Question

a). Calculate the fraction of the weak base in each of Adiprotic weak base (B) has pks values of 3.294 pm) and 5.765 ( p its
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Answer #1

pKa1 + pKa2 = 14

pKb1 = 3.294 ---------------> pKa2 = 10.706

pKb2 = 5.765--------------------> pKa1 = 8.235

Ka1 = 5.82 x 10^-9

Ka2 = 1.97 x 10^-11

PH = 9.883

[H+] = 1.309 x 10^-10 M

[H+]^2 = 1.714 x10^-20

Ka1 Ka2 = 1.146 x 10^-19

Ka1 [H+] = 7.62 x 10^-19

[H+]^2 + Ka1 Ka2 + Ka1 [H+]   = 8.936 x 10^-19

B = Ka1 Ka2 / [H+]^2 + Ka1 Ka2 + Ka1 [H+] = 0.128

BH+   = Ka1 [H+] / [H+]^2 + Ka1 Ka2 + Ka1 [H+]    = 0.853

BH2+ = [H+]^2 / [H+]^2 + Ka1 Ka2 + Ka1 [H+] = 0.0192

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