[HP04 ], [H2P04], and [H3P04] may be eliminated by expressing them in terms of the K constants using Eqs. (33) to (35). The results are:

[ HP0^-] = {HPO22-} = {po4-}{h+} = 7i-o,7nip04 11 ii'1 (39)


m pr>2-] = {H2PO-} = {HP°4-}{H+} = M rH[P°4-]||H+]2

K H3 P04 K H3 P04 K H2P04 k HP04

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