153A_MIDTERM_F02_KEY

153A_MIDTERM_F02_KEY - Seat Last initial Chemistry and...

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Seat # ________ Last initial ________ Chemistry and Biochemistry 153A Fall 2002 Midterm Examination Question # Value Score 1 10 2 22 3 8 4 5 5 6 6 24 7 11 8 8 9 32 10 19 TOTAL 145 _________ KEY ______________ Print your full name ( last name first) ______________________________ Student ID Number ______________________________ Print your TA's name ______________________________ Discussion section # [or time & day] I have read the instructions below. ______________________________ Signature INSTRUCTIONS READ EACH QUESTION CAREFULLY! SHOW YOUR CALCULATIONS! WRITE YOUR LAST NAME IN THE SPACE PROVIDED ON EVERY PAGE OF THE EXAM. NO CREDIT WILL BE GIVEN FOR ANYTHING WRITTEN ON THE BACK OF A PAGE on ANYWHERE OTHER THAN THE SPACE PROVIDED FOR EACH ANSWER. ANSWERS MUST BE BRIEF AND TO THE POINT. WHERE A WORD LIMIT IS SPECIFIED, NO CREDIT WILL BE GIVEN FOR THE PART OF THE ANSWER IN EXCESS OF THE LIMIT. Numbers in parentheses represent the point values.
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Name ________ KEY ______________ 2 1. (10) a. 1) (3) You have a beaker containing l liter of pure water. The molecular weight of water is 18.016. Calculate the molarity of water in your beaker. Show your calculations and reasoning. (1g water/ml X 1000ml/liter) /18.016g/mol = 55.5 mol/liter = 55.5M 2) (2) If 200 grams of water evaporate from the beaker in a. (1) above, what will be the molarity of the remaining water. Still 55.5 M b. (3) The expression for the equilibrium constant for the ionization of water is frequently written as follows. [H + ] [OH - ] [H 2 O] K eq = Given that the equilibrium constant for the ionization of water the expression written above is 1.8 x 10 -16 M, calculate the K w , the “ion product” of water. (Show all of your work.) [H + ] [OH - ] [H 2 O] [H + ] [OH - ] [H + ] [OH - ] K eq = l.8 x 10 -16 M= = 55.5 M K w = ( l.8 x 10 -16 M ) ( 55.5 M) = K w = 10 -14 M 2 c. (2) Calculate the pK a of a weak acid if the concentration of the protonated form is 200 times that of the deprotonated form at pH 8. (Show all of your work.) pH = pK a + log ([A - ]/[HA]) pK a = pH - log ([A - ]/[HA]) pK a = 8 - log (1/200) = 8 – (-2.3) = 8 + 2.3 = 10.3 pK a = 10.3
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Name ________ KEY ______________ 3
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Name ________ KEY ______________ 4 2. (22) One of the titration curves shown below represents a single five-carbon α -amino acid, and the other represents a homodimer of a three-carbon α -amino acid. pH equivalents of OH - 1 3 4 5 6 7 8 9 10 11 1 1.5 2 2.5 3 3.5 4 .5 12 14 2 pH equivalents of OH - 1 3 4 5 6 7 8 9 10 11 1 1.5 2 2.5 3 3.5 4 .5 12 14 2 a -carboxyl g -carboxyl a -carboxyl terminal a -amino a -amino terminal sulfhydryls of cys side chain ___ Glutamic Acid __or___(Glutamate) _______ __ di-Cysteine on Cysteinyl-Cysteine______ a. (4) Give the name of each compound on the line directly below its titration curve. b. (6) On both titration curves above, assign each buffering zone to a specific
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153A_MIDTERM_F02_KEY - Seat Last initial Chemistry and...

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