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# 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 – find 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 find 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. From the formulas of the two compounds, you can expect that they will react in a one-to-one fashion. So our first order of business will be to determine how many moles of each compound we have. For HClO 4 , we have (32 . 0 mL) parenleftbigg 1 L 1000 mL parenrightbiggparenleftbigg 0 . 200 mol 1 L parenrightbigg = 0 . 00640 mol HClO 4 Likewise, for NaOH, we have (65 . 0 mL) parenleftbigg 1 L 1000 mL parenrightbiggparenleftbigg 0 . 150 mol 1 L parenrightbigg = 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 buffered 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 buffer with a strong base. 15.0 mL are needed to reach the first 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. Cu Cu 2+ + 2 e 2. Cu 2+ + 2 e Cu 3. Zn Zn 2+ + 2 e correct 4. Zn 2+ + 2 e Zn Explanation: 005 10.0 points Ferrous hydroxide is a slightly soluble base. In which of the following would Fe(OH) 2 be most soluble? It is not necessary to know K sp for Fe(OH) 2 . 1. 0.1 M Ca(OH) 2 2. 0.1 M FeCl 2 3. 0.1 M HCl correct 4. 0.1 M KOH Explanation: 006 10.0 points What molar ratio of sodium acetate to acetic acid (NaAc/HAc) should be used in preparing a buffer having a pH of 4.35? ( K a = 1 . 8 × 10 5 for acetic acid.) 1. 1 . 0 : 0 . 40 2. 2 . 0 : 1 . 0
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Exam 3 - Version 050 Exam 3 Holcombe(52460 This print-out...

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