-9 M
[A2-] = 8.2*10-9
answers =>
pH = 3.27
[H2A] = 0.06947 M
[A2-] = 8.2*10-9 M
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