Chem 112, Test 3 -- March 18, page 1 Name

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Chem 112, Test 3 March 18, 2005 page 1 Name Please enter your name legibly. You are allowed 2 consecutive hours (no cutting classes) for this test. You may use your calculator but NO extra scratch paper. Be sure to sign the pledge at the end of the test. I: Multiple Choice (28 pts) 2 points each. Choose the best answer and write your answer legibly to the left of the question. (No circling or filling in the choices; too confusing) Given the chemical reaction: 2Fe(s) + 3H 2 O(l) = Fe 2 O 3 (s) + 3H 2 (g) ; ΔH rxn = 764 KJ The next 4 questions pertain to the reaction given above: 1. What is the system here? a) the whole reaction b) the reactants c) the products d) only the solid substances 2. What is the criterion by which we decide whether or not this reaction is spontaneous a) ΔH rxn must be positive (+) b) ΔG rxn must be positive (+) c) ΔS system must be positive (+) d) ΔG rxn must be negative () e) ΔS surr must be positive (+) f) ΔH rxn must be negative () 3. What would you predict would be the sign to be attributed to ΔS surr? a) positive (+) b) negative () c) no change d) there is not enough information here to make this prediction 4. Which category fits this reaction when predicting its spontaneity? a) Spontaneous because entropy of system and entropy of surroundings both increase b) Not spontaneous because entropy of system and entropy of surroundings both decrease c) Maybe spontaneous, depending on temperature, because entropy of system decreases, but entropy of surroundings increases d) Maybe spontaneous, depending on temperature, because entropy of system increases, but entropy of surroundings decreases. 5. Given these values of ΔH and ΔS for 4 chemical reactions, which of the following will NOT be spontaneous at constant T and P? ) ΔH = +25 kj, ΔS = 5.0 J/K, T = 300 K ) ΔH = +25 kj, ΔS = +100.0 J/K, T = 300 K ) ΔH = 10 kj, ΔS = +5.0 J/K, T = 298 K ) ΔH = 10 kj, ΔS = 40 J/K, T = 260 K ) b and c ) a and d ) all of the above are spontaneous

Chem 112, Test 3 March 18, 2005 page 2 6. The unit associated with Water Hardness is ppm CaCO 3. One ppm CaCO 3 is a) 1 mg CaCO 3 per Liter of solution b) 1 g CaCO 3 per Liter of solution c) 1 millimole CaCO 3 per Liter of solution d) the same as 1 N CaCO 3 7. What is the greatest advantage of a polydentate ligand over a monodendate ligand in a complexformation titration? a) Increased solubility b) Increased specificity of reaction c) Increased speed of reaction d) Increased achievability of stoichiometry of titration reaction e) All of the above 8. Given a solution of the salt of a weak base and a weak acid (such as Ammonium Benzoate, NH 4 OBz ), which of the following reactions is the defining reaction, the reaction which ultimately sets the ph of the solution? + a) NH 4 + H 2 O = H 3 O + + NH 3 b) OBz + H 2 O = HOBz + OH + c) NH 4 + OBz = NH 3 + HOBz d) Since you have the salt of a Weak Base and a Weak Acid, the ph will be set by the K w reaction since you have a solution of a salt in water. The ph will turn out to be 7.00 9. In your Hardness of Water determination, the buffer action reaction which actually occurs during the titration is: a) OH + H 2 Y 2 = H 2 O + HY 3 b) Ca +2 + H 2 Y 2 = CaY 2 + 2H + + c) NH 4 + H 2 O = NH 3 + H 3 O + d) NH 3 + H 3 O + = + NH 4 + H 2 O e) none of the above 10. If your Hard Water sample happens to have a great deal of Iron (in the form of Fe 3+ ) in it, you will have a great deal of trouble with the titration. The reason for this will be: a) An uncertain stoichiometry, as Iron is +3 in charge, while the EDTA titrant (written as H 2 Y 2 ) is only 2 in charge. You ll need 3 EDTA s for 2 Irons and this is not easy to achieve under laboratory conditions, even if it is easy to write on the blackboard. b) The IronIndicator complex will be tied so tightly that you ll not be able to release the final indicator color c) The IronIndicator complex will be tied too loosely to allow you to completely titrate all the hard ions before the color changes d) The presence of Fe 3+ interferes with the NH 4 + /NH 3 buffering system, making the color change hard to control e) Such added presence of an additional hard ion (other than the expected Ca 2+ and Mg 2+ ) along with the low concentration of EDTA will cause you to have to refill the buret. f) 2 of the above g) none of the above

