H09- Buffers 1-solutions

H09- Buffers 1-solutions - patteson (dmp786) H09: Buffers 1...

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Unformatted text preview: patteson (dmp786) H09: Buffers 1 McCord (52405) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which of the five solutions below is most ba- sic? 1. 0.10 M aq NH 3 and 0.20 M NH 4 Cl 2. 0.10 M CH 3 COOH and 0.20 M NaCH 3 COO 3. 0.10 M aq NH 3 and 0.10 M NH 4 Cl cor- rect 4. 0.10 M CH 3 COOH and 0.10 M NaCH 3 COO 5. 0.10 M HClO and 0.10 M NaClO Explanation: All are buffer where pH = p K a + log parenleftbigg [base] [acid] parenrightbigg 002 10.0 points Explain why the salt of a weak acid, as well as the acid itself, must be present to form a buffer solution. 1. The cation from the salt is needed to partially neutralize added base. 2. Actually, a weak acid by itself is a buffer; no salt is needed. 3. The anion from the salt is needed to partially neutralize added base. 4. The anion from the salt is needed to partially neutralize added acid. correct 5. The cation from the salt is needed to partially neutralize added acid. Explanation: The salt of the acid provides the anion which is the conjugate base of the buffer sys- tem: HA + H 2 O A- + H 3 O + This anion A- reacts with any added acid (H 3 O + ) to prevent any appreciable change in pH. 003 10.0 points What is the pH of an aqueous solution that is 0.018 M C 6 H 5 NH 2 ( K b = 4 . 3 10- 10 ) and 0.12 M C 6 H 5 NH 3 Cl? 1. 5.46 2. 4.02 3. 3.81 correct 4. 2.87 5. 10.19 6. 4.63 7. 9.37 8. 8.54 Explanation: 004 10.0 points A buffer solution is made by dissolving 0.45 moles of a weak acid (HA) and 0 . 13 moles of KOH into 850 mL of solution. What is the pH of this buffer? K a = 1 . 6 10- 6 for HA. Correct answer: 5 . 40467 pH. Explanation: n HA = 0 . 45 mol n KOH = 0 . 13 mol K a = 1 . 6 10- 6 for HA You must substract the 0 . 13 moles of KOH from the 0.45 moles of HA because the strong base will neutralize the weak acid. You there- fore would make . 13 moles of A- and be left with 0 . 32 moles of HA. You can now plug this ratio into the equilibrium equation or in the Henderson-Hasselbalch equation to get pH....
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This note was uploaded on 09/28/2009 for the course GE 206E taught by Professor staff during the Spring '09 term at University of Texas at Austin.

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H09- Buffers 1-solutions - patteson (dmp786) H09: Buffers 1...

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