Download Exam Review Packet Table of Contents

Exam Review Packet Table of Contents Directions ..................................................................................................................................................... 3 Periodic Table and Equations / Constants Sheet ......................................................................................... 4 Big Idea 1: The chemical elements are fundamental building materials of matter, and all matter can be understood in terms of arrangement of atoms. These atoms retain their identity in chemical reactions 7 Concepts -­‐ ATOMIC THEORY, BONDING AND PERIODIC TRENDS ............................................................ 7 Free Response Questions ........................................................................................................................ 7 Multiple Choice Questions ..................................................................................................................... 11 Big Idea 2: Chemical and physical properties of materials can be explained by the structure and arrangement of atoms, ions, or molecules and the forces between them ............................................... 12 Concepts -­‐ BONDING, LEWIS STRUCTURES AND INTERMOLECULAR FORCES ....................................... 12 Free Response Questions ...................................................................................................................... 12 Multiple Choice Questions ..................................................................................................................... 22 Big Idea 3: Changes in matter involve the rearrangement and / or reorganization of atoms and / or the transfer of electrons .................................................................................................................................. 24 Concepts – ELECTROCHEMISTRY ........................................................................................................... 24 Free Response Questions ...................................................................................................................... 25 Multiple Choice Questions ..................................................................................................................... 31 Big Idea 4: Rates of chemical reactions are determined by details of the molecular collisions ............... 34 Concepts List – KINETICS ........................................................................................................................ 34 Free Response Questions ...................................................................................................................... 35 Multiple Choice Questions ..................................................................................................................... 42 Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the direction of changes in matter ............................................................................................................ 46 Concepts – THERMODYNAMICS ............................................................................................................ 46 Free Response Questions ...................................................................................................................... 46 Multiple Choice Questions ..................................................................................................................... 52 Big Idea 6: Any bond or intermolecular attraction that can be formed can be broken. These two processes are in a dynamic competition, sensitive to initial conditions and external perturbations. ...... 55 Concept List – EQUILIBRIUM ................................................................................................................. 55 1 Free Response Questions ...................................................................................................................... 62 Multiple Choice Questions ..................................................................................................................... 77 Laboratory ................................................................................................................................................. 80 Concepts – LABORATORY QUESTIONS – contained in several Big Ideas and the Science Practices ...... 80 Free Response Questions ...................................................................................................................... 80 Multiple Choice ...................................................................................................................................... 90 2 Directions Multiple Choice Section In section 1, there are 60 multiple choice questions. These questions represent the knowledge and skills students should know, understand, and be able to apply. Students will be given a periodic table and an equations and constants list to use during this section. For all questions, assume that the temperature is 298 K, the pressure is 1.00 atmosphere, and solutions are aqueous unless otherwise specified. Free Response section Section II Directions: Questions 1 through 3 are long constructed response questions that should require about 20 minutes to answer. Questions 4 through 7 are short constructed response questions that should require about 7 minutes each to answer. Read each question carefully and write your response in the space provided following each question. Your responses to these questions will be scored on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Specific answers are preferable to broad, diffuse responses. For calculations, clearly show the method used and the steps involved in arriving at your answers. