Lec-31-Exam3Review

Lec-31-Exam3Review - Lec-33-Exam3Review 11/14/2011 Lec-31:...

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Lec-33-Exam3Review 11/14/2011 Acid-Base Equilibrium, Buffers, and Titrations. Solubility. Electrochemistry Chaps. 16,17.5, and 18 1 Lec-31: Third Midterm Review 1 Exam 3, Thursday, November 17 th , 8:30 PM – 10:00 PM Room Assignments Solar ID Room 000000000 – 107757371 Javits 100 Main Floor 107757372 – 999999999 Javits 100 Balcony Review Sessions Monday, Nov. 14, Harriman Hall 137, 7-9 PM Wednesday, Nov. 16, Javits 110, 7-9 PM Acid-Base Calculations             23 2 2 2 0 2 0 HNO aq H O aq HNO HO NO aq 0 0 HNO xx x        2 3 22 0 2 2 0 HNO HNO H NO a x K x x x x   HNO 2 has a ionization constant of K a =4.5×10 –4 in water at 25°C. What is the pH of a solution that is 0.010 mol L –1 in HNO 2 ? If we assume that   2 0 HNO x 2
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Lec-33-Exam3Review 11/14/2011 Acid-Base Equilibrium, Buffers, and Titrations. Solubility. Electrochemistry Chaps. 16,17.5, and 18 2 Acid-Base Calculations     22 00 HNO HNO a K xx x The approximation then gives       31 2 0 4 HNO 0.10 6.71 10 mol L pH 2. 4.5 1 17 0 a x K 3 Acid-Base Reactions of Salts - Hydrolysis 0.200 moles of NaBrO are dissolved in 1.00 L of water. What is the pH of the resulting solution? The ionization constant for HBrO is K a =2.06×10 –9 .         2 0 0 BrO aq B OH aq H HB rO rO aq O BO 0 r 0 x x x    Considering the ionization equilibrium constant of the acid and of its conjugate base:                 2 3 2 HBrO aq H O aq HBrO aq BrO aq BrO aq H O Oa H H q O a b K K  4
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Lec-33-Exam3Review 11/14/2011 Acid-Base Equilibrium, Buffers, and Titrations. Solubility. Electrochemistry Chaps. 16,17.5, and 18 3 Acid-Base Reactions of Salts 0.200 moles of NaBrO are dissolved in 1.00 L of water. What is the pH of the resulting solution? The ionization constant for HBrO is K a =2.06×10 –9 .     3 H O HBrO HBr BrO OH Br O O ab w K KK   Then:   14 6 9 1.0 10 4.85 10 2.06 1 OH H 0 BrO BrO b a w K K K  Replacing the expressions for the concentrations at equilibrium 5 Acid-Base Reactions of Salts we have   0 2 BrO HB O B r rO b x x K   the equation reduces to 2 0 0 BrO BrO bb x K x K  On the assumption that 0 BrO 6
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Lec-33-Exam3Review 11/14/2011 Acid-Base Equilibrium, Buffers, and Titrations. Solubility. Electrochemistry Chaps. 16,17.5, and 18 4 Acid-Base Reactions of Salts 0.200 moles of NaBrO are dissolved in 1.00 L of water. What is the pH of the resulting solution? The ionization constant for HBrO is K a =2.06×10 –9 .
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This note was uploaded on 03/06/2012 for the course CHE 132 taught by Professor Hanson during the Fall '08 term at SUNY Stony Brook.

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Lec-31-Exam3Review - Lec-33-Exam3Review 11/14/2011 Lec-31:...

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