Homework #5 Solutions - purdy (sbp456) Homework 5 flowers...

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Unformatted text preview: 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 . 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 . 400 400 - 50- 50 50 50 fin 350 450 450 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) . 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) . 01 L . 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 . 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 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)?...
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This note was uploaded on 11/06/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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Homework #5 Solutions - purdy (sbp456) Homework 5 flowers...

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