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. 3 Periodic Table and Equations / Constants Sheet 4 5 6 Big Idea 1: The chemical elements are fundamental building materials of matter, and all matter can be understood in terms of arrangement of atoms. These atoms retain their identity in chemical reactions Concepts -­‐ ATOMIC THEORY, BONDING AND PERIODIC TRENDS 1. Quantum Numbers, electron configurations, Hund’s rule, orbital diagrams 2 Trends of the periodic table a) size for atoms and ions b) size of ions c) IE, EA, EN 3. Effective nuclear charge (Zeff ) increases as more protons added to same energy level Zeff is a comparison tool. Coulomb’s Law F=kqq/d2 4. Effective nuclear charge (Zeff ) decreases as more shielding electrons are present. 5. When students talk about EN differences they are talking about bonds (within a molecule), we need them to talk about IMF (between molecules ) 6. Students often talk about atoms “wanting to gain/lose electrons”, being happy, Full, rather than having a stable octet, complete energy level. 7. Correct use of spectroscopy (UV, IR, VIS, PES ) 8. Vocabulary IE (ionization energy) EN (electronegativity) EA (electron affinity) Free Response Questions 1987 Q5 5. Use the details of modern atomic theory to explain each of the following experimental observations. a. Within a family such as the alkali metals, the ionic radius increases as the atomic number increases. b. The radius of the chlorine atom is smaller than the radius of the chloride ion, Cl¯ˉ. (Radii: Cl atom = 0.99 Å; Cl-­‐ ion = 1.81 Å) c. The first ionization energy of aluminum is lower than the first ionization energy of magnesium. (First ionization energies: 12Mg = 7.6 ev, 13Al = 6.0 ev) 7 d. For magnesium, the difference between the second and third ionization energies is much larger than the difference between the first and second ionization energies. (Ionization energies, in electron-­‐volts, for Mg: 1st = 7.6, 2nd = 14, 3rd = 80) 5) Special rules for this question: An ambiguous 2 pt. response receives only 1 pt An incorrect statement in an otherwise correct 2 pt response will result in a score of 1 pt The answers labeled (i) below received two points; (ii) received one point. a) two points -­‐ The radii of the alkali metal ions increase with increasing atomic number because: (i) the principle quantum number (or shell or energy level) increases (ii) there is an increase in shielding (or the number of orbitals increases) b) two points -­‐ The chloride ion is larger than the chlorine atom because: (i) electron-­‐ electron repulsion increases (or shielding increases or the electron-­‐proton ratio increases or the effective nuclear charge decreases) (ii) an extra electron generally increases the size c) two points -­‐ The ionization energy of Mg is greater than that for Al because: (i) the 3p orbital is at a higher energy than the 3s orbital (or the electron in Al is shielded from the nucleus more completely by the 3s electron than the 3s electrons shield one another from the nucleus) (ii) a 3p electron is easier to remove than a 3s electron d) two points -­‐ The much greater difference between the 2nd and 3rd ionization energies in Mg (relative to the difference between the 1st and 2nd) is due to the 3rd electron being removed from the 2p subshell after the first 2 were removed from the 3s subshell. 1987 Q9 9. Two important concepts that relate to the behavior of electrons in atomic system are the Heisenberg uncertainty principle and the wave-­‐particle duality of matter. a. State the Heisenberg uncertainty principle as it relates to determining the position and momentum of an object. b. What aspect of the Bohr theory of the atom is considered unsatisfactory as a result of the Heisenberg uncertainty principle? c. Explain why the uncertainty principle or the wave nature of particles is not significant when describing the behavior of macroscopic objects, but is very significant when describing the behavior of electrons. a) three points -­‐ It is impossible to determine (or measure) both the position and momentum of any particle (or object, or body) simutaneously. 8 OR The more exactly the position of a particle is known, the less exactly the position or velocity of the particle is known. OR (Δx) (Δp) [greater than or equal to] h-­‐bar (or h / 2π) h = Planck's constant Δx = uncertainty in position Δp = uncertainty in momentum Notes: 1 point is given for the notion of simultaneous determination. (A number of students give the first sentence but omit the word simultaneously. (They got 2 out of 3.) If the second or third versions of the answer are given, simultaneity is understood. If they give the equation, they must have a > sign, not just an = sign, or they lose 1 pt. A student who gives a correct answer but adds erroneous material gets one point deducted. b) five points Bohr postulated that the electron in a H atom travels about the nucleus in a circular orbit and has a fixed angular momentum. With a fixed radius of orbit and a fixed momentum (or energy), (Δx) (Δp) < h/4pi and violates the uncertainty principle. Students receive 2 pts total for the above. If they say only "the electron travels in circular orbits", they get 1 point. They also get credit for saying elliptical orbits. To receive full credit, they must describe an aspect of the Bohr theory. The following portion of the answer is worth three points. The wavelength of a particle is given by the deBroglie relation: λ = h/mv For masses of macroscopic objects, h/m is so small for any v that λ is so small as to be undetectable. For an electron, m is so small that h/mv yields a detectable λ. OR They may say the product of the uncertainties in postion and velocity depends on h/m and since h is so small (h = 6.63 x 10¯ˉ34 J s) unless m is very small, as for an electron, the product of the uncertainties is too small to be detected. 