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-0- Rules in My Class 1. Cheating on a test will result in a zero for a major exam. Plagiarizing a lab report will result in a zero for that lab report. 2. Cutting class will result in a zero for any test or lab report expected for that class day. In addition, I will require a note from your parents informing me that they are aware you've cut class and have received a zero for any test or lab report expected for that class day. 3. Going beyond the allotted time on a test will result in loss of all credit for that question/problem. 4. Work missed due to an illness must be made up within a reasonable period of time on returning to school. 5. Lab reports will not be accepted late, incomplete, or if they do not meet guidelines set by me. 6. You are not allowed to eat in class. -1- Grading Policy First Semester: Tests comprise 54% of the semester grade. Tests consist of free response which may require several concepts pieced together to provide a complete response or math problems which require setups that lead to the answer. Lab Work comprises 36% of the semester grade. Lab grades are built around work in the lab, analyzing data, and interpreting what occurred during the lab. Lab reports turned in late receive a zero. Final Exam comprises 10% of the semester grade. Second Semester: 60% for tests 40% for lab work Assignments are usually begun in class. Those students needing additional time finish the work outside of class. The AP Chem Website provides an overview of each unit, assignments, handouts along with accompanying answers to the text and handouts. http://www.mro-chemweb.com If you have questions and/or concerns, and you e-mail those to me BEFORE 8:00 P.M., I will try to get back to you the same evening. [email protected] -2- Assignments The Mole Concept & Stoichiometry Readings: Questions: Experiments: Chapter 2 & 3 2.5c, 78, 79, 84, 85; 3.119, 120, 123, 126 – 128, 135 (For Fun: 136, 140, 141) Basic Laboratory Techniques Identification of Substances by Physical Properties Separation of the Components of a Mixture Chemical Reactions of Copper & Percent Yield Gravimetric Analysis of a Chloride Salt Atomic Structure & The Periodic Table Powerpoint: Readings: Questions: Experiments: AP Questions: Atomic Theory & Structure 2.2, 2.3; Chapter 7 2.57 7.33, 36, 65, 66, 69, 71, 72, 85, 87, 88, 95, 97, 103, 104, 111, 112, 141, 142, 145 (For Fun: 138) Beer's Law: Determining the Concentration of a Solution 2007: #2, #6; 2006: #5, #7; 2002: #6a,b Bonding & Molecular Geometry Powerpoint: Readings: Questions: Handouts: Video: AP Questions: Bonding & Molecular Geometry Chapter 8 & 9 8.3 – 5, 8, 22, 24, 37, 46, 47, 51, 54, 69, 70, 82, 89 – 92, 94 – 100, 103, 113, 116, 118, 121 9.13, 14, 25, 27, 34, 39 – 43, 50 – 61, 77, 82, 86, 89, 93, 95 – 97, 99 Formal Charge Worksheet Drawing Lewis Structures Worksheet #1, #2 Hybrid Orbital Identification Worksheet Review of Lewis Structures, Hybrid Orbitals & Formal Charges Bonding Worksheet Signals From Within 2011: #6; 2006: #6; 2005: #8a – c; 2002: #6c Gas Laws Powerpoint: Readings: Gas Laws Chapter 10 -3- Questions: Handouts: Experiments: AP Questions: 10.4, 19, 30, 35 – 37, 39, 41, 43, 47, 49, 51, 55, 57c, 59, 60, 63 – 65, 67 – 73, 75, 79 – 81, 97, 99 Gas Law Problems Worksheet Dalton’s Law Worksheet Behavior of Gases: Molar Mass of a Vapor Determination of R: The Gas Law Constant 2011: #2; 2009: #3; 2005: # 6; 2004: #2; 2002: #3 Liquids, Solids & Changes of State Powerpoint: Readings: Questions: Handouts: AP Questions: Intermolecular Attractive Forces Chapter 11 11.2, 5 – 7, 9 – 11, 16, 17, 20 – 24, 26, 29, 30, 32, 33, 36, 39 – 41, 43 – 44, 75 – 87, 102 – 110, 115 IMF Worksheet 2005: # 8d,c; 2002: #6d Reactions in Aqueous Solutions & Physical Properties of Solutions Powerpoint: Readings: Questions: Handouts: Experiments: AP Questions: Solution Chemistry Chapter 4 & 12 4.11, 16, 17, 21, 33, 35, 40, 45, 49, 52, 55, 57 – 60, 63, – 66, 67a,b, 68a,c, 69, 71 – 79, 81b, 82b, 83b, 84a, 85, 87, 89, 92, 93, 95c,d, 97 – 99, 102, 104, 106, 109 – 114, 117, 120, 5.30, 49, 55, 56, 65, 67 – 70, 75, 76, 82 12.3, 6, 7, 9 – 12, 15, 16, 22, 23, 26, 33, 35, 36, 39, 44, 45, 47, 48, 50, 54 – 58, 61 – 63, 66, 72 Balancing Redox Reactions Solution Chemistry Worksheet #1 – #3 Net Ionic Equations #1 – #3 Chemical Reactions Solubility & Fractional Crystallization Colligative Properties: Freezing Point Depression & Molar Mass 20101: #3; 2008: #3, 5; 2005: #5 Thermochemistry, & The Spontaneity of Chemical Reactions Powerpoint: Readings: Questions: Chemical Thermodynamics Chapter 6 & 18 6.3a-c, 11, 14, 15, 18, 19, 21 – 24, 27, 32 – 34, 38, 39, 45, 48 – 51, 55, 57, 59, 61 – 64, 67, 68, 71, 74, 76, 78, 84 11.45 – 49 18.4, 7, 9, 12, 13, 18 – 20, 23, 26, 27, 34 – 37, 44 – 46, 55, 56, 58, 60c-e, 66 – 68, 70, 74 – 76, 79, 88 – 90, 92, 94 – 96 -4- Handouts: Experiments: AP Questions: Thermochemical Worksheet Heats of Reaction & Bond Enthalpies Worksheet Heat of Fusion Hess’s Law 2011: #3; 2009: #5; 2008: #6; 2006: #3; 2005: #7; 2004: #7; 2002:#5, #8 Kinetics Powerpoint: Readings: Questions: Handouts: Experiments: AP Questions: Kinetics Chapter 13 13.14, 15, 17 – 20, 25, 27, 29, 31, 33, 34, 37, 38, 40 – 42, 51, 53, 55, 56, 59, 63, 64, 67, 68, 71, 73 – 76, 79, 83, 84, 91, 93, 94, 95, 97 Kinetics Worksheet Rates of Chemical Reactions: A Clock Reaction 2010: #6; 2009: #2; 2008: #2; 2006: #3; 2005: #3; 2004: #3; 2002:# 7 Chemical Equilibria, Solubility & Complex Ion Equilibria Powerpoint: Readings: Questions: Handouts: Experiments: AP Questions: Chemical Equilibrium Chapter 14 & 17 14.12, 13, 15, 18a,b, 21, 23d, 25, 26, 28, 29, 31, 35, 36, 39, 40, 43a,d, 44b,c, 54, 55, 57, 59, 61, 65 17.2, 3, 16a,c,d, 18 – 21, 32, 36, 38, 42, 53, 55 18.39 – 42, 80 – 82, 84, 87 Equilibrium Worksheet Determination of the Ksp for a Sparingly Soluble Salt 2011: #1; 2010: #1; 2009: #1; 2008: #1; 2007: #1; 2006: #1; 2005: #1; 2004: #1, #8; 2002:# 1 Acid-Base Concepts Revisited & Acid-Base Equilibria Powerpoint: Readings: Questions: Handouts: Experiments: Acid-Base Chemistry Chapter 15 & 16 15.4, 6, 9, 10, 14, 17, 19, 21, 25, 29, 31, 33, 39, 40, 43b,c, 44c,d, 59a, 60c, 63a, 64a, 67, 77, 78 16.2b, 5, 6, 8a, 9, 11, 12, 14, 15, 18, 20, 22, 24, 28, 29, 32, 36, 39, 40, 44 – 46, 49, 52, 56, 58, 62, 64, 88, 89 Acid Base Worksheet Acid – Base Reactions Worksheet #1, #2 Buffers & Salts Worksheet Ka for a Weak Acid -5- AP Questions: Titration Curves of Polyprotic Acids 2011: #5; 2010: #5; 2007: #5; 2004: #5 Electron Transfer Reactions, Electricity & Chemical Change Powerpoint: Readings: Questions: Experiments: AP Questions: Electrochemistry Chapter 5 & 19 19.2, 4, 5, 7, 8, 10, 11, 13, 15, 17 – 20, 24, 26, 28, 29, 50, 52, 55, 56a,d, 57a,d, 58, 59, 62, 63, 70, 78, 81, 83, 84, 87, 88 Activity Series of Metals Electrolysis, the Faraday, & Avogadro's Number Electrochemical Cells & Thermodynamics 2010: #2; 2009: #6; 2007: #3; 2006: #2; 2005: #2; 2004: #6; 2002: #2 -6- Determining the Concentration of a Solution: Beer’s Law Introduction The purpose of this lab is to become acquainted with how to use a colorimeter and how it is used to determine the concentration of a solution. Equipment & Materials LabQuest Vernier colorimeter pipet pump 6 cuvettes 5 - 20 × 150 mm test tubes 1 - 10 mL pipet 2 - 100 mL beakers 0.40 M CuSO4(aq) CuSO4(aq), unknown solution distilled water test tube rack stirring rod Kimwipe Procedure 1. Label five clean, dry, test tubes 1–5. Use pipets to prepare five standard solutions according to the chart below. Thoroughly mix each solution with a stirring rod. Clean and dry the stirring rod between uses. Trial 1 2 3 4 5 CuSO4(aq) (mL) 2 4 6 8 10 Distilled H2O (mL) 8 6 4 2 0 Concentration (M) 0.08 0.16 0.24 0.32 0.40 2. Connect a colorimeter and your computer to the LabQuest. 