I am stuck on one question of my home work: Vitamin C has a pk of 4.2. according to the henderson hasselbach's equation, how much of vitamin C molecules in human blood (pH = 7.4) are protonated? How would i go about solving this?? I need it in a simple sort of way, if possible!
Chemistry - 1 Answers
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Lancenigo di Villorba (TV), Italy HASSELBACH equation treats onto WEAK ACIDs (or WEAK BASEs) DISSOCIATION IN AQUEOUS MEDIA. Well, this equation comes from LOGARITHMIC ARRANGEMENT OF CHEMICAL EQUILIBRIUM OF IONIC DISSOCIATION for a weak acid as Vitamin C is. So, Vitamin C may be named as C6H8O6 ; IT PLAYS LIKE AN ACID ONCE IT TOOK PART TO THE EQUILIBRIUM C6H8O6(aq) <---> C6H7O6-(aq) + H+(aq) K = 6.0E-5 = |C6H7O6-| * |H+| / |C6H8O6| HASSELBACH LOGARITHM will be the equation you look for pH = pK - LOG[|C6H8O6| / |C6H7O6-|] and once you play at pH=7.4, I MAY CALCULATE THE CHEMICAL RATIO ABOUT VITAMIN C |C6H8O6| / |C6H7O6-| = ALOG[pK - pH] = ALOG[4.2 - 7.4] = = 6.0E-4 THE CALCULATION SAYS YOU THAT A PERCENT FRACTION 6.0E-4 / (6.0E-4 + 1) = 0.06% HAS TO RESULT THE PROTONATED PART OF VITAMIN C IN A MEDIUM HAVING pH=7.4. I hope this helps you.
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