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