3. In Logger Pro, under the file menu, open the Advanced Chemistry Folder, and then open file 17 - Colorimeter. 4. Calibrate the colorimeter. a) Open the Experiment menu, click on Calibrate Colorimeter, click on the Calibrate Tab, and click Calibrate Now. -7- b) c) d) 5. Prepare a blank by filling an empty cuvette 3⁄4 full with distilled water. Place the blank in the cuvette slot of the colorimeter and close the lid. Turn the wavelength knob to the “0% T” position. When the voltage reading stabilizes, type 0% in the first data box, and click . Move the knob to the Red LED position (635 nm), and when the reading stabilizes, type 100% in the data box, and click , then click Done. To collect absorbance-concentration data for the five standard solutions: a) Remove the cuvette from the colorimeter and pour out the distilled water. Using the solution in Test Tube 1, rinse the cuvette twice with ~1 mL amounts, and then fill it 3⁄4 full. Wipe the outside with a tissue, place it in the colorimeter, and close the lid. b) Click c) When the absorbance readings stabilize, click edit box, and press the ENTER key. d) Discard the cuvette contents, and repeat the procedure for test tubes 2, 3 and 4. Trial 5 is the original 0.40 M CuSO4 solution. e) When you finish testing the standard solutions, click f) Click the Linear Regression button, and a best-fit linear regression line will be shown for your five data points. . , type 0.080 in the . 6. Record the absorbance values, for each of the five trials, in your data table. 7. Determine the absorbance value of the unknown CuSO 4(aq). a) Rinse a cuvette with distilled water, followed by rinsing twice with the unknown solution. Fill the cuvette about 3⁄4 full with the unknown. Wipe the outside of the cuvette, place it into the colorimeter, and close the lid. b) Record the absorbance value in your data table. c) Dump the solutions into the sink, rinse the cuvettes, and turn these upside down so they can drain. -8- DATA TABLE Trial 1 2 3 4 5 Unknown # Concentration (M) 0.080 0.16 0.24 0.32 0.40 Absorbance Questions 1. For a given substance, the amount of light absorbed depends on what four factors? 2. How are percent transmittance and absorbance related algebraically? 3. State the Beer-Lambert Law. 4. What is the purpose of preparing a calibration curve? 5. Why is the line on the cuvette always aligned with that of the sample holder? -9- - 10 - - 11 - - 12 - - 13 - - 14 - Using Formal Charge to Predict Best Lewis Structure Developing Possible Lewis Structures ● ● ● ● Add up valence electrons Add to or take away electrons based on the charge on a radical Place a pair of electrons between central atom and surrounding atoms With the exception of hydrogen, the remainder of the electrons are placed around the surrounding atoms till four pairs of electrons surround them. ● Place any remaining electrons on central atom Calculating Formal Charge (FC) FC = Valence Electrons –Non-Bonding Electrons – Bonding Pairs Determining The Best Lewis Structure IF central atom has a FC of zero or a charge equal to its normal charge, the Lewis dot structure is correct. IF NOT, begin forming multiple bonds with the surrounding atom that has the most negative FC; this will lead to a reduction in the FC, and to a lower potential energy state. ● For a central atom in the second period, continue this process until the central atom as an octet. ● For a central atom in the third period or lower, continue this process until the FC on central atom is reduced to zero. What if the choice is 2+ 2- versus 1+ 1- ? More potential energy is required to separate larger charges than smaller charges; English . . . larger charges are less favored. What other Considerations Are There? Negative FC on atoms that are more electronegative, and Positive FC on atoms that are less electronegative atoms are preferred. - 15 - Exercises Involving Formal Charge http://www.cem.msu.edu/~reusch/VirtualText/Questions/General/resnce1.htm http://www.usm.maine.edu/~newton/Chy251_253/Lectures/Formal%20Charge/FC Exercises.html - 16 - Formal Charge Worksheet Which structures satisfy the following conditions? 1. No formally charged atoms are present in the structure. 2. At least one nitrogen has a (+) formal charge 3. At least one nitrogen has a (-) formal charge 4. At least one oxygen has a (+) formal charge 5. At least one oxygen has a (-) formal charge 6. At least one carbon has a (+) formal charge 7. At least one carbon has a (-) formal charge 8. At least one sulfur has a (+) formal charge. - 17 - Drawing Lewis Structures Worksheet #1 PH4 + CH2O SiH4 SeF6 SeF4 ICl3 NO + NO2 - HCN AsCl5 ICl2 - XeF4 BF4 - SCl2 PCl4 + - 18 - ClF3 SeCl4 PCl5 ClO3 – SeO3 2- OSCl2 Draw resonance structures for: SO3 SO2 CH3COO - N2O4 - 19 - Drawing Lewis Structures Worksheet #2 Obey Octet Rule: AsBr3 H3CCN SeF2 O2SF2 TeF4 ClF5 CS2 NI3 H2CO3 PF6 - SbCl6 - XeF2 SiCl4 CN - SnCl4 - 20 - FCl2 + PBr3 AsCl4 + IO3 - ICl4 - BrF4 - SeO3 C2O4 2- Draw resonance structures for: HN3 - 21 - Hybrid Orbital Identification Worksheet 1. For each structure shown below, identify the hybrid orbital on each atom, any missing electrons, and the molecular geometry around each atom. - 22 - 2. Draw the complete Lewis structures for the following molecules. a) HC(NH2)HCOOH c) CH3C(NH2)HCOOH b) CH3CH(CH3)COOH d) CH2CHCH2CH2Cl - 23 - Review of Lewis Structures, Hybrid Orbitals & Formal Charges 1. Draw Lewis structures for each of the following molecules/radicals, and then identify: a) b) c) d) the geometry non-bonded electron pairs on each atom formal charges on each atom hybrid orbital on central atom SO2 ClF5 O3 SnCl2 CH2O SOF4 CO3 2- NO2 - SO3 2- SnCl5 - - 24 - PH4 + 2. In the following molecules a) which has the most polar bond? b) which has a net dipole greater than zero? BH3 3. CH4 H2O HF Write complete Lewis structures for each of the following compounds: CH3C = N = O CH2Cl2 CH2OHCHOHCH = O H3CCN - 25 - CH3OCH3 CH3NH2 Bonding Worksheet 1. Briefly discuss the relationship between the dipole moment of a molecule and the polar character of the bonds within it. Account for the difference between the dipole moments of CH2F2 and CF4. 2. Suppose that a molecule has the formula AB 3. Sketch and name two different molecular geometries this molecule may have. For each of the two molecular geometries, give an example of a known molecule that has that shape, and then identify the molecular geometry of each shape. 