9 Students may discuss the fact that measuring the position and momentum requires having a photon strike the particle. The photon has an energy comparable to that of an electron but small compared to that of a macroscopic object. They must stress mass rather than size as the important distinction to get full credit. A student who says λ or (Δx) (Δp) depends on "size" and not mass gets 2 points if discussion is otherwise okay. 1994 Q9 9. Use principles of atomic structure and/or chemical bonding to answer of each of the following. a. The radius of the Ca atom is 0.197 nanometers; the radius of the Ca2+ ion is 0.099 nanometer. Account for this difference. b. The lattice energy of CaO(s) is -­‐3,460 kilojoules per mole; the lattice energy for K2O(s) is -­‐2,240 kilojoules per mole. Account for this difference. Ionization Energy First Second K 419 3,050 Ca 590 1,140 c. Explain the difference between Ca and K in regard to: i. ii. 9. their first ionization energies. their second ionization energies. a) The valence electrons in a calcium atom are the 4s2. In a calcium ion these electrons are absent and the highest energy electrons are 3p, which has a much smaller size because the (-­‐) / (+) charge ratio is less than 1 causing a contraction of the electron shell. b) Lattice energy can be represented by Coulomb’s law: lattice energy = k Q1 Q2 / r, where Q1 and Q2 are the charges in the ions, in CaO these are +2 and -­‐2 respectively, while in K2O they are +1 and -­‐2. The r (the distance between ions) is slightly smaller in CaO, combined with the larger charges, thus accounts for the larger lattice energy. c) Electron arrangements: K = [Ar] 4s1, Ca = [Ar] 4s2 Potassium has a single 4s electron that is easily removed to produce an [Ar] core, whereas, calcium has paired 4s electrons which require greater energy to remove one. A K+ ion has a stable [Ar] electron core and requires a large amount of energy to destabilize it and create a K2+ ion. Ca+ has a remaining 4s1 electron that is more easily removed than a core electron, but not as easily as its first 4s electron. 10 d) Electron arrangements: Mg = [Ne]3s2, Al = [Ne] 3s2 3p1 It is easier to remove a shielded, single, unpaired 3p electron from the aluminum than to remove one electron from a paired 3s orbital in magnesium. Multiple Choice Questions 18. 22. Which of the following elements has the largest first ionization energy? a. Li b. Be c. B d. C e. N How many protons, neutrons, and electrons are in an 5626Fe atom? Protons Neutrons Electrons a. b. c. d. e. 23. 30 56 26 26 82 26 26 30 26 56 1s2 2s22p6 3s23p1 1s2 2s22p6 3s23p5 1s2 2s22p6 3s2 1s2 2s22p6 3s1 1s2 2s22p6 3s13p1 Which of the following lists Mg, P, and Cl in order of increasing atomic radius? a. b. c. d. e. Which of the following is the electron configuration of an excited atom that is likely to emit a quantum of energy? a. b. c. d. e. 57. 26 26 30 56 56 Cl < P < Mg Cl < Mg < P Mg < P < Cl Mg < Cl < P P < Cl < Mg 11 Big Idea 2: Chemical and physical properties of materials can be explained by the structure and arrangement of atoms, ions, or molecules and the forces between them Concepts -­‐ BONDING, LEWIS STRUCTURES AND INTERMOLECULAR FORCES 1. 2. 3. 4. 5. 6. Ionic bonds Covalent bonds, Lewis structures, geometric shapes, bond polarity, molecular polarity, resonance, hybridization, London dispersion forces (LDF), inter vs. intramolecular forces Intermolecular Forces (IMF) are between molecules and help explain differences in FP, BP, solids, liquids, gases, and solubilities. a. ion – ion b. dipole – dipole with H bonding c. dipole – dipole d. London dispersion forces ( LDF ) When students talk about EN differences they are talking about bonds (within a molecule), we need them to talk about IMF (between molecules ) Molecular polarity depends on bond polarity and shape of the molecule Vocabulary IE (ionization energy) EN (electronegativity) EA (electron affinity) Free Response Questions 2013 Q5 5. Methanamide, CH3NO, is a liquid at 25 oC. a. The complete Lewis electron dot diagram for methanamide is shown below. 12 i. In the molecule, angle x is not 180o. Estimate the observed angle. Justify your answer. ii. In the molecule, angle y is not 90o. Explain why in terms of electron domains (VSEPR model). b. Consider a molecule with the formula CH2O2. The structure of this molecule has a geometry around the carbon atom similar to the geometry around carbon in methanamide. In the box provided below, draw the complete Lewis electron dot diagram for the molecule. a. i. Angle x is approximately 120o. Three electron domains around the carbon atoms will maximally separate the electrons and minimize the energy when the bond angles are 120o. 1 point is earned for the correct angle with justification. Note: accept 120o ± 10o for the angle (i.e. 110 to 130o). Also accept steric number (SN) = 3 or trigonal planar geometry for the justification in part I only. a. ii. Angle y is 109.5o Four electron domains around the nitrogen atom will maximally separate the electrons and minimize the energy when the bond angles are 109.5o OR Angle y is approximately 120o. Considering possible resonance structures involving