Info

[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)

HPO4 HPO4KHPO4 HPO4KHPO4

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

K H3 P04 K H3 P04 K H2P04 k HP04

Was this article helpful?

0 0
Healthy Chemistry For Optimal Health

Healthy Chemistry For Optimal Health

Thousands Have Used Chemicals To Improve Their Medical Condition. This Book Is one Of The Most Valuable Resources In The World When It Comes To Chemicals. Not All Chemicals Are Harmful For Your Body – Find Out Those That Helps To Maintain Your Health.

Get My Free Ebook


Post a comment