3. For CO3 2-, CO2, and CO, a) draw the Lewis structures for all possible resonance structures. b) which of the three species has the shortest C-O bond length? Explain the reason for your answer. c) predict the molecular geometry of the three species. Explain how you arrived at your predictions. - 26 - 4. 5. Explain the following trends in lattice energy a) CaF2 > BaF2 b) NaCl > RbBr c) BaO > KF Calculate the lattice energy (and diagram the Born-Haber cycle) for CaH2 using the following information: Electron affinity for H Ionization energy for Ca Ionization energy for Ca 1+ ∆Hsubl for Ca Bond dissociation energy for H2 ∆Hform CaH2 - 27 - - 72.8 kj/mol + 589.8 kJ/mol + 1145 kJ/mol + 178.2 kJ/mol + 435.9 kJ/mol - 186.2 kJ/mol 6. For each of the following molecules, PH3 SF2 NI3 a) draw the best Lewis structure b) identify the molecular geometry c) identify the hybridization on the central atom d) state whether the molecule is polar or nonpolar - 28 - OF2 Gas Law Problems Worksheet 1. Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide, NaN3, to decompose explosively according to the following unbalanced equation: NaN3(s) Na(s) + N2(g) What mass of NaN3(s) must react to inflate an air bag to 70.0 L at STP? 2. Urea, NH2CONH2 is a nitrogen fertilizer that is manufactured from ammonia and carbon dioxide, identified in the following unbalanced equation: NH3(g) + CO2(g) NH2CONH2(aq) + H2O(l) 0.908 g of ammonia reacts with 178 mL of CO 2 at a pressure of 35 C and 1.50 atm. How many grams of NH2CONH2 will be produced? 3. When methane, CH4, reacts with molecular chlorine, hydrogen chloride and carbon tetrachloride are produced. If 49.62 mL of methane reacts at a temperature of 465.° C and at a pressure of 1400. torr, with 75.00 mL of molecular chlorine, how many grams of carbon tetrachloride will be produced? - 29 - 4. Calculate the volume of H2(g), measured at 26 C and 751 mm Hg, required to react with 28.5 L CO2(g), measured at 0 C and 760 mm Hg, in the following unbalanced equation: 3 CO(g) + 10 H2(g) 5. C3H8(g) + 6 H2O(l) Hydrogen cyanide is prepared commercially by the reaction of methane, CH4(g), ammonia, and molecular oxygen, at high temperatures. The other product is gaseous water. What volume of HCN(g) can be obtained from 20.0 L CH4(g), 20.0 L NH3(g), and 20.0 L O2(g)? 2 CH4(g) + 2 NH3(g) + 3 O2(g) 2 HCN(g) + 6 H2O(g) 6. A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750. torr, contains 0.800 g of the gaseous compound. Identify the molecular formula. - 30 - 7. A gas consisting of only carbon and hydrogen has an empirical formula of CH2. The gas has a density of 1.65 g/L at 27 C and 734 torr. Determine the molar mass and molecular formula of the gas. 8. The percent composition of a SFx compound is 29.69% S and 70.31% F. At 20 C, 0.100 g of the gaseous compound occupies a volume of 22.1 mL and exerts a pressure of 1.02 atm. What is the molecular formula of this gas? 9. An anesthetic contains 64.9% C, 13.5% H, and 21.6% O, by mass. At 120 C and 750 mm Hg, 1.00 L of the gaseous compound weighs 2.29 g. What is the molecular formula of the compound? - 31 - Dalton’s Law Worksheet 1. A mixture of 4.0 g of molecular hydrogen and 10.0 g of He(g) in a 4.3 L flask is maintained at 0 C. What is the partial pressure of each gas? What is the total pressure? 2. A 2.00 L flask is filled with Ar(g) at 752 mm Hg and 35 C. A 0.728 g sample of benzene, C6H6(g) is then added. 3. a) What is the total pressure in the flask? b) What is the mole fraction of each gas inside the flask? A 1.65 g sample of Al reacts with excess HCl(aq), producing aluminum chloride and H2(g). The molecular hydrogen is collected over water at 25 C at a barometric pressure of 744 mm Hg. What volume of dry H 2(g), in L, is collected? (PH2O at 25 C = 23.8 mm Hg) - 32 - 4. Concentrated hydrogen peroxide solutions, H2O2(aq), are explosively decomposed by traces of transition metal ions: 2 H2O2(aq) 2 H2O(l) + O2(g) What volume of pure O2(g), collected at 27 C and 746 torr, would be generated by decomposition of 125. g of hydrogen peroxide solution? 5. On warming, formic acid, HCOOH, decomposes into: HCOOH(l) H2O(l) + CO(g) If 3.85 L of carbon monoxide was collected over water at 25 C and 689 mm Hg, how many grams of formic acid reacted? - 33 - IMF Worksheet Draw Lewis structures and identify all intermolecular and intramolecular attractive forces. C2H4 NCl3 HCN H2S N2H4 NH2Cl SiH4 CO (CH3)2CHCl CH3F CH3NH2 CH3CHO NH2CH2COOH CH3COCH3 HSCH2CH2SH - 34 - The Mystery of Excessive Perspiration While Jogging Ronald Delorenzo Middle Georgia College While jogging or running several miles, perspiration may be produced in such large quantities that it flows down the body and drips onto the ground. Since the purpose of perspiration is to produce a cooling effect by evaporation, why does the human body not produce just enough perspiration to keep the skin surface moist? The production of perspiration requires relatively large amounts of energy, yet wasted effort is uncharacteristic of nature. In addition, excessive perspiration can lead to dehydration and death. What is the advantage to the human body of producing more perspiration than would be needed to keep the skin surface just moist? Perspiration contains solutes. If the human body were to produce just enough perspiration to keep its skin moist, the solute concentration of the residual sweat would increase as the perspiration evaporates. As the residual sweat increases, the vapor pressure of this remaining concentrated solute solution decreases. This, in turn, decreases the evaporation rate and consequently the cooling rate. Therefore, it appears that one possible advantage of excessive sweating may be that excess sweat rinses away concentrated solute solution from the skin surface and thereby increases evaporation and cooling rates. - 35 - Fire Walking, Temperature & Heat Ronald Delorenzo Middle Georgia College For thousands of years people have marveled at and been puzzled by the ability of fire walkers to walk across beds of glowing coals without apparent harm. Many scientists, having never observed fire walking firsthand, simplistically explained it away with assumptions such as "fire walkers have toughened and calloused feet." Today, tens of thousands of Americans without the benefits of toughened and calloused feet engage in fire walking. How then do they do it? In chemistry classes, teachers explain to students that there is a difference between the amount of heat objects contain and the temperature of objects. For example, sparks given off by a cigarette lighter are very hot (~ 1000° C), but when sparks from a cigarette lighter touch one's hands or feet, one does not usually detect any warmth. Likewise, there is a difference between the amount of heat that glowing thousand-degree coals contain and their temperature. The dark footprints that momentarily form after firewalkers step on previously red coals indicate that the coals cool quickly. This cooling results in part by an ash compression and oxygen depletion at the glow front