Chem 112, Test 3 March 18, 2005 page 3 The next four (4) questions all refer to the Phosphoric Acid system, H 3 PO 4 (pk 1 = 2.12, pk 2 = 7.21, and pk 3 = 12.33) 11. Which, if any, of the following formulas, when mixed together, will cause a major reaction to occur? a) H 3 PO 4 and H 2 PO 4 b) H 3 PO 4 and PO 4 3 c) H 3 PO 4 and H 3 O + d) H 2 PO 4 and HPO 4 2 e) H 3 PO 4 and HPO 4 2 f) PO 4 3 and H 3 O + g) 3 of the above h) 2 of the above i) none of the above 12. If it was desired to prepare a buffer of ph value = 7.00 using the appropriate members of the Phosphoric acid family, which member of the Phosphoric acid family should be used? j) H 3 PO 4 and H 2 PO 4 k) H 3 PO 4 and PO 4 3 l) H 3 PO 4 and HPO 4 2 m) H 2 PO 4 and HPO 4 2 n) H 3 PO 4 and HPO 4 2 o) PO 4 3 and HPO 4 2 p) Only H 2 PO 4 q) Only HPO 4 2 r) Only PO 4 3 13. Soft drinks, such as CocaCola, were originally known as phosphates as they were prepared with members of the phosphoric acid family, so it is actually quite acidic. By actual measurement CocaCola is said to have a ph of about 3.0. If such is the case, which member of the phosphoric acid family is present in largest amount in CocaCola? s) H 3 PO 4 and H 2 PO 4 in equal amounts t) H 3 PO 4 u) H 2 PO 4 and HPO 2 4 in equal amounts v) H 2 PO 4 2 w) HPO 4 3 x) PO 4 3 2 y) PO 4 and HPO 4 in equal amounts 14. We ve introduced the notion of acidic protons being neutralizable but nottitratable. Which, if any, of the phosphoric protons would fit in this category of being neutralizable but not titratable? a) the pk 1 proton b) the pk 2 proton c) the pk 3 proton d) both the pk 1 and pk 2 protons e) the pk 2 and pk 3 protons f) All Phosphoric Acid s protons are neutralizable but not titratable g) None of Phosphoric Acid s protons are titratable

Chem 112, Test 3 March 18, 2005 page 4 II (51 points) Short Answer: Answer each question in the space provided. The Hard Water Analysis (6 pts) 1. In your standardization of the EDTA titrant, you used a strategy that differed from the KHP case. In the KHP analysis you simply weighed out by difference the primary standard directly into the titration flasks for each standardization titration. In the Hard Water case you made up a master solution of CaCO 3 primary standard, from which you drew 25.00 ml pipetfuls for standardization purposes. a) What is the significant advantage that you gained in the Hard Water case by employing the standardization strategy described above? b) What is the disadvantage associated with the Hard Water standardization strategy? In other words, why didn t you carry out the KHP standardization in the same manner you did in the hard water case? Thermodynamics. (45 pts) 1. What is the Clausius definition of entropy? 2. What is the Boltzman definition of entropy? 3. For a box of gas: Describe a macrostate : Describe a microstate: 4. Consider a hot body at a temperature T h and a cold body at a temperature T c. They are put in thermal contact long enough for a small quantity of heat q to be transferred. Assume that q is too small to significantly change the temperatures. a) What is the entropy change of the hot body? b) What is the entropy change of the cold body? c) What is the total entropy change for this process? Is it always positive? Why? 5.a) If you put a partition a box of gas into two equal parts, what happens to the number of microstates for the system?