a double bond between the C and N atoms, there are three electron domains around the N atom. In this case, minimization of energy leads to bond angles of approximately 120o. 1 point is earned for the justification of an angle different than 90o. Note, the justification need not give a specific bond angle, but it must mention the repulsion of 4 electron domains (or 3 electron domains if resonance structures are mentioned). b. 2 points are earned for a correct Lewis electron dot structure for formic acid. Notes: 1 point is earned for the correct skeletal structure for formic acid with the C=O double bond (i.e., containing all five bonding pairs) but missing one or more lone pairs. Also 1 point is earned for a Lewis electron dot diagram representing i) the correct molecular formula, CH2O2, ii) with three electron domains and at least three bonded pairs of electrons around the carbon atom, with no more than three bonded pairs of electrons around any oxygen atom, and iii) the proper distribution of all 18 electrons in accordance with the octet rule. 2013 PR Q6 13 6. The structures of a water molecule and a crystal of LiCl(s) are represented above. A student prepares a 1.0 M solution by dissolving 4.2 g of LiCl(s) in enough water to make 100 mL of solution. a. In the space provided below, show the interactions of the components of LiCl(aq) by making a drawing that represents the different particles present in the solution. Base the particles in your drawing on the particles shown in the representations above. Include only one formula unit of LiCl and no more than 10 molecules of water. Your drawing must include the following details. Identity of ions (symbol and charge) The arrangement and proper orientation of the particles in the solution The sketch should clearly show 1. A clear representation of at least 1 Li+ ion and 1 Cl-­‐ ion separated from each other, labeled and charged. 2. Each ion surrounded by at least 2 H2O molecules; and 3. H2O molecules with the proper orientation around each ion (i.e., the oxygen end of the H2O molecule closer to the Li+ ion and the hydrogen end of the water molecules closer to the chloride ion) 1 point is earned for a correctly drawn and labeled particulate representation of the ions . (Representation must indicate that the smaller ion is Li+. Representations that include more than one formula unit of LiCl (dissolved or undissolved) are acceptable as long as at least one of the formula units is separated into ions and the ions are correctly labeled with their respective identities and charges.) 1 point is earned for a correctly drawn particulate representation of waters of hydration surrounding the ions. 1 point is earned for correctly representing the orientation of the water molecules of hydration with the proper polarity. 14 2013 PR Q7 7. HIn(aq) + H2O(l)  In-­‐(aq) + H3O+(aq) Yellow blue The indicator HIn is a weak acid with a pKa value of 5.0. It reacts with water as represented in the equation above. Consider the two beakers below. Each beaker has a layer of colorless oil (a nonpolar solvent) on top of a layer of aqueous buffer solution. In beaker X the pH of the buffer solution is 3, and in beaker Y the pH of the buffer solution is 7. A small amount of HIn is placed in both beakers. The mixtures are stirred well, and the oil and water layers are allowed to separate. a. What is the predominant form of HIn in the aqueous buffer in beaker Y, the acid form or the conjugate base form? Explain your reasoning. b. In beaker X the oil layer is yellow, whereas in beaker Y the oil layer is colorless. Explain these observations in terms of both acid-­‐base equilibria and interparticle forces. a. The conjugate base form, In(aq), is the predominant form of the indicator in the aqueous pH 7 buffer in beaker Y. This is because the pH is greater than the pKa of HIn, causing the equilibrium to form a significant amount of products, In-­‐(aq) and H3O+(aq) 1 point is earned for correctly identifying In (aq) as the predominant form in the aqueous layer in beaker Y because the solution is not acidic (may be implicit). 1 point is earned for stating that pH > pKa and that this causes the equilibrium to favor the products. b. At pH 3, the acid form HIn(aq) predominates in the aqueous layer of beaker X because pH < pKa. Since HIn(aq) is a neutral molecule, some of it can dissolve in the oil layer of 15 beaker X because of London dispersion interactions with the oil, causing the oil layer to be yellow. Since In-­‐(aq) is charged, it will preferentially dissolve in the aqueous layer of beaker Y because of ion-­‐dipole interactions with the water, leaving the oil layer colorless. 1 point is earned for explaining the yellow color in the oil layer of beaker X in terms of acid-­‐base equilibrium and interparticle forces between HIn molecules and oil molecules. 1 point is earned for explaining the colorless oil layer of beaker Y in terms of interparticle forces between In-­‐ ions and water molecules. 2010 Q5 5. Use the information in the table below to respond to the statements and questions that follow. Your answers should be in terms of principles of molecular structure and intermolecular forces. a. Draw the complete Lewis electron dot diagram for ethyne in the appropriate cell in the table above. The lower right cell contains two possible structures H-­‐CC-­‐H and the equivalent dot structure. 1 point for the correct structure. b. Which of the four molecules contains the shortest carbon-­‐carbon bond? Explain. Ethyne, which contains a triple bond, has the shortest C to C bond. The other molecules have single C to C bonds, and triple bonds are shorter than single bonds. 