and by the heat flow from coals to feet. The specific heat of feet, which have a high water content, is much greater than the specific heat of coals. If red-hot coals were replaced by aluminum with a larger specific heat, the firewalkers would have problems! A second factor to be considered is the conductivity of coal. Coal and coal ash are insulators. Once human feet cool the surface of glowing coals, the heat transfer from the internally hot coal to the cooler coal surface is relatively slow. Again, if a heat conductor such as copper or aluminum replaced coals, firewalkers would have additional problems. A third factor to be considered is the Leidenfrost effect, which is used to explain why water droplets spend such an unexpectedly long amount of time on hot griddles before vaporizing. The bottoms of water droplets on hot griddles vaporize very quickly, allowing the drops to float on insulating layers of vapor. Firewalkers also dip there feet in ice water beforehand, or stand on damp grass at one end of the coal bed before their walk. Additional factors to be considered include: (1) coal beds are only 3 meters long, (2) coal beds are traversed rapidly with only two quick steps, and (3) embers are spread thinly. - 36 - Heat Calculations Involving Heating & Cooling Curves Heat capacity is the amount of heat needed to change the temperature of a substance by 1° C. Specific Heat Capacity, s, is the amount of heat needed to change the temperature of 1 g of a substance by 1° C. The amount of heat lost or gained during a temperature change is: ∆H = (∆T)(m)(s) The amount of heat lost or gained during a phase change is: Heat of fusion, ∆Hfus is the amount of energy needed to melt, fuse, one gram of a substance. q = (∆Hfus)(m) Heat of vaporization, ∆Hvap is the amount of energy needed to vaporize one gram of a substance. q = (∆Hvap)(m) How much heat is needed to convert 46 g of ice at - 20° C to 138° C? - 20° C to 0° C fusion 0° C to 100° C vaporization 100° C to 138° C (2.092 J/g-°C)(20° C)(46 g) (334.72 J/g)(46 g) (4.184 J/g-°C)(100° C)(46 g) (2259.36 J/g)(46 g) (2.017 J/g-°C)(38° C)(46 g) - 37 - 1.925 kJ 15.397 kJ 19.264 kJ 103.931 kJ 3.526 kJ Thermochemical Worksheet 1. How much energy is needed to melt 86.8 g of HCl? (∆Hfus = 117.15 J/g) 2. How much energy is used to increase the temperature of 48.06 g of nickel from - 14 °C to 1020 °C? M.P. = 1200° C B.P. = 1850° C ∆Hfus = 17.47 kJ/g ∆Hvap = 370.40 kJ/g Cp(s) = 1.67 J/g-° C Cp(l) = 2.47 J/g-° C Cp(g) = 3.64 J/g-° C 3. How much energy is lost as 45 g of cesium is cooled from 880° C to 28° C? M.P. = 28.59° C B.P. = 690° C ∆Hfus = 2.09 kJ/g ∆Hvap = 67.74 kJ/g - 38 - Cp(s) = 30.04 J/g-° C Cp(l) = 33.47 J/g-° C Cp(g) = 20.79 J/g-° C Heats of Reaction & Bond Enthalpies Worksheet Based on average bond enthalpies, determine the heat of reaction for the following chemical reactions: C4H8 + O2 CH3CHCH2 + HCl CH3COCH3 + HCN CO2 + H2O - 39 - CH3CHClCH3 CH3C(OH)(CN)CH3 Heat of Fusion Introduction The purpose of this experiment is to identify the heat of fusion, ∆Hfus, of water. Equipment & Materials ice cubes beaker, 150 or 250 mL thermometer calorimeter with lid balance crucible tongs hot hands Procedure 1. Heat about 150 mL of water to around 50° C. 2. Record the mass of a calorimeter and lid. 3. Pour about 100 mL of the heated water into the calorimeter and then secure with the lid. 4. Record the mass of the calorimeter and its contents. 5. Place a thermometer through the lid and record the temperature after 20 seconds. 6. Put 4 – 6 ice cubes into the calorimeter, and stir the mixture until the temperature is constant. Note: If the temperature does not drop to between 0 – 5° C, additional ice cubes should be added. 7. Remove any unmelted ice with crucible tongs, draining off any liquid on the ice into the calorimeter. 8. Measure the mass of the calorimeter and its contents along with the temperature. - 40 - Processing the Data 1. Determine the change in temperature. 2. Calculate the number of joules transferred from the hot water, Htotal, to the ice: (s = 4.184 J / g-° C) 3. Calculate the number of joules that melted one gram of ice, ∆H fus. 4. Percent Error: (Theoretical (∆Hfus is 334.72 J/g.) Conclusions 1. What does heat of fusion represent? 2. In terms of potential and/or kinetic energy, what happens to ice at its melting point? 3. Discuss the type of attractions/bonds that exist between molecules of ice, and what happens to these as melting occurs. 4. What was your experimental heat of fusion? 5. What was the percent error? 6. Identify causes for this error. - 41 - Hess’s Law Introduction The purpose of this experiment is to use Hess's Law to determine the heat of neutralization when an acid is reacted with a base. Equipment & Materials NaOH electronic balance 0.5 M HCl NaOH Erlenmeyer flask, 250 or 500 mL thermometer Procedure Part I. Solid NaOH + H2O 1. Record the mass of a flask. 2. Add 100.0 mL of tap water to the flask, and record the mass. 3. Place a thermometer in the water, and after 20 seconds record the temperature. 4. Measure out 2.00 g of NaOH. Caution: NaOH is caustic. If you get any on your hands or clothing wash it off with soap and water. 5. Add the NaOH to the flask, stirring gently with the thermometer. Record the highest temperature. Part II. Aqueous NaOH + Aqueous HCl 1. Add 50 mL of HCl solution to a second flask. 2. Record the temperature of the NaOH solution. - 42 - 3. Using the NaOH solution you prepared in Part I, pour 50. mL of NaOH into the HCl solution. Record the highest temperature. Part III. Solid NaOH + Aqueous HCl 1. Add 50 mL of HCl solution to a flask, and record the temperature. 2. Add 2.00 g of NaOH to the flask, gently stirring. temperature. Part I Part II Record the highest Part III ΔT Heat Absorbed by Solution Heat Absorbed by Flask Mol NaOH Total Heat/mol NaOH Processing the Data 1. Calculate the heat absorbed by the solution: (sH2O = 4.184 J/g-° C) 2. Calculate the heat absorbed by the flask: (sflask = 0.92 J/g-° C) 3. Total heat absorbed: 4. Moles of NaOH: Part I or Part III Part II - 43 - 5. Total heat per mole of NaOH: 6. Percent Error: (Part III = T.Y., Parts I + II = E.Y.) Conclusions 1. What does the heat of a reaction represent? 2. Identify the net ionic equations that correspond to the reactions that took place in Part I, II, and III. 3. Using these net ionic equations, show how the changes that occurred in Part I and Part II are the same as the changes that occurred in Part III. 4. What is Hess’s Law? 5. How do the results in this lab illustrate Hess's Law? - 44 - Balancing Redox Reactions Worksheet Acidic Solutions Mn 2+ + BiO3 - ClO3 - + Cl - MnO4 - + Bi 3+ Cl2 + ClO2 P + Cu 2+ Cu + H2PO4 - PH3 + I2 H3PO2 + I - NO2 NO3 - + NO - 45 - Basic Solutions MnO4 - + C2O4 2- ClO2 Zn MnO2 + CO2 ClO2 - + ClO3 - Cu(NH3)4 2+ + S2O4 2- Zn + NO3 - SO3 2- + Cu + NH3 Zn(OH)4 2- + NH3 Zn(OH)4 2- + H2 - 46 - Solution Chemistry Worksheet #1 1. How many grams of sodium hydroxide are required to neutralize 1.965 g of H2S in each of the following unbalanced equations? NaOH + H2S NaOH + H2S NaHS + H2O Na2S + 2 H2O 2. How many grams of lithium hydroxide are required to neutralize 75.0 g of phosphoric acid if two its three hydrogens has been neutralized? 