Isoionic point [H+] = √ [(K1K2F + K1Kw) / K1 + F)]
[H+] = √ [(10-2.37 x 10-9.33 x 0.03479) + 10-2.37 x 1.0 x 10-14) / (10-2.37 + 0.03479)] = 0.00000133357
pH = -log[H+] = -log[ 0.00000133357 ] = 5.87
pH = 5.87 for isoionic point.
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Isoelectric point pI = 1/2 (pKa1 + pKa2) = 1/2 (2.37 + 9.33) = 5.85
Isoelectric pH = 5.85
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