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exam2key - Chemistry 41 Exam 2 June 2, 2010 Honor Pledge: I...

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Chemistry 41 Exam 2 June 2, 2010 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 10 15 25 30 10 100 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. (10 pts) What is the pH of a 2.08 x 10 -8 M solution of HNO 3 , a strong acid? [H + ] = [NO 3 - ] + [OH - ] = 2.08 x 10 -8 + K w /[H + ] [H + ] 2 – 2.08 x 10 -8 [H + ] - K w = 0 solving the quadratic equation [H + ] = 1.11 x 10 -7 pH = 6.955
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3. (15 pts) When 100.0 mL of a weak acid was titrated with 9.381 x 10 -2 M NaOH, 27.63 mL was required to reach the equivalence point. At this point the pH was 10.99. What was the pH when 19.47 mL of NaOH had been added?
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exam2key - Chemistry 41 Exam 2 June 2, 2010 Honor Pledge: I...

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