1 point is earned for the correct choice. 1 point is earned for the correct explanation. 16 c. A Lewis electron dot diagram of a molecule of ethanoic acid is given below. The carbon atoms in the molecule are labeled x and y, respectively. Identify the geometry of the arrangement of atoms bonded to each of the following. i. Carbon x Trigonal planar. 1 point is earned for the correct geometry. ii. Carbon y Distorted tetrahedral, tetrahedral, or trigonal pyramidal. 1 point is earned for the correct geometry. d. Energy is required to boil ethanol. Consider the statement “As ethanol boils, energy goes into breaking C-­‐C bonds, C-­‐H bonds, C-­‐O bonds, and O-­‐H bonds.” Is the statement true or false? Justify your answer. The statement is false. All of the bonds described are intramolecular; these bonds are not broken during vaporization. When ethanol boils, the added energy overcomes intermolecular, not intramolecular forces. One point for the correct choice with justification. e. Identify a compound from the table above that is nonpolar. Justify your answer. Either ethane or ethyne can be identified as nonpolar. The ethane / ethyne molecule is nonpolar because the bond dipoles in the molecule cancel. OR The ethane / ethyne molecule is nonpolar because the molecule is symmetric. Note: Explanation must refer to the shape of the molecule. Statements such as “All hydrocarbons are nonpolar.”, “The carbons are surrounded by hydrogens”, or “there are no lone pairs” do not earn the point. 1 point is earned for a correct choice with justification. 17 f. Ethanol is completely soluble in water, whereas ethanethiol has limited solubility in water. Account for the difference in solubilities between the two compounds in terms of intermolecular forces Ethanol is able to form strong hydrogen bonds with water whereas ethanethiol does not have similar capability. The formation of hydrogen bonds increases the attraction between molecules of ethanol and molecules of water, making them more soluble in each other. Note: The answer must clearly focus on the solute-­‐solvent interaction. Just the mention of hydrogen bonding does not earn the point. One point is earned for the correct explanation. 2011B #6 6. Use principles of molecular structure, intermolecular forces, and kinetic molecular theory to answer the following questions. a. A complete Lewis electron dot diagram of a molecule of ethyl methanoate is given below. i. Identify the hybridization of the valence electrons of the carbon atoms labeled Cw. sp2 ii. Estimate the numerical value of the Hy – Cx – O bond angle in an ethyl methanoate molecule. Explain the basis of your estimate. The Cx is the central atom in a tetrahedral arrangement of bonding electron pairs; thus the angle would be approximately 109.5o. 1 point is earned for the correct angle with an appropriate explanation. 1 point is earned for the correct answer b. Ethyl methanoate, CH3CH2OCHO, is synthesized in the laboratory from ethanol, C2H5OH, and methanoic acid, HCOOH, as represented by the following equation. C2H5OH(l) + HCOOH(l)  CH3CH2OCHO(l) + H2O(l) i. In the box below, draw the complete Lewis electron dot diagram of a methanoic acid molecule. 18 ii. In the box below, draw the complete Lewis electron dot diagrams of a methanoic acid molecule and a water molecule in an orientation that allows a hydrogen bond to form between them. 2011 A Q5 5. Hydrazine is an inorganic compound with the formula N2H4. a. In the box below, complete the Lewis electron-­‐dot diagram for the N2H4 molecule by drawing in all the electron pairs. The correct electron dot diagram showing all electron pairs has one pair of electrons between each of the N and H’s, a single pair of electrons between the two N’s, and a lone pair on each of the N’s. 19 1 point for the correct Lewis diagram. b. On the basis of the diagram you complete in part (a), do all six atoms in the N2H4 molecule lie in the same plane? Explain. No, they do not. The molecular geometry surrounding both nitrogen atoms is trigonal pyramidal. Therefore, the molecule as a whole can not have all the atoms in the same plane. 1 point is earned for the correct answer with a valid explanation. c. The normal boiling point of N2H4 is 114 oC, whereas the normal boiling point of C2H6 is -­‐89 oC. Explain, in terms of the intermolecular forces present in each liquid, which the boiling point of N2H4 is so much higher than that of C2H6. N2H4 is a polar molecule with London dispersion forces, dipole-­‐dipole, and hydrogen bonding between molecules, whereas C2H6 is nonpolar and only has London dispersion forces between molecules. It takes more energy to overcome the stronger IMF’s in hydrazine, resulting in a higher boiling point. 1 point is earned for correct reference to the two different types of IMF’s. 1 point is earned for a valid explanation based on the relative strengths of the types of IMF’s. d. Write a balanced chemical equation for the reaction between N2H4 and H2O that explains why a solution of hydrazine in water has a pH greater than 7. N2H4 + H2O à N2H5+ + OH-­‐ 1 point is earned for a valid equation. N2H4 reacts in air according to the equation below. ΔHo = -­‐534 kJ mol-­‐1 N2H4(l) + O2(g) à N2(g) + 2 H2O(g) e. Is the reaction an oxidation-­‐reduction, acid-­‐base, or decomposition reaction? Justify your answer. The reaction is an oxidation-­‐reduction reaction. The oxidation state of Nitrogen changes from -­‐2 to 0 while that of O changes from 0 to -­‐2. 