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Homework #5 Solutions

# Homework #5 Solutions - purdy(sbp456 Homework 5 owers(53885...

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purdy (sbp456) – Homework 5 – flowers – (53885) 1 This print-out should have 37 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Homework 5. Due 21 October 2008 7PM 001 10.0 points A buffer (pH 3.70) was prepared by mixing 1.00 mole of formic acid and 1.00 mole of sodium formate to form an aqueous solution with a total volume of 1.00 L. To 400 mL of this solution was added 50.0 mL of 1.00 M NaOH. What is the pH of this solution? 1. 3.81 correct 2. 4.39 3. 4.25 4. 4.52 5. 3.63 Explanation: [HF] = 1 M [NaOH] = 1 M [F ] = 1 M pH ini = 3.70 Initial condition (ini): n HF = 400 × 1 . 0 = 400 mmol n NaOH = 50 . 0 × 1 . 0 = 50 mmol n Na + = 400 × 1 . 0 = 400 mmol n F - = 400 × 1 . 0 = 400 mmol HF + NaOH Na + + F + H 2 O ini 400 50 . 0 400 400 Δ - 50 - 50 50 50 fin 350 0 450 450 Na + is a spectator ion. HF / F is a buffer system. Since [HF] = [F ] in the original buffer p K a = pH ini = 3 . 70, and pH fin = p K a + log parenleftBigg bracketleftbig F bracketrightbig [HF] parenrightBigg = 3 . 70 + log parenleftbigg 450 350 parenrightbigg = 3 . 80914 002 (part 1 of 2) 10.0 points Find the Q sp when 7 drops of 0 . 018 M Na 2 CO 3 (aq) are added to 10 mL of 0 . 005 M AgNO 3 (aq). The solubility product of Ag 2 CO 3 is 6 . 2 × 10 12 . Assume 20 drops per milliliter. Correct answer: 1 . 42056 × 10 8 . Explanation: V AgNO 3 = 10 mL = 0 . 01 L m Na 2 CO 3 = 0 . 018 M m AgNO 3 = 0 . 018 M V Na 2 CO 3 = (7 drops) 0 . 001 L 20 drops = 0 . 00035 L V tot = 0 . 00035 L + 0 . 01 L = 0 . 01035 L The reaction is 2 Ag + (aq) + CO 2 3 (aq) Ag 2 CO 3 (s) [Ag + ] = (0 . 005 M) · 0 . 01 L 0 . 01035 L = 0 . 00483092 M n CO 2 - 3 = (0 . 00035 L K 2 CO 3 )(0 . 018 M) × 1 mol CO 2 3 1 mol K 2 CO 3 = 6 . 3 × 10 6 mol CO 2 3 [CO 2 3 ] = 6 . 3 × 10 6 mol 0 . 01035 L = 0 . 000608696 M Q sp = [Ag + ] 2 [CO 2 3 ] = (0 . 00483092) 2 (0 . 000608696) = 1 . 42056 × 10 8 003 (part 2 of 2) 10.0 points Determine whether a precipitate form in the solution. 1. Yes correct 2. No Explanation: K sp = 6 . 2 × 10 12 A precipitate will form because Q sp > K sp . 004 10.0 points The acid form of an indicator is yellow and

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purdy (sbp456) – Homework 5 – flowers – (53885) 2 its anion is blue. The K a of this indicator is 1 × 10 5 . What will be the approximate pH range over which this indicator changes color? 1. 5 < pH < 7 2. 8 < pH < 10 3. 3 < pH < 5 4. 9 < pH < 11 5. 4 < pH < 6 correct Explanation: The p K a of this indicator is 5, so the indi- cator will change colors around pH 5. Thus you would expect a color change between pH 4 and pH 6. 005 10.0 points Which of the indicators A: thymol blue B: methyl red C: bromoscresol green could you use for a titration of 0.20 M NH 3 (aq) with 0.20 M HCl(aq)? 1. None of the indicators 2. Only B and C correct 3. Only A 4. Only C 5. A, B, and C 6. Only A and C 7. Only A and B 8. Only B Explanation: The titration converts NH 3 into NH + 4 : NH 3 + HCl NH + 4 + Cl . Then the NH + 4 hydrolyzes via NH + 4 (aq) + H 2 O( ) NH 3 (aq) + H 3 O + (aq) At the stoichiometric point the volume of the solution will have doubled, so the concen- tration of NH + 4 will reduce to 0.10 M. Analyzing the reaction using concentra- tions, NH + 4 + H 2 O( ) NH 3 + H 3 O + 0.10 - 0 0 - x - + x + x 0 . 10 - x - x x K a = K w K b = 1 × 10 14 1 . 8 × 10 5 = 5 . 55556 × 10 10 K a = [NH 3 ] [H 3 O + ] [NH + 4 ] = x 2 0 . 10 - x x 2 0 . 10 x 2 0 . 10 = 5 . 55556 × 10 10 x = 7 . 45356 × 10 6 = [H 3 O + ] pH = - log(7 . 45356 × 10 6 ) = 5 . 12764 The pH range for the indicators is A) thymol blue: 1.2 to 2.8 or 8.0 to 9.6
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Homework #5 Solutions - purdy(sbp456 Homework 5 owers(53885...

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