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Henderson-Hasselbach equation 
Fadail Al Haddad
Whatis PH ? 
• [H] = .000 000 04 EQL 
• [H] = 40 n eql 
• -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]
Henderson-Hasslbach equation
Body produce high [H]
Buffer and pk
Henderson-Hasselbach equation
Benefit of this equation
Thank you 
reference ; Guyton&Hall

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Henderson-Hasslbach equation