Chem 112, Test 3 March 18, 2005 page 5 b) If you double the volume of a box of gas, what happens to the number of microstates? c) Is the entropy change for each of these two processes positive or negative? 6. The entropy of a system increases during an endothermic process. Under what conditions will this process be spontaneous? 7. A chemical reaction in a closed container is at equilibrium. Assuming we add nothing else to the container, what can you say about the free energy change for any process that moves us away from this equilibrium? 8. Assume we add some products to the container. Using the relations between the free energy, Q, and K, explain what happens and why that process is spontaneous. III. PROBLEMS (61 pts) SHOW ALL WORK. NO MAGIC PERMITTED. (Watch quality of all processed values!) 1. You are given a Coke Can (355 ml) which contains a pint sample of hard water to be analyzed. You are told that it s known to be very hard. Your analysis goes as follows: you pipet out a 100.0 ml aliquot from the Coke can and dilute up to 250.0 ml in a volumetric flask. You then take 100.00 ml of the resulting diluted solution and titrate it with your standardized EDTA titrant, whose molarity is known to be 0.01005 M. You find it takes 45.23 ml to reach the Calmagite end point. What is the hardness of the water, expressed in ppm CaCO 3? (MW = 100.0) 2. You prepare a buffer solution and measure 200.0 ml of this solution in a ph meter, obtaining a ph value of 6.13. You decide to check the buffer capacity of this solution toward added base by adding dropwise a solution of 3.0M NaOH (MW = 40.00). a) At what value of ph do you intend to stop adding NaOH, as you will have reached the buffering capacity of this buffer? b) Suppose you find you need to add 56 drops of the 3.0 M NaOH solution to reach the value in part a). How many millimoles of hydroxide ion will you have added (Assume that 20 drops = 1.0 ml)

Chem 112, Test 3 March 18, 2005 page 6 c) What is the buffer capacity of this buffer in units of milligrams of Sodium Hydroxide per ml of buffer? 3. Hess s Law: a) Hydrogen can be generated from water and coal by the reaction: C + H 2 O CO + H 2 (This reaction will be called the main reaction) Determine Hº for the main reaction from the following heats of reaction data: C + ½ O 2 CO C + O 2 CO 2 CO + ½ O 2 CO 2 H 2 + ½ O 2 H 2 O Hº= 111 kj Hº= 394 kj Hº= 283 kj Hº= 242 kj b) Some of the equations above give the heat of formation for different compounds. Using those equations, what are the heats of formation for H 2 O, CO, and CO 2? c) Using any results you have obtained so far, show that we can calculate the heat of reaction for the main reaction using heats of formation.

Chem 112, Test 3 March 18, 2005 page 7 4. Consider the hypothetical tetraprotic organic acid, H 4 ORG (MW = 236.0, pk 1 = 3.17, pk 2 = 6.88, pk 3 = 7.60, and pk 4 = 12.10). Assume you start with 10.00 ml of a 0.10 M solution of this acid in water: a) Which members of this family would be expected to be present in major amounts in a solution whose ph = 7.0? Which SINGLE member of this family would be present in largest amount? b) Sketch the ph curve which would result when a complete buretful (i.e. 50 ml) of 0.10 M NaOH titrant were added to this solution. Be sure to identify on your titration curve all equivalence points (and their ph values) that could be achieved by a titration process. Also show on your titration curve where pk values are to be found. Make your identifications clear, using ph and either mmoles or ml values as coordinate values on the titration curve. The following table will prove useful in all of the upcoming Acid/Base characterizations. Use of the ph line is encouraged. Name Formula MW Other Formic Acid HCOOH 46.00 pk a = 3.77 Acetic Acid HOAc 60.00 pk a = 4.74 Phosphoric Acid H 3 PO 4 98.00 pk 1 = 2.12, pk 2 = 7.21, pk 3 = 12.33 Sodium Chloride NaCl 58.50 Sodium Phosphate Na 3 PO 4 164.0 Sodium Hydroxide NaOH 40.00 Ammonia NH 3 17.00 pk b = 4.75 Methyl Amine CH 3 NH 2 31.00 pk b = 3.36 Sodium Formate HCOONa 68.00 Sodium Acetate NaOAc 82.00 Hydrochloric Acid HCl 36.50 Methyl Ammonium Formate (CH 3 NH 3 ) + (HCOO) 77.00

Chem 112, Test 3 March 18, 2005 page 8 5. Determine the ph of the following systems: a) The system that results when 75.00 ml of 0.20 M Methyl Amine, CH 3 NH 2, has added to it 100.0 ml of 0.15 M HCl b) The system that results when 50.00 ml of 0.20 M Methyl Ammonium Formate, (CH 3 NH 3 ) + (HCOO) is diluted to 200 ml with distilled water. c) The system that results when 240.0 ml of 0.1500 M NaOH is added to 100.0 ml of 0.10 M Phosphoric Acid, H 3 PO 4 d) The system that results when 180 ml of 0.1500 M NaOH is added to 100.0 ml of 0.10 M Phosphoric Acid, H 3 PO 4 e) The system that results when 100 ml of 0.1000 M Na 3 PO 4, Sodium Phosphate, is added to 100.0 ml of 0.10 M Phosphoric Acid, H 3 PO 4 I have neither given nor received any unacknowledged aid on this test. Signed