1 point is earned for a correct choice with a valid justification. f. Predict the sign of the entropy change, ΔS, for the reaction. Justify your prediction. The entropy change for the reaction is expected to be positive. There are three moles of gas produced from one mole of liquid and one mole of gas. The net increase of two moles of gas results in a greater entropy of products relative to the entropy of reactants. 20 1 point is earned for a correct prediction with a valid justification. g. Indicate whether the statement written below is true or false. Justify your answer. The large negative ΔHo for the combustion of hydrazine results from the large release of energy that occurs when the strong bonds of the reactants are broken. The statement is false on two counts. First, the energy is released when the bonds are broken, but rather when they are formed. Second, the bonds in the reactants are relatively weak compared to the bonds in the products. 1 point is earned for correctly identifying that the statement is false along with a valid justification. 2012 Q5 (2nd half) An experiment is performed to compare the solubilities of I2(s) in different solvents, water and hexane (C6H14). A student adds 2 mL of H2O and 2 mL of C6H14 to a test tube. Because H2O and C6H14 are immiscible, two layers are observed in the test tube. The student drops a small, purple crystal of I2(s) into the test tube, which is then corked and inverted several times. The C6H14 layer becomes light purple, while the H2O layer remains virtually colorless. d. e. i. Explain why the hexane layer is light purple while the water layer is virtually colorless. Your explanation should reference the relative strengths of interactions between molecules of I2 and the solvents H2O and C6H14, and the reasons for the differences. The hexane layer is purple because most of the I2 is dissolved in it. The entrance of the I2 into water requires disruption of the hydrogen bonds in water, which are much stronger than the London dispersion forces in hexane. Meanwhile, the London dispersion forces between I2 and hexane would be stronger than the London dispersion forces between I2 and water. (Water and I2 can also interact through a dipole-­‐induced dipole force, but this attraction is insufficient to overcome the other differences noted above.) 1 point is earned for recognizing from the experimental observations that the iodine dissolved in the hexane. 1 point is earned for a correct explanation referencing the differences between water and hexane in their interactions with I2. The student then adds a small crystal of KI(s) to the test tube. The test tube is corked and inverted several times. The I-­‐ ion reacts with I2 to form the I3-­‐ ion, a linear species. In the box below, draw the complete Lewis electron-­‐dot diagram for the I3-­‐ ion. 1 point is earned for a correct Lewis diagram. 21 ii. In which layer, water or hexane, would the concentration of I3-­‐ be higher? Explain. I3− would be more soluble in water because of the ion-­‐dipole interactions that would occur between the ions and the polar water molecules. No such interactions are possible in the nonpolar hexane. 1 point is earned for the correct choice and explanation. Multiple Choice Questions Questions 4-­‐7 refer to the following species. a. H2O b. NH3 c. BH3 d. CH4 e. SiH4 4. Has two lone pairs of electrons – a is correct response 5. Has a central atom with less than an octet of electrons – c is correct response 6. Is predicted to have the largest bond angle – c is correct response 7. Has a trigonal-­‐pyramidal molecular geometry – b is correct response 42. Which of the following lists the substances F2, HCl, and HF in order of increasing boiling point? a. b. c. d. e. 47. HF < HCl < F2 HF < F2 < HCl HCl < F2 < HF HCl < HF < F2 F2 < HCl < HF Which of the following is an isomer of CH3OCH3? a. CH3CH3 b. CH3COOH c. CH3CH2OH d. CH3CH2CH3 e. CH3CH2OCH2CH3 49. Which of the following substances has the greatest solubility in C5H12(l) at 1 atm? a. SiO2(s) 22 54. 71. b. NaCl(s) c. H2O(l) d. CCl4(l) e. NH3(g) High solubility of an ionic solid in water is favored by which of the following conditions? I. The existence of strong ionic attractions in the crystal lattice II. The formation of strong ion-­‐dipole attractions III. An increase in entropy upon dissolving a. I only b. I and II only c. I and III only d. II and III only e. I, II, and III Which of the following molecules contains exactly three sigma (σ) bonds and two pi (π) bonds? a. b. c. d. e. 73. Resonance is most commonly used to describe the bonding in molecules of which of the following? a. b. c. d. e. C2H2 CO2 HCN SO3 N2 CO2 O3 H2O CH4 SF6 23 Big Idea 3: Changes in matter involve the rearrangement and / or reorganization of atoms and / or the transfer of electrons Concepts – ELECTROCHEMISTRY 1. 2. 3. 4. 5. 6. 7. 8. oxidation / reduction – balancing equations (review) galvanic cells – {positive, Red Cat} electrolytic cells cathode anode current, charge, Faradays, (voltage / EMF) (amps, coulombs and volts – unit problem) cell notation salt bridge – “balance of charge” not electron balance, Good salt bridge materials are soluble salts, not easily oxidized or reduced, doesn’t interfere with given redox reaction, ie complex ion formation or precipitation 9. Eo and thermodynamically favored 10. ΔGo = -­‐ n F Eo 11. E = Eo – (0.059 / n) log Kc 12. Vocabulary – Galvanic Voltaic CONCENTRATION UNITS OF SOLUTIONS / COLLIGATIVE PROPERTIES 1. Molarity M = mole of solute/ L of solution 2. mole fraction = xa = mole of a /total moles in solution Colligative Properties (Conceptual Only) 24 1. 2. 3. 4. ∆ FP ↓ = (kf ) ( m ) ( i ) freezing point depression ∆ BP ↑ = ( kb ) (m ) ( i ) boiling point elevation ∏ = ( M ) ( R ) ( T ) ( i ) osmotic pressure Vapor Pressure Lowering = VPL = (x solvent) VP pure solvent i = Van’t Hoff factor for organic solutes nonelectrolytes i = 1 for electrolytes i = 2,3,4… NaCl i = 2 H2SO4 i = 3 AlCl3 i = 4 Free Response Questions 2013 PR Q1 1. A student performs an experiment in which the conductivity of a solution of Ba(OH)2 is monitored as the solution is titrated with 0.10 M H2SO4. The original volume of the Ba(OH)2 solution is 25.0 mL. A precipitate of BaSO4 (Ksp = 1.0 x 10-­‐10) formed during the titration. The data collected from the experiment are plotted in the graph above. a. As the first 30.0 mL of 0.10 M H2SO4 are added to the Ba(OH)2 solution, two types of chemical reactions occur simultaneously. On the lines proved below, write the balanced net ionic equations for (i) the neutralization reaction, and (ii) the precipitation reaction. i) Equation for neutralization reaction 25 _______________________ ii) Equation for precipitation reaction Ba2+(aq) + SO42-­‐(aq) à BaSO4(s) H+(aq) + OH-­‐(aq) à H2O(l) _______________________ 1 point is earned for each correct product. 1 point is earned for correct reactants with atoms and charges balanced in both reactions. b. The conductivity of the Ba(OH)2 solution decreases as the volume of added 0.10 M H2SO4 changes from 0.0 mL to 30.0 mL. i. Identify the chemical species that enable the solution to conduct electricity as the first 30.0 mL of 0.10 M H2SO4 are added. ii. On the basis of the equations you wrote in part a, explain why the conductivity decreases. Ba2+(aq) and / or OH-­‐(aq) 1 point is earned for either ion. As the titration approaches the equivalence point, Ba2+ (aq) ions are removed from the solution by the precipitation reaction, and OH-­‐ (aq) ions are removed from the solution by the neutralization reaction. 1 point is earned for each correct explanation. Note: Response must refer to both reactions for full credit. c. Using the information in the graph, calculate the molarity of the original Ba(OH)2 solution. moles Ba(OH)2 = moles H2SO4 (at equivalence point) moles H2SO4 = (0.10 moles / L) x 0.030 L = 0.0030 moles [BaSO4] = 0.0030 moles / volume of original solution = 0.0030 / 0.025 = 0.12 M One point is earned for the correct determination of the number of moles of titrant added at the equivalence point (can be implicit). One point is earned for the correct calculation of the original concentration of Ba(OH)2(aq). d. Calculate the concentration of Ba2+(aq) in the solution at the equivalence point (after exactly 30.0 mL of 0.10 M H2SO4 are added). Ksp = [Ba2+] [SO42-­‐] = 1.0 x 10-­‐10 with [Ba2+] [SO42-­‐] [Ba2+] = (1.0 x 10-­‐10)1/2 = 1.0 x 10-­‐5 26 1 point is earned for the correct calculation based on Ksp e. The concentration of Ba2+(aq) in the solution decreases as the volume of added 0.10 M H2SO4 increases from 30.0 mL to 31.0 mL. Explain. Because of the common ion effect, adding sulfate ions to the equilibrium reaction involving sulfate ions will cause the reaction to consume the added ions as a new equilibrium is established. Consequently, more BaSO4(aq) is formed, causing the [Ba2+(aq)] concentration to decrease. 1 point is earned for a correct explanation, which must use an equilibrium argument (for example, citing the common ion effect or LeChatelier’s principle) rather than a stoichiometric argument. 2013 PR Q 6 b. The student passes a direct current through the solution and observes that chlorine gas is produced at the anode. Identify the chemical species produced at the cathode and justify your answer using the information given in the table below. Half reaction Li+(aq) + e-­‐ à Li(s) 2 H2O(l) + 2 e-­‐ à H2(g) + 2 OH-­‐(aq) b. H2(g) and OH-­‐(aq). The hydrogen atoms in H2O are reduced to H2 at the cathode because this reaction has a higher (more favorable or less negative) standard reduction potential than the reduction of Li+ (aq) ions to Li(s) Standard reduction potential at 25 oC (V) -­‐3.05 -­‐0.83 1 point is earned for correctly identifying either of the chemical species produced at the cathode with the correct justification. Note: The justification must clearly indicate that “higher” means “less negative”. A “lower magnitude” negative value also earns the point. 27 2010 B Q2. 5 Fe2+(aq) + MnO4 -­‐ (aq) + 8 H+(aq) à 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) A galvanic cell and the balanced equation for the spontaneous cell reaction are shown above. The two reduction half reactions for the overall reaction that occurs in the cell are shown in the table below. Half reaction Eo (V) at 298 K Fe3+(aq) + e-­‐ à Fe2+(aq) + 0.77 V MnO4-­‐(aq) + 8 H+(aq) + 5 e-­‐ à Mn2+(aq) + 4 H2O(l) a. On the diagram, clearly label the cathode. b. The electrode in the beaker on the right should be labeled. One point is earned for correct identification of the cathode. Calculate the value of the standard potential, Eo, for the spontaneous cell reaction. + 1.49 Ecell = 1.49 − 0.77 = 0.72 V One point is earned for the correct numerical answer. c. How many moles of electrons are transferred when 1.0 mol of MnO4-­‐(aq) is consumed in the overall cell reaction? 5.0 moles of electrons are transferred. One point is earned for the correct numerical answer. d. Calculate the value of the equilibrium constant, Keq, for the cell reaction at 25 oC. Explain what the magnitude of Keq tells you about the extent of the reaction. log Keq = nE / 0.0592 = 5 x 0.72 / 0.0592 = 61 Keq = 6.5 × 1060 28 Because the magnitude of Keq is very large, the extent of the cell reaction is also very large and the reaction goes essentially to completion. One point is earned for the correct substitution. One point is earned for the correct numerical answer. One point is earned for an explanation. Three solutions, one containing Fe2+(aq), one containing MnO4-­‐(aq) and one containing H+(aq), are mixed in a beaker and allowed to react. The initial concentrations of the species in the mixture are 0.60 M Fe2+(aq), 0.10 M MnO4-­‐(aq), and 1.0 M H+(aq). e. f. When the reaction mixture has come to equilibrium, which species has the higher concentration, Mn2+(aq) or MnO4-­‐(aq)? Explain. [Mn2+(aq)] will be greater than [MnO4−(aq)] because: (1) as indicated in part (d), the reaction essentially goes to completion, and (2) there is more than sufficient Fe2+ and H+ to react completely with the MnO4−. [MnO4−(aq)] at equilibrium is essentially zero. One point is earned for the choice of Mn2+ with the explanation including only item (1). One point is earned for including item (2) in the explanation. When the reaction mixture has come to equilibrium, what are the molar concentrations of Fe2+(aq) and Fe3+(aq)? At equilibrium, [Fe2+(aq)] = [Fe2+(aq)]initial − 5[MnO4−(aq)]reacted = 0.60 − 5(0.10) = 0.10 M [Fe3+(aq)] = 5 × [MnO4−(aq)]reacted = 5 (0.10) = 0.50 M One point is earned for a correct setup (including a correct setup for an equilibrium calculation). One point is earned for correct numerical answers. 2012 Q2 2. A sample of a pure, gaseous hydrocarbon is introduced into a previously evacuated rigid 1.00 L vessel. The pressure of the gas is 0.200 atm at a temperature of 127 oC. a. Calculate the number of moles of the hydrocarbon in the vessel. n = P V / R T = (0.200 atm) (1.00 L) / (0.0821 L atm mol-­‐1 K-­‐1) (400 K) = 6.09 x 10-­‐3 mol 1 point is earned for the setup. 1 point is earned for the numerical answer. b. O2(g) is introduced into the same vessel containing the hydrocarbon. After the addition of the O2(g), the total pressure of the gas mixture in the vessel is 1.40 atm at 127 oC. Calculate the partial pressure of O2(g) in the vessel. 29 P(O2) = 1.40 atm – 0.200 atm = 1.20 atm 1 point is earned for the correct pressure. The mixture of the hydrocarbon and oxygen is sparked so that a complete combustion reaction occurs, producing CO2(g) and H2O(g). The partial pressures of these gases at 127 oC are 0.600 atm for CO2(g) and 0.800 atm for H2O(g). There is O2(g) remaining in the container after the reaction is complete. c. Use the partial pressures of CO2(g) and H2O(g) to calculate the partial pressure of the O2(g) consumed in the combustion. . . . CxHy + . . . O2 à . . . CO2 + . . . H2O 0.200 atm 0 atm 1.20 atm ? atm -­‐ 0.600 atm -­‐ 0.800 atm before rxn: after rxn: 0.600 atm CO2 (1 atm O2 / 1 atm CO2) = 0.600 atm O2 0.800 atm H2O (1 atm O2 / 2 atm H2O) = 0.400 atm O2
Total O2 consumed = 1.000 atm OR, based on PV = nRT and mole calculations: n(H2O) = P V / R T = (0.800 atm) (1.00 L) / (0.0821 L atm mol-­‐1 K-­‐1) (400 K) = 0.0244 mol H2O 0.244 mol H2O (1 mol O2 / 2 mol H2O) = = 0.0122 mol O2 n(CO2) = P V / R T = (0.600 atm) (1.00 L) / (0.021 L atm mol-­‐1 K-­‐1) (400 K) = 0.0183 mol CO2 (1 mol O2 / 1 mol CO2) = = 0.0183 mol O2 Total moles O2 = = 0.0305; P = n R T / V = (0.0305 mol) (0.0821 L atm mol-­‐1 K-­‐1) (400 K) / 1.00 L = 1.00 atm O2 1 point is earned for the correct stoichiometry in O2 consumption. 1 point is earned for the calculated result. d. On the basis of your answers above, write the balanced chemical equation for the combustion reaction and determine the formula of the hydrocarbon. The partial pressures occur in the same proportions as the number of moles. Phydrocarbon : P(O2) : P(CO2) : P(H2O) = 0.200 atm : 1.00 atm : 0.600 atm : 0.800 atm = 1 : 5 : 3 : 4 C3H8 + 5 O2 à 3 CO2 + 4 H2O 30 n(H2O) = P V / R T = (0.800 atm) (1.00 L) / (0.0821 L atm mol-­‐1 K-­‐1) (400 K) = 0.0244 mol H2O (2 mol H / 1 mol H2O) = 0.0487 mol H n(CO2) = P V / R T = (0.600 atm) (1.00 L) / (0.0821 L atm mol-­‐1 K-­‐1) (400 K) = 0.0183 mol CO2 (1 mol C / 1 mol CO2) = 0.0183 mol C (0.0487 mol H / 0.0183 mol C) = (2.66 mol H / 1 mol C) (3 / 3) = 8 mol H / 3 mol C = C3H8 C3H8 + 5 O2 à 3 CO2 + 4 H2O e. Calculate the mass of the hydrocarbon that was combusted. mass = (number of moles)(molar mass) = (6.09 x 10-­‐3 mol)(44.1 g/mol) = 0.269 g 1 point is earned for using the number of moles combusted from part (a). 1 point is earned for the calculated mass. f. As the vessel cools to room temperature, droplets of liquid water form on the inside walls of the container. Predict whether the pH of the water in the vessel is less than 7, equal to 7, or greater than 7. Explain your prediction. The pH will be less than 7 because CO2 is soluble in water, with which it reacts to form H+ ions. 1 point is earned for the correct choice and explanation. Multiple Choice Questions 1. Contains an element in a +1 oxidation state a. b. c. d. e. CO2 PbO2 CaO N2O5 Cu2O Questions 8-­‐9 refer to the chemical reactions represented below. 8. a. 2 C6H6(l) + 15 O2(g) à 12 CO2(g) + 6 H2O(l ) b. C2H3O2-­‐(aq) + H3O+(aq) à HC2H3O2(aq) + H2O(l ) c. 4 H+(aq) + 4 Co2+(aq) + O2(g) + 24 NH3(aq) à 4 Co(NH3)63+(aq) + 2H2O(l ) d. CaCO3(s) à CaO(s) + CO2(g) e. 2 H2O2(l ) à O2(g) + 2 H2O(l ) The reaction between a Brønsted-­‐Lowry acid and a Brønsted-­‐Lowry base – correct response is c. 31 9. The reaction in which a single species is both oxidized and reduced – correct response is e. 21. C3H8(g) + 5 O2(g) à 3 CO2(g) + 4 H2O(l) In the reaction represented above, what is the total number of moles of reactants consumed when 1.00 mole of CO2(g) is produced? a. