exam2B

# exam2B - Chemistry 41 Exam 2 Honor Pledge: I have neither...

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Chemistry 41 Exam 2 Honor Pledge: I have neither given or received help on this exam, nor have I seen anyone else do so. __________________________________________(Signed) 1. 2. 3. 4. 5. 6. Total _________ 10 15 10 10 15 40 100 SHOW YOUR WORK CIRCLE NUMERICAL ANSWERS 1. (10 pts) The pH of a 0.10 M solution of B is 10.80. What is K b for this base? B + H 2 O BH + + OH - pH 10.80 [H + ]=1.6 x 10 -11 [OH - ]=6.3 x 10 -4 M init .10 equil. .10-x x x (= 6.3 x 10 -4 ) K b = (6.3 x 10 -4 ) 2 /.1 = 4.0 x 10 -6 2. (15 pts) How many milliliters of 0.246 M HNO 3 should be added to 213 mL of 6.66 x 10 -3 M ethylamine (pK b = 3.364) to make a solution with a pH = 10.520. B + H + BH + B= (.213L)(6.66 x 10 -3 M) = 1.42 x 10 -3 moles init 1.42 x 10 -3 x pK a = 14-pK b =10.636 equil. 1.42 x 10 -3 -x 0 x [Acid] [Base] log + pK = pH a [x] ] x - 10 x 1.42 [ log + 636 . 10 = 52 . 10 -3 X = 8.04 x 10 -4 8.04 x 10 -4 moles/ .246 M = 3.27 x 10 -3 L = 3.27 mL

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3. (10 pts) What is the pH of a 5.62 x 10 -8 M solution of HNO
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## This note was uploaded on 03/24/2009 for the course CHEM 241 taught by Professor Tiani during the Summer '08 term at UNC.

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exam2B - Chemistry 41 Exam 2 Honor Pledge: I have neither...

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