Henderson-Hasselbach equation
Fadail Al Haddad
• [H] = .000 000 04 EQ\L • [H] = 40 n eq\l• -log [H] = 7.4• Pour of [H] = 7.4• Ph= 7.4
SO INCREASE PH = DECREASE [H] because( – log) ALSO one different in ph means 10 different in [H]
Whatis PH?
Body produce high [H]
Buffer and pk
Benefit of this equation
Thank you
reference ; Guyton&Hall
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