Chem works - No 103(260 Panjwani Sameer Exam 3 Due Nov 6 2007 5:00 pm Inst James Holcombe This print-out should have 25 questions Multiple-choice

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No 103 (260) Panjwani, Sameer Exam 3 Due: Nov 6 2007, 5:00 pm Inst: James Holcombe This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. The pKa values for phosphoric acid (H3PO4) are: pK1=2.148; pK2=7.199; pK312.15 pKa for acetic acid (CH3COOH) = 4.74 pKa for benzoic acid (C6H5COOH) = 4.20 001 (part 1 of 1) 10 points If a small amount of a strong base is added to a buffer made up of a weak acid HA and the sodium salt of its conjugate base NaA, the pH of the buffer solution does not change appreciably because 1. the strong base reacts with A- to give HA, a weak acid. 2. no reaction occurs. 3. the strong base reacts with HA to give A- , a weak base. correct 4. the strong base reacts with HA to give AOH and H+ . 5. the Ka of HA is changed. Explanation: 002 (part 1 of 1) 10 points What is the pH of an aqueous solution that is 0.20 M HNO2 ( Ka = 4.3 × 10 -4 ) and 0.20 M NaNO2 ? 1. 10.63 2. 3.67 3. 3.37 correct 4. 4.39 5. 2.37 Explanation: 2 003 (part 1 of 1) 10 points If you wish to increase the solubility of silver benzoate, a preservative, you would 1. add sodium benzoate. 2. decrease the pH . correct 3. add sodium hydroxide. 4. add silver nitrate. 5. add sodium acetate. Explanation: 004 (part 1 of 1) 10 points Consider the Ksp values of the following salts and indicate which of these is least soluble in water. You really don't need to use a calculator to solve this problem. 1. AB2 2. AB3 3. AB4 4. AB Ksp = 1 × 10 -15 Ksp = 1 × 10 -20 Ksp = 1 × 10 -25 correct Ksp = 1 × 10 - 10 Explanation: 005 (part 1 of 1) 10 points Which of the following pairs of compounds could be used to produce a buffer? I) NaOH and HCl II) HF and NaF III) NH3 and NH4 Cl 1. III only 2. I only 3. I and II only 4. II only 5. II and III only correct Explanation: A buffer must contain a weak acid/base Panjwani, Sameer Exam 3 Due: Nov 6 2007, 5:00 pm Inst: James Holcombe conjugate pair. They can be made by mixing a weak acid or a weak base with a salt containing the conjugate of that weak acid or base or by adding a strong acid/base to the weak base/acid which reacts to produce the conjugate. NaOH and HCl are both strong and will not, together, produce a buffer. However, buffers are produced by the weak acid HF and its salt NaF, and the weak base (NH3 ) and its salt NH4 Cl. 006 (part 1 of 1) 10 points 50.0 mL of 0.0018 M aniline (a weak base) is titrated with 0.0048 M HNO3 . How many mL of the acid are required to reach the equivalence point? 1. Bad titration since HNO3 is not a strong acid. 2. 133 mL 3. Need to know the Kb of aniline. 4. 4.21 mL 5. 18.8 mL correct Explanation: Vaniline = 50 mL [Aniline] = 0.0018 M [HNO3 ] = 0.0048 M Aniline is a monobasic base (i.e., it produces one OH- in solution ). Thus you can expect that aniline and HNO3 will react in a one-to-one fashion. With this ratio, we can determine
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This note was uploaded on 09/25/2010 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.

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Chem works - No 103(260 Panjwani Sameer Exam 3 Due Nov 6 2007 5:00 pm Inst James Holcombe This print-out should have 25 questions Multiple-choice

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