Study Questions MT1

# Study Questions MT1 - BIS 102 Prof C Gasser 36 SOLUTIONS TO...

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BIS 102, Prof. C. Gasser 36 SOLUTIONS TO : SAMPLE STUDY PROBLEMS - This set of problems is provided as a study aid and is not to be turned in. These questions are representative of the types of questions to expect on the midterms and final exam and are derived from exams given in this course by a number of different instructors in past years. These sets of practice problems are not meant to reflect the actual length of exams, and represent only a subset of the material presented in each section. Be sure to practice showing your work clearly and concisely to help in obtaining full credit or partial credit on the real exams! BioSci 102, Sample Study Problems for First Midterm 1. What two factors determine how "good" (i. e. effective) a particular buffer solution is for maintaining a given pH? a) How close the given pH is to the pK a of the buffering agent (the closer the better). b) The concentration of the buffering agent. 2. The following question has two parts. a) What is the final pH of a solution obtained by mixing 250 ml of 0.3 M acetic acid with 300 ml of 0.2 M KOH? (pK b of acetate = 9.24). (Show your work!) Moles of acetic acid = 0.25 l X 0.3 M = 0.075 moles Adding 0.3 l X 0.2 M = 0.06 moles of OH- to this solution will convert 0.06 moles of acetic acid to 0.06 moles of acetate, 0.015 moles of protonated acetic acid will remain. pK a = 14 - pK b = 14 - 9.24 = 4.76 pH = pK a + log [CH 3 CO 2 - ] [CH 3 CO 2 H] = 4.76 + log 0.06 moles / 0.55 l 0.015 moles / 0.55 l = 4.76 + log 0.06 moles 0.015 moles = 5.36 pH = 5.36 b) What is the net charge on the acetic acid/acetate molecules at this pH? Since acetic acid is neutral, the net charge is determined by the fraction of the moleucles in the negatively charged acetate form. moles of acetate total mole of acetate and acetic acid = 0.06 moles 0.075 moles = 0.8 net charge = 0.8 X -1 = -0.8

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BIS 102, Prof. C. Gasser 37 3. Explain how you would make 2 liters of a 0.015 M buffer at pH 4.4 using crystalline disodium glutamate (MW = 191) and 0.9 M HCl. (use pK a values: α -COOH = 2.2, side chain = 4.2, and α - NH 2 = 9.7). Glu
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## This note was uploaded on 09/08/2010 for the course BIS BIS 102 taught by Professor Hilt during the Spring '09 term at UC Davis.

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Study Questions MT1 - BIS 102 Prof C Gasser 36 SOLUTIONS TO...

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