Ch 302 hw 6 solution - reyes (mr29667) H06: Buffers and...

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Unformatted text preview: reyes (mr29667) H06: Buffers and more Mccord (90540) 1 This print-out should have 28 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 CH 3 COOH and 0.20 M NaCH 3 COO 2. 0.10 M HClO and 0.10 M NaClO 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 aq NH 3 and 0.20 M NH 4 Cl 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 acid. 2. The cation from the salt is needed to partially neutralize added base. 3. Actually, a weak acid by itself is a buffer; no salt is needed. 4. The anion from the salt is needed to partially neutralize added base. 5. The anion from the salt is needed to partially neutralize added acid. correct 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. 2.87 2. 5.46 3. 9.37 4. 10.19 5. 8.54 6. 4.02 7. 4.63 8. 3.81 correct Explanation: 004 10.0 points A buffer solution is made by dissolving 0.45 moles of a weak acid (HA) and 0 . 33 moles of KOH into 730 mL of solution. What is the pH of this buffer? K a = 4 . 2 10- 6 for HA. Correct answer: 5 . 81608 pH. Explanation: n HA = 0 . 45 mol n KOH = 0 . 33 mol K a = 4 . 2 10- 6 for HA You must substract the 0 . 33 moles of KOH from the 0.45 moles of HA because the strong base will neutralize the weak acid. You there- fore would make . 33 moles of A- and be left with 0 . 12 moles of HA. You can now plug this ratio into the equilibrium equation or in the Henderson-Hasselbalch equation to get pH. 005 10.0 points reyes (mr29667) H06: Buffers and more Mccord (90540) 2 Which of the following mixtures will be a buffer when dissolved in a liter of water? 1. 0.2 mol HBr and 0.1 mol NaOH 2. 0.3 mol NaCl and 0.3 mol HCl 3. 0.1 mol Ca(OH) 2 and 0.3 mol HI 4. 0.4 mol NH 3 and 0.4 mol HCl 5. 0.2 mol HF and 0.1 mol NaOH correct Explanation: Eliminate answers that are obviously incor- rect. The choice with 0.2 mol HBr and 0.1 mol Ca(OH) 2 are strong acids and strong bases respectively; therefore, NOT buffers. The choice with 0.3 mol NaCl is a combina- tion of spectator ions and a strong acid; this does not form a buffer. Remaining for cal- culation are choices with 0.4 mol NH 3 and 0.2 mol HF. Now perform the neutralizaton calculations on the remaining possibilities: Choice with 0.4 mol NH 3 NH 3 + H + NH...
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This note was uploaded on 08/07/2010 for the course CH 90540 taught by Professor Mccord during the Summer '10 term at University of Texas at Austin.

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Ch 302 hw 6 solution - reyes (mr29667) H06: Buffers and...

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