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The concept of buffers in organic chemistry and how the henderson-hasselbalch equation is used to describe the relationship between the ph of a buffered solution, the pka value of a weak acid, and the ratio of the weak acid to its conjugate base. The document also includes examples of how to use the henderson-hasselbalch equation to solve for various components and understand the behavior of functional groups in different ph environments.
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SCHE 2060: Principles of Organic Chemistry Vermont Tech 1
Biological organism and solutions are buffered , meaning that they are able to maintain a stable pH when challenged by addition of acid or base. When added to unbuffered solutions, acid would cause the pH to drop while base would cause the pH to increase. Buffered solutions contain a weak acid and a salt of that weak acid. The acid provides protons to counter any base added to the buffered solution. The salt of the weak acid is a metal cation ionically bound to the anion of the weak salt and that anion counters (or absorbs) the protons of any acid added to the buffered solution. Weak acids create good buffers for pH values close to their pKa values. The Henderson-Hasselbalch equation describes the relationship between:
SCHE 2060: Principles of Organic Chemistry Vermont Tech 2 Examples: A. A buffer is made with 30 mM acetic acid (pKa 4.76) and 40 mM sodium acetate. What is the pH of the buffer? pH = pKa + log (40 mM/30 mM) = 4.76 + 0. 1 25 = 4. B. A protein with exposed aspartic acid side chains is placed in a solution with a pH of 7.0. What percentage of the aspartic acid side chains are protonated? Use the Henderson- Hasselbalch equation! 7.0 = 4.76 + log [carboxylate ion] à 2.24 = log [carboxylate ion] [carboxylic acid] [carboxylic acid] To solve, take the antilog of both sides (see reminders or pointers above). 10 (2.24)^ = [carboxylate ion] à 174 :1 à (174/175)(100) = 99.5% deprotonated [carboxylic acid] C. How much sodium formate do you need to add to 400 mL of 1.00 M formic acid to make a buffer with pH 3.50? (MW sodium formate is 68.01; Ka of formic acid is 1.77 E-4). pKa = - log (1.77 E-4) = 3. 3.50 = (-log 1.77 E-4) + log [sodium formate] à - 0.25 = log [sodium formate] [1.00] To solve, take the antilog of both sides à 10