Exam 3 - Version 050 Exam 3 Holcombe (52460) This print-out...

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Version 050 – Exam 3 – Holcombe – (52460) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Calculate the pH oF the solution resulting From the addition oF 32.0 mL oF 0.200 M HClO 4 to 65.0 mL oF 0.150 M NaOH. 1. 7.00 2. 1.29 3. 1.46 4. 13.18 5. 12.54 correct 6. 0.82 7. 12.71 Explanation: V HClO 4 = 32.0 mL [HClO 4 ] = 0.200 M V NaOH = 65.0 mL [NaOH] = 0.150 M Here it’s important to fnd out which oF these two species (HClO 4 and NaOH) is in ex- cess. The one that is in excess will determine the pH oF this solution. ±rom the Formulas oF the two compounds, you can expect that they will react in a one-to-one Fashion. So our frst order oF business will be to determine how many moles oF each compound we have. ±or HClO 4 , we have (32 . 0 mL) p 1 L 1000 mL Pp 0 . 200 mol 1 L P = 0 . 00640 mol HClO 4 Likewise, For NaOH, we have (65 . 0 mL) p 1 L 1000 mL Pp 0 . 150 mol 1 L P = 0 . 00975 mol NaOH So when HClO 4 and NaOH react, all oF the HClO 4 will be consumed (it’s the limiting reagent) and 0 . 00975 mol - 0 . 00640 mol = 0 . 00335 mol oF NaOH will remain. This 0.00335 mol ex- cess oF NaOH will determine the pH oF this solution. The solution now is 32 . 0 mL + 65 . 0 mL = 97 . 0 mL and, since NaOH is a strong base ( i.e. , it is completely dissociated), it contains 0.00335 mol OH . [OH ] is then [OH ] = 0 . 00335 mol 0 . 0970 L = 0 . 0345 M , which means that the pOH oF this solution is pOH = - log [OH ] = - log 0 . 0345 = 1 . 46 However, we wanted pH. We can use the equa- tion that relates pH to pOH to get the pH: pH = 14 - pOH = 14 - 1 . 46 = 12 . 54 002 10.0 points You have a solution that is bu²ered at pH = 2.0 using H 3 PO 4 and H 2 PO 4 (p K a1 = 2 . 12; p K a2 = 7 . 21; p K a3 = 12 . 68). You decide to titrate this bu²er with a strong base. 15.0 mL are needed to reach the frst equivalence point. What is the total volume oF base that will have been added when the second equivalence point is reached? 1. < 30 mL 2. 30 mL 3. A second equivalence point in the titra- tion will never be observed. 4. > 30 mL correct Explanation: 003 10.0 points You add a small amount oF HCl to a solution oF 0.20 M HBrO and 0.20 M NaBrO. What do you expect to happen? 1. The H + ions will react with the HBrO molecules.
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Version 050 – Exam 3 – Holcombe – (52460) 2 2. The [OH ] will decrease slightly. correct 3. The K a for HBrO will increase. 4. The pH will increase slightly. Explanation: 004 10.0 points Consider the cell Zn | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu What reaction is occurring at the anode? 1.
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This note was uploaded on 09/27/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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Exam 3 - Version 050 Exam 3 Holcombe (52460) This print-out...

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