Henderson-Hasslbach equation
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Henderson-Hasselbach equation Fadail Al Haddad
Transcript of Henderson-Hasslbach equation
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
Henderson-Hasselbach equation
Benefit of this equation
Thank you
reference ; Guyton&Hall
![E:HandoutsChemisty Key Equations and Conversionscollision theory) K, and (14.4) pH Scale (15.5) pH = -log[H30+] Henderson—Hasselbalch Equation (16.2) Ptotal pa pc Mole Fraction (5.6)](https://static.fdocuments.net/doc/165x107/5b029df37f8b9af1148ff41a/ehandoutschemisty-key-equations-and-conversions-collision-theory-k-and-144.jpg)
![Control in the laboratory and the dyehouse · By rearranging equation (1) we obtain the pH of a buffer solutio n: pH = pK a ' + log 10 [A?] / [HA] (Henderson-Hasselbach Equation).](https://static.fdocuments.net/doc/165x107/5e44f214b3071435ef46331e/control-in-the-laboratory-and-the-dyehouse-by-rearranging-equation-1-we-obtain.jpg)