3. What is the molar mass of a monoprotic acid (HX), 0.1248 g of which required 0.2475 g of potassium hydroxide for neutralization? - 47 - 4. Determine the oxidation state for tin in the following tin oxide compounds: oxide containing 88.12% Sn oxide containing 78.77% Sn 5. Balance the following redox equations: MnO2 + Fe 2+ + H + Cr2O7 2- + Sn 2+ + H + - 48 - Mn 2+ + Fe 3+ + H2O Cr 3+ + Sn 4+ + H2O Solution Chemistry Worksheet #2 1. A chemist synthesizes a new organic acid. After dissolving 0.500 g of the acid in water she finds that it requires 15.73 mL of 0.437 M sodium hydroxide for neutralization to occur. If the acid contains three carboxylic acid groups, what is the molar mass of this acid? 2. 500. mL containing 1.00 g of potassium permanganate is used in titrating a reducing agent. During the reaction, KMnO4 is reduced to Mn 2+. What is the molarity of the KMnO4 solution? 3. 31.25 mL of 0.100 M sodium oxalate (acidified) is titrated with 17.38 mL of potassium permanganate solution of unknown strength. What is the molarity of the potassium permanganate? C2O4 2- + MnO4 - + H + - 49 - CO2 + Mn 2+ + H2O 4. Titration of 1.500 g of an diprotic acid required 47.00 mL of 1.20 M sodium hydroxide to reach the endpoint. What is the molar mass of the acid? 5. A chemist synthesizes a substance which he believes is barbituric acid (MM = 128.0), a precursor for many sleeping tablets. Barbituric acid has one acidic hydrogen. To help with the identification, a 0.500 g crystalline sample is titrated with 0.100 M sodium hydroxide. The equivalence point is reached after 39.10 mL of base was added. Is this sample barbituric acid? 6. What mass of sodium carbonate is required to neutralize 4.89 g of hydrochloric acid in the following non-balanced equation? HCl + Na2CO3 NaCl - 50 - + H2O + CO2 Solution Chemistry Worksheet #3 1. Hydrochloric acid reacts with zinc, resulting in the production of zinc chloride and hydrogen gas. The mass of hydrogen gas that evolved was 0.0572 g. How many grams of sodium hydroxide are required to neutralize this acid? 2. 3.245 g of cadmium was dissolved in a dilute nitric acid solution, producing 0.0532 g of hydrogen gas. Based on this data, identify the oxidation state of cadmium. 3. Solid potassium hydrogen phthalate, KHC8H4O4, is often used to standardize a base. KHC8H4O4 has one ionizable H + per formula unit. How many moles of acidic hydrogen are contained in 0.7325 g of this salt? - 51 - 4. When FeCl3 oxidizes KI, I2, FeCl2 and KCl are produced. How many grams of iodine will form when 13.20 g of iron (III) chloride reacts with an excess of potassium iodide? 2 Fe 3+ + 2 I - 5. I2 + 2 Fe 2+ Analysis shows that 16.20 g of an iron oxide compound contains 11.33 g of iron. a) What is the formula for this iron oxide compound b) What is the oxidation state of iron in this oxide? - 52 - Reaction Type & Net Ionic Equations Precipitation – the solubility rules need to be memorized, and applied. Acid-Base Reactions fall into several categories: (1) Typical Acid-Base Reaction: H + + OH – CH3COOH + OH – (2) Mg(OH)2 Non-Metal Oxides Added to Water: SO2 + H2O (4) H2SO3 Formation of a Weak Acid or Base: F-+ H+ (5) CH3COO – + H2O Metal Oxides Added to Water: MgO + H2O (3) H2O HF Nonmetal oxide Added to a Basic Solution: SO2 + 2 OH - SO3 2- + H2O SO2 dissolves in the solution, producing H2SO3,.which then reacts with the base. H2SO3 SO3 2- + 2 SO2 + H2O H2SO3 + 2 OH (6) H2O Metal oxides Added to an Acidic Solution: Na2O + 2 H + - 53 - 2 Na + + H2O Na2O dissolves in the solution, producing NaOH, which then reacts with the acid. Na2O + H2O 2 Na + + 2 OH - Na + + OH - + H + Na + + H2O (7) Salts Added to Water: NaHSO4 + H2O HSO4 - ↔ KC2H3O2 + H2O C2H3O2 – + H2O Na + + HSO4 – H + + SO4 2- ↔ K + + C2H3O2 – ↔ HC2H3O2 + OH – Redox Reactions http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/oxred_3.php (1) Free metal or non-metal undergoing a chemical change: NaOH Br - F - + Na + H2O F2 + (2) + H2 Br2 An oxidizing agent grabs electrons from another source. The higher the electronegativity, the greater its ability to absorb electrons. These tend to include substances with higher oxidation states (MnO 4 -, CrO4 2-, Cr2O7 2-, HNO3, H2SO4). (3) A reducing agent releases electrons. The lower the electronegativity, the lower the ionization energy, the easier for valence electrons to be lost. Formation of a Gas You are expected to know common chemical reactions that produce CO 2, SO2, and NH3. - 54 - Net Ionic Equations Precipitate NH4OH(aq) + FeCI3(aq) NH4CI (aq) + Fe(OH)3(aq) NH4 +(aq) + OH -(aq) + Fe 3+(aq) + CI -(aq) Fe(OH)3(s) + NH4 +(aq) + Cl -(aq) 3 OH -(aq) + Fe 3+(aq) Fe(OH)3(s) NaI(aq) + Pb(NO3)2(aq) PbI2(aq) + NaNO3(aq) Na +(aq) + I -(aq) + Pb 2+(aq) + NO3 -(aq) PbI2(s) + Na +(aq) + NO3-(aq) 2 I+ Pb 2+ PbI (aq) (aq) 2(s) Redox Cl2(g) + NaBr(aq) Br2(g) + NaCl(aq) Cl2 + Na +(aq) + Br -(aq) Cl + 2 Br 2 (aq) Br2(l) + Na +(aq) + Cl -(aq) Br + 2 Cl 2(l) (aq) Cu(s) + H2SO4(aq) CuSO4(aq) + H2(g) Cu(s) + 2 H +(aq) + SO4 2-(aq) Cu 2+(aq) + SO4 2-(aq) + H2(g) Cu + 2 H+ Cu 2+ + H (s) (aq) (aq) 2(g) Acid - Base + KOH(aq) K2S(aq) + H2O(l) H2S(aq) + K +(aq) + OH -(aq) K +(aq) + S 2-(aq) + H2O(l) H S + 2 OH S 2+ 2 H O H2S(aq) 2 (aq) NaOH(aq) (aq) (aq) + NH4Cl(aq) NH4OH(aq) NH4 +(aq) Na +(aq) + OH -(aq) + NH4 +(aq) + Cl -(aq) OH -(aq) + - 55 - 2 (l) + NaCl(aq) NH3(aq) + H2O(l) + Na +(aq) + Cl -(aq) NH3(aq) + H2O(l) HCl(aq) + K2CO3(aq) KCl(aq) + H2CO3(aq) H +(aq) + Cl -(aq) + K +(aq) + CO3 2-(aq) K +(aq) + Cl -(aq) + H2O(l) + CO2(g) 2 H +(aq) + CO3 2-(aq) H2O(l) + CO2(g) - 56 - Net Ionic Equations Worksheet #1 Precipitate AgNO3(aq) + Na2CrO4(aq) H2SO4(aq) + BaCl2(aq) Na3PO4(aq) + CaCl2(aq) Redox Bi(s) + O2(g) Ca(s) + H2O(l) Pb(s) + H2SO4(aq) Acid - Base HC2H3O2(aq) + NaHCO3(s) (NH4)2SO4(aq) + KOH(aq) NaF(aq) + HCl(aq) - 57 - Net Ionic Equations Worksheet #2 Precipitate H2S(aq) + Pb(NO3)2(aq) H2SO4(aq) + CaF2(aq) AgNO3(aq) + LiBr(aq) Redox Mg(s) + FeCl3(aq) KClO3(s) CS2(l) + O2(g) Acid - Base HCl(aq) + K2SO3(aq) CO2(g) + NaOH(aq) H2SO4(aq) + LiHCO3(s) - 58 - Net Ionic Equations Worksheet #3 Sulfur dioxide gas is bubbled into distilled water. Ammonia gas is mixed with hydrogen chloride gas. Solid sodium bicarbonate is strongly heated. A solution of potassium hydroxide is added to solid ammonium chloride. Solutions of strontium nitrate and sodium sulfate are mixed. Solid ammonium nitrate is heated to temperatures above 300° C - 59 - - 60 - Fractional Crystallization Fractional crystallization involves separating a mixture by using differences in solubility between components of a mixture. As a heated solution cools, the less soluble component of the mixture comes out of solution. The crystals are then separated by filtration, and dried. Suppose we have a mixture that consists of two amino acids, and the composition of this mixture is 80% glycine (Gly) and 20% alanine (Ala). How much pure glycine could be recovered from 100 g of this mixture? The solubilities of Ala and Gly are as follows: Alanine Glycine g solute/100 g H2O 0° C 100° C 13 37 14 67 The mixture contains 20 g of Ala, and the minimum amount of water required to keep the Ala dissolved at 0° C is 20 g Ala x (100 g H2O/13 g Ala) ~ 154 g H2O At 100° C the solubility of Gly is 67 g/100 g of H2O: (67 g Gly/100 g H2O) x 154 g H2O ~ 103 g Gly The 154 g of hot water needed to dissolve the Ala at 0° C is more than enough to dissolve all of the Gly as well. The amount of pure Gly that will separate from the solution after it has been cooled to 0° C is the difference between the amount of Gly in the original mixture, 80 g, and the amount that will remain in the solution at 0° C. The solubility of Gly at 0° C is 14 g/100 g H2O and therefore in 154 g of water at 0° C we have (14 g Gly / 100 g H2O) x 154 g H2O ~ 22 g Gly The quantity of solid Gly that is precipitated is 80 g - 22 g = 58 g. This represents the maximum amount of pure Gly that can be recovered from this mixture. - 61 - Solubility & Fractional Crystallization In this experiment we will be taking advantage of the differences in solubility of solid substances in different kinds of liquid solvents to separate the components of a mixture in essentially pure form. This technique is called fractional crystallization. You will be given a sample containing silicon carbide, potassium nitrate, and copper sulfate. Silicon carbide, SiC, is a black, very hard material; it is the classic abrasive, and completely insoluble in water. Potassium nitrate, KNO 3, and copper sulfate, CuSO4 • 5 H2O, are water-soluble ionic substances, with different solubility at different temperatures. Copper sulfate is blue in its crystalline form and in solution. The solubility of the hydrate increases fairly rapidly with temperature. Potassium nitrate is a white solid, colorless in solution. Its solubility increases about twenty fold between 0° C and 100° C. The solid KNO3 that is recovered is contaminated by a small amount of copper sulfate. Redissolving the solid in a minimum amount of hot water, cooling and then recrystallizing the KNO3, increases its purity markedly. The purity of the KNO3 can be established by the intensity of the color produced by the copper impurity when treated with ammonia, NH3. Procedure 1. Add the sample mixture to a beaker filled with 50-mL of distilled water. 2. Warm the solution, stirring continuously, until KNO3 and CuSO4 • 5 H2O are in solution. 3. Pour the warm solution through a Büchner funnel with the aid of a rubber police person, transferring as much of the mixture as is possible. 4. Set the SiC aside to dry, and transfer the filtrate to a beaker. 5. Rinse the filter flask and Büchner funnel with tap water. 6. Place the Büchner funnel in the ice chest to chill for several minutes. 7. Add 15 drops of 6 M HNO3 to the filtrate, and heat the filtrate until white crystals of KNO3 begin to form. Note: If the liquid begins bumping, gently move the flame back-and-forth below the beaker to control the heating. - 62 - 8. Add 5-mL of distilled water to the solution to re-dissolve any crystals clinging to the walls of the beaker. 9. Cool the solution in an ice bath to force white KNO 3 crystals come out of solution. 10. Reassemble the filtering apparatus, and filter the slurry. 11. Place a filter paper over the crystals and, with the aid of a rubber stopper, press-dry the crystals. 12. Visually inspect the crystals. If CuSO4 is still present, spray the crystals with a jet of ice-cold distilled water, and re-press the crystals. Caution: Using too much water will dissolve the KNO3 as well as the CuSO4. 13. Allow the crystals to air dry. - 63 - Kinetics Worksheet 1. Distinguish between average rate and instantaneous rate. Which of the two rates gives us an unambiguous measurement of reaction rate? Why? 2. Suggest experimental means by which the rates of the following reactions could be followed: a) CaCO3(s) CaO(s) + CO2(g) b) Cl2(g) + 2 Br –(aq) Br2(aq) + 2 Cl –(aq) 3. What are the units for the rate constants of zero order, first order, and second order reactions? 4. On which of the following properties does the rate constant of a reaction depend? Briefly explain why. a) b) c) reactant concentration nature of the reactants temperature. - 64 - 5. Consider the reaction X + Y Z From the following data, obtained at 360 K, initial rate of disappearance of X (M/s) 0.053 0.127 1.02 0.254 0.509 6. [X] [Y] 0.10 0.20 0.40 0.20 0.40 0.50 0.30 0.60 0.60 0.30 a) determine the order of the reaction b) determine the initial rate of disappearance of X when the concentration of X is 0.30 M and that of Y is 0.40 M Consider the reaction A B The rate of the reaction is 1.6 x10-2 M/s when the concentration of A is 0.35 M. Calculate the rate constant if the reaction is first order in A and second order in A. - 65 - 7. The rate constant for the second order reaction 2 NOBr(g) 2 NO(g) + Br(g) is 0.80 M-s at 10 C. 8. a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. b) Calculate the half-lives when [NOBr]0 = 0.072 M and [NOBr]0 = 0.054 M The rate constant for the second order reaction 2 NO2(g) 2 NO(g) + O2 (g) is 0.54 M-s at 300 C. How many seconds would it take for the concentration of NO2 to decrease from 0.62 M to 0.28 M? 9. The burning of methane in oxygen is a highly exothermic reaction. Yet a mixture of methane and oxygen gas can be kept indefinitely without any apparent change. Explain. - 66 - 10. The rate law for the reaction 2 NO(g) + Cl2(g) 2 NOCl(g) is given by rate = k [NO][Cl2]. a) What is the order of the reaction? b) A mechanism involving the following steps has been proposed for the reaction NO(g) + Cl2(g) NOCl2(g) + NO(g) NOCl2(g) 2 NOCl(g) If this mechanism is correct, what does it imply about the relative rates of these two steps? 11. For the reaction X2 + Y + Z XY + XZ it is found that doubling the concentration of X2 doubles the reaction rate, tripling the concentration of Y triples the rate, and doubling the concentration of Z has no effect. a) What is the rate law for this reaction? b) Why is it that the change in the concentration of Z has no effect on the rate? - 67 - c) Suggest a mechanism for the reaction that is consistent with the rate law. 12. The decomposition of N2O to N2 and O2 is a first order reaction. At 730 C the half-life of the reaction is 3.58 x103 min. If the initial pressure of N2O is 2.10 atm at 730 C, calculate the total gas pressure after one half-life. Assume that the volume remains constant. - 68 - Equilibrium Worksheet 1. What is the value of Kc at 225 C? CO(g) + 2 H2(g) CH3OH(g) Kp = 6.3 x 10-3 2. At 773 C the equilibrium shown above has a Kc of 0.40. What is the Kp, in torr, at this temperature? 3. If 0.100 mol of HCl and solid I2 are placed in a 3.00 L container, at 25 C, and the following equilibrium occurred: 2 HCl(g) + I2(s) Cl2(g) + 2 HI(g) Kc = 1.6 x 10-34 What will be the equilibrium concentration of HI and Cl2 in the container? - 69 - 4. Based on the following equilibrium: SO3(g) + NO(g) NO2(g) + SO2(g) Kc = 0.500 if 0.240 mol of SO3 and 0.240 mol of NO are placed in a 2.00 L vessel and allowed to react, what will be the equilibrium concentration of each gas? 5. Based on the equilibrium shown below: 2 HCl(g) Cl2 (g) + H2(g) Kc = 3.2 x 10-34 if a flask initially contains 0.0500 M of HCl which then reacts to reach equilibrium, what will be the concentration of H2 and Cl2? 6. The equilibrium system PCl5 PCl3 + Cl2 Kp = 1.05 The system is found to have pressures for PCl5, PCl3, and Cl2 of 0.177 atm, 0.223 atm and 0.111 atm, respectively. Is the system at equilibrium? Prove with a calculation and predict which way it will shift if it is not already at equilibrium. - 70 - 7. When 1.05 moles of Br2 are put into a 0.980 L flask, 1.20% of the Br 2 dissociates into Br atoms. Calculate the Kc for the following reaction Br2(g) 8. Ammonium carbamate, equilibrium: 2 Br(g) NH4CO2NH2 NH4CO2NH2(s) decomposes into the following 2 NH3(g) + CO2(g) Starting with only the solid, it is found at 40 C the total gas pressure is 0.363 atm. Calculate the Kp for the reaction. 9. Consider the reaction 2 SO2(g) + O2(g) 2 SO3(g) ∆H = - 198.2 kJ if the reaction has established equilibrium and then the following stresses added, what would happen to the molarity of SO 3? a) increase the temperature b) remove some O2 - 71 - c) move the reaction mixture to a larger volume container d) add some SO2 e) add a catalyst - 72 - Acid-Base Worksheet 1. Determine the Molarity of H2S solution that produces a pH of 4.00? (Ka = 1.10 x10-7) H2S(g) + H2O(l) H +(aq) + HS -(aq) 2. Determine the [OH -] when 0.270 g of C5H5N is dissolved in 100. mL of water? (Kb = 1.70 x10-9) 3. Determine the pH and equilibrium concentrations of both products and reactants of a 2.00 M NH3 solution. (Kb = 1.80 x10-4) 4. Determine the pOH when 0.500 M CH3COOH ionizes 1.36%. CH3COOH + H2O - 73 - H + + CH3COO - 5. 3.42 g of Mg(OH)2 is added to enough water to produce 350.0 mL of solution. This solution is then diluted to 658. mL. What is the Molarity of the final solution? 6. How many mL of 0.250 M H2SO4 is required to neutralize 0.0500 mol OH - in Al(OH)3? 7. How many moles of Sr(OH)2 is required to neutralize 235. mL of 0.0560 M H3PO4? 8. How many g of HC2H3O2 are needed to neutralize 50.0 mL of 0.100 M Ca(OH)2? - 74 - Acid-Base Reactions Worksheet #1 1. 0.250 L of 0.100 M Mg(OH)2 is required to neutralize 4.63 g of a monoprotic acid. What is the molar mass of this acid? 2. 25.0 mL of 0.100 M NaOH is reacted with 75.0 mL of 0.0200 M H2SO4. What is the [OH -] of this solution? 3. 3.50 L of a 0.0400 M H3PO4 reacts with 0.500 L of a 0.160 M Ba(OH)2. Show that the resulting concentration is 0.0217 M H3PO4. - 75 - 4. 38.0 mL of 0.300 M Al(OH)3 reacts with 20.0 mL of 0.0300 M HClO4. What is the [H3O +] of this solution? 5. A dye mixture contains an unknown amount of H 2SO4. 60.0 g of this sample neutralizes 36.50 mL of 0.250 M Ca(OH)2. Determine the percent, by mass, of sulfuric acid in the sample. 6. 45.76 mL of 0.250 M hydroiodic acid is added to 2.09 g of barium hydroxide. Is the solution acidic or basic? What is the concentration of hydronium and hydroxide ions in this solution? 7. How many grams of carbonic acid are required to neutralize 35.00 mL of an aluminum hydroxide solution with a pH of 10.60? - 76 - 8. The empirical formula of ascorbic acid is HC3H3O3. A 0.200 g sample required 17.48 mL of 0.0650 M Ba(OH)2 to reach neutralization. Assuming the acid is diprotic, calculate its molar mass and its molar formula. 9. If 45.0 mL of 0.490 M Ba(OH)2 is added to 36.00 mL of 0.330 M HNO 3, will the resulting solution be acidic or basic? Identify the final [H +] and [OH -] concentration? 10. Acetic acid, HC2H3O2, gives vinegar its sour taste. A 0.110 g sample of vinegar requires 4.62 x10-3 L of 0.0150 M KOH to completely neutralize it. What is the percent by mass of acetic acid in vinegar? - 77 - 11. 6.80 mL of a monoprotic acid is diluted to 12.85 mL. 4.00 mL of this is reacted with 5.86 mL of 0.160 M solution of NaOH. What is the original Molarity? 12. How much water must be added to 0.650 M solution of HNO 3 to produce 500.00 mL of 0.330 M HNO3? - 78 - Acid-Base Reactions Worksheet #2 1. In water, hydrazoic acid, HN3, is a weak acid that has an equilibrium constant, Ka, equal to 2.80 x10-5 at 25° C. A 0.300 L sample of 0.0500 M solution of the acid is prepared. a) Write the expression for the equilibrium constant, K a, for hydrazoic acid. b) Calculate the pH of this solution at 25° C. c) To 0.150 L of this solution, 0.800 g of sodium azide, NaN 3, is added. The salt dissolves completely. Calculate the pH of the resulting solution at 25° C if the volume of the solution remains unchanged. d) To the remaining 0.150 L of the original solution, 0.0750 L of 0.100 M of NaOH solution is added. Calculate the [OH -] for the resulting solution at 25° C. - 79 - 2. Ammonia is a weak base that dissociates in water. dissociation constant, Kb, for NH3 is 1.80 x10-5. NH3 + H2O At 25° C, the base NH4 + + OH - a) Determine the hydroxide ion concentration and the percentage dissociation of a 0.150 M solution of ammonia at 25° C. b) Determine the pH of a solution prepared by adding 0.0500 mol of solid ammonium chloride to 100. mL of a 0.150 M solution of ammonia. c) If 0.0800 mol of solid magnesium chloride, MgCl2, is dissolved in the solution prepared in part (b) and the resulting solution is well stirred, will a precipitate of Mg(OH)2 form? (Assume the volume of the solution is unchanged). The solubility product constant for Mg(OH)2 is 1.50 x10-11. - 80 - Identifying pH in a Titration A 50.0 mL sample of 0.100 M HCN, hydrogen cyanide, is titrated with 0.100 M NaOH. a) 0 mL of NaOH has been added: HCN(aq) H + (aq) + CN - (aq) Ka = 6.2 x10-10 Ka = [H +][CN -]/[HCN] 6.2 x10-10 [X]2/[0.100] [X] = 7.87 x10-6 pH = - log10 [7.87 x10-6] = 5.10 b) 8.00 mL of 0.100 M NaOH has been added: [H +] = Ka [HCN]/[CN -] [H +] = 6.2 x10-10 [7.24 x10-2]/[1.38 x10-2] [H +] = 3.25 x10-9 c) pH = 8.49 at the half-equivalence point: [H +]/Ka = [HCN]/[CN -] [H +] = Ka = 6.2 x10-10 d) pH = 9.21 at the equivalence point: CN - + H2O HCN + OH - Kb = Kw/6.2 x10-10 = 1.6 x10-5 Kb = [HCN][OH -]/[CN -] = 1.6 x10-5 = [X][ X]/[5.0 x10-2] X = [OH -] = 8.9 x10-4 pOH = 3.05 pH = 10.95 - 81 - Choosing an Indicator http://www.psigate.ac.uk/newsite/reference/plambeck/chem1/p01172.htm Indicator pH range methyl violet thymol blue methyl yellow methyl orange bromocresol green methyl red chlorophenol red bromothymol blue phenol red cresol purple thymol blue phenolphthalein thymolphthalein alizarin yellow R indigo carmine 0.0- 1.6 1.2- 2.8 2.9- 4.0 3.1- 4.4 3.8- 5.4 4.2- 6.2 4.8- 6.4 6.0- 7.6 6.4- 8.0 7.4- 9.0 8.0- 9.6 8.0- 9.8 9.3-10.5 10.1-12.0 11.4-13.0 pKa Acid Form 0.8 1.6 3.3 4.2 4.7 5.0 6.0 7.1 7.4 8.3 8.9 9.7 9.9 11.0 12.2 yellow red red red yellow red yellow yellow yellow yellow yellow colorless colorless yellow blue Base Form blue yellow yellow yellow blue yellow red blue red purple blue red blue red yellow A chemical indicator is a compound which changes color indicating when the endpoint of a titration has been reached. The color of the indicator changes are the result of the concentrations of ions in the solution. An acid-base indicator is an weak acid-weak base conjugate pair, in which the two forms have different colors. The indicator is added in low concentration to minimize any change to the pH of the system. Ka = [H3O +][In -]/[HIn] When [In -]/[HIn] = 1 [H3O +] = Ka pH = pKa In order to choose an indicator that will work for an acid – base reaction, you must have an approximate idea where the equivalence point would be. Using the expected pH range when neutralization occurs, the indicator with a pK a value within ±1 pH unit of the equivalence point will be the correct indicator. - 82 - - 83 - Buffers & Salts Worksheet 1. How does a basic buffer differ from a base that dissolves in distilled water? 2. Working with the buffer shown below: 0.350 M 0.200 M NH3 NH4 + + OH – a) is the concentration of OH – be the same as NH4 +? choice. b) what is the starting pH of this buffer? c) when 0.150 M HNO3 is added to this buffer, identify the balanced net ionic equation showing how the buffer neutralizes the acid. d) calculate the equilibrium concentrations of the buffer after the acid has been neutralized? - 84 - Explain your 3. e) calculate the pH of the solution when the acid has been neutralized? f) write the net ionic equation identifying what change occurs at the equivalence point. A 25.00 mL sample of a 0.500 M HN3 solution is titrated with a 0.350 M solution of KOH. Ka = 2.80 x10-5 a) What is the pH before any base is added? b) Is the [H +] and [N3 -] the same BEFORE any base is added? Explain your choice. c) BEFORE base is added, is the HN3 solution a buffer? Explain your choice. d) Calculate the pH after 15.25 mL of a 0.350 M KOH solution is added. - 85 - e) Identify the pH at the ½ equivalence point. f) How does a ½ equivalence point differ from an equivalence point? g) Is a buffer is created when the ½ equivalence point is reached? Explain. h) How many mL of the KOH solution must be added to neutralize the HN 3 solution? i) When neutralization has occurred, what is the pH of the solution? - 86 - Essential Information for the AP Chemistry Exam Strong Acids: HCl, HBr, HI, HNO3, H2SO4, HClO4 Strong Bases: LiOH, NaOH, KOH, Sr(OH)2, Ba(OH)2 Soluble Salts: lithium, sodium, potassium, ammonium cations, nitrate, acetate; chlorides except Ag, Pb, Hg(I), sulfates except Pb, Ca, Sr, Ba. All other salts should be considered only slightly soluble Complex Ion Formation Ag(NH3)2 + Al(OH)4- or AgCl + NH3 Al(OH)3 + OH - + Cl Al(OH)6 3- [Co(NH3)6] 2+ [CoCl4] 2- [Ni(NH3)6] 2+ [NiCl4] 2- [Zn(NH3)6] 2+ [Zn(OH)4] 2- [Ag(NH3)] 2+ [Cu(NH3)4] 2+ [Cu(H2O)6] 2+ [Al(OH)4] – [Al(OH)6] 3- Hydrolysis Reactions Three types of salts exist; acidic, basic, and neutral. When acidic and basic salts are added to water the number of hydrogen and hydroxide ions change, defining what the pH of the solution will be. Neutral Salts When neutral salts dissociate in water, both the cation and anion are spectator ions; neither ion has the ability to tie up or contribute H + in the solution. Ex: KNO3, NaCl Acidic Salts Acidic salts dissociate in water and increase the number of H + in the solution. NH4Cl: NH4 + + H2O - 87 - NH3 + H3O + Basic Salts Basic salts dissociate in water and decrease the number of H + in the solution. C2H3O2 - NaC2H3O2: HC2H3O2 + OH - Reactions Involving a Change in Oxidation State Reactions between oxidizers and reducers can usually be predicted from the following list of reagents: Important Reducers Product halide ions free metals sulfite ions nitrite ions free halogens (dil, OH -), Cl2 free halogen metal ions sulfate ions nitrate ions hypohalite ions, ClO - free halogens (conc, OH -), Cl2 metallous ions halate ions, ClO3 metallic ions Important oxidizers Product MnO4 - (H +) Mn 2+ MnO2 (H +) Mn 2+ MnO4 - (neutral/OH -) Cr2O7 2- (H MnO2 +) Cr 3+ Cr2O7 2- (OH -) HNO3 (con) HNO3 (dilute) H2SO4 (hot, conc) metallic ions free halogens, X2, (F2, Cl2, etc) Na2O2 CrO4 2NO2 NO SO2 metallous ions halide ions: F -, Cl -, I NaOH HClO4 Cl - - 88 - Basic Organic Chemistry Oxidation: complete oxidation CO2 + H2O incomplete oxidation CO + H2O Addition: involves the breaking of a multiple bond through the addition of a mole of a reactant. Addition reactions occur with H2 (catalyst Ni, Pt, or Pd), halogens, hydrogen halide, water. C2H4 + Br2 Substitution: C2H4Br2 Hydrocarbons usually lacking multiple bonds. The hydrogen atom leaves organic compound, becomes bonded to an inorganic reactant, and is replaced by an atom/radical from inorganic reactant producing two products C2H6 + Br2 Esterification: C2H5Br + HBr Organic acid reacts with an alcohol to produce an ester and water. CH3COOH + CH3OH CH3COO - CH3+ Being able to identify the names of organic compounds with 1-6 carbons is helpful. meth but C1 C4 eth pent C2 C5 - 89 - prop hex C3 C6 Functional Groups - Knowledge of functional groups is essential. Isomers – Organic compounds with the same number of the same type of atoms, set in different positions. E. Basics in Radioactivity A basic knowledge of changes in the nucleus of a radioactive isotope is required. alpha particles 4 2 - 90 - beta particles 0 1 gamma radiation 0 0 no change in nuclear charge or mass number - 91 - The AP Chemistry Exam Section I: You have 90 minutes to answer 75 multiple-choice questions; 72 seconds per question. The first time through the multiple choice, focus only on questions that can be answered within 20 - 30 seconds, then return to work on those questions that require more time. There are 10 questions you will not be able to answer. Accept that concept. Total scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers. The majority of students who get a 4 or 5 are able to correctly answer at least 50 of the 75 questions. A simplified periodic table is provided. Section II: Part A: You have 95 minutes to answer six questions. You have 55 minutes to respond to 3 problems. Calculators can ONLY be used in this part of the exam. ● One point is subtracted for arithmetic errors or not calculating the answer. ● Clearly define all unknowns, show reasoning and each formula used. ● Clearly show all work that leads to the numerical answer. Make sure the grader can follow your work. ● If you do not have time to complete the arithmetic set-up, identify the expected unit and number of significant figures in the answer. It ensures you don’t lose points. ● If you are unsure how to respond to a question, move to the next question and write, "Assuming the answer to the previous question was . . .", and continue with the problem. - 92 - Part B You have 40 minutes to answer 3 questions. Question 4 requires balanced net ionic equations for three reactions, along with an answer to a question about each reaction. Questions 5 and 6 test your understanding of theory. Answer questions 5 and 6 clearly, logically, and in a sequential order so the grader can follow your work. Make sure you have addressed each point in the question, but DO NOT write to write. A simplified periodic table and a list of equations and constants are provided. Multiple choice and free response portions of the test are equally weighted. - 93 -