Exam 2-solutions[1]

Exam 2-solutions[1] - Version 199 – Exam 2 – vanden...

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Unformatted text preview: Version 199 – Exam 2 – vanden bout – (51640) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When mixed together which two solutions would yield a buffer with the largest capacity? 1. 100 mL of 1M HCl and 200 mL of 1M NaF correct 2. 10 L of 0.2M HCl and 20L of a 0.2M NaF 3. 100 mL of 1M HCl and 100 mL of 1M NaF 4. 1L of 1M HCl and 1L of 1M NaCl 5. 100 mL of 0.1M HCl and 100 mL of 0.1 M NaF 6. 100 mL of 2M HCl and 200 mL of 2M NaCl Explanation: A buffer requires substantial amounts of both the protonated and deprotonated form of a weak acid. In this case the solution should be made up of a weak base (NaF) that is partially neutralized with a strong acid (HCl). The highest buffer capacity will come from the solution with the largest concentrations. This is 100 mL of 1M HCl and 200 mL of 1M NaF. Mixtures of HCl and NaCl will not form a buffer since Cl − is not a weak base. 002 10.0 points Which of I) HCl II) HF III) LiOH IV) HClO 2 V) HNO 3 are strong acids or strong bases in water? 1. I, III, and V only correct 2. I, II, IV, and V only 3. All of the compounds 4. I, III, IV, and V only 5. I, II, III, and V only Explanation: 003 10.0 points Consider the titration of 50.0 mL of 0.0200 M HClO(aq) with 0.100 M NaOH(aq). What is the formula of the main species in the solution after the addition of 10.0 mL of base? 1. ClOH 2. HClO 2 3. ClO − correct 4. NaOH 5. ClO 2 Explanation: 004 10.0 points How much acetic acid ( K a = 1 . 8 × 10 − 5 ) must be used to prepare a liter of solution having a pH of 2.75? 1. 0.435 moles 2. 5.69 moles 3. 0.178 moles correct 4. 9.88 moles 5. 0.0101 moles Explanation: pH =- log[H + ] = 2 . 75 so [H + ] = log − 1 (- 2 . 75) = 1 . 778 × 10 − 3 mol HA ⇀ ↽ H + + A − Substitute [H + ] = [A − ] = 1 . 778 × 10 − 3 into the equation for K a : Version 199 – Exam 2 – vanden bout – (51640) 2 K a = [H + ][A − ] [HA] [HA] = [H + ][A − ] K a = ( 1 . 778 × 10 − 3 ) 2 1 . 8 × 10 − 5 ≈ . 178 mol 005 10.0 points What would be the pH of a solution pre- pared from 200 mL of 5 M HOBr and 200 mL of 1 M NaOBr? The K a of hypobromous acid is 2 × 10 − 9 . 1. 8 correct 2. 6 3. 4 4. 10 5. 7 Explanation: [H + ] = K a ( C a /C b ) = ( 2 × 10 − 9 ) (2 . 5 M /. 5 M) = 10 − 8 pH = 8 006 10.0 points For a triprotic acid, H 3 A, with pK a values of 2.5, 6.5 and 10.5, what species is present in highest concentration when 1M H 3 A is buffered at pH 7? 1. A 3 − 2. H 3 A 3. H 2 A − 4. H + 5. HA 2 − correct Explanation: Since the pH is greater than pK a 2 , but much closer to pK a 2 than to pK a 3 , the most abun- dant species in solution will be the second deprotonated species, HA 2 − ....
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This note was uploaded on 03/27/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.

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Exam 2-solutions[1] - Version 199 – Exam 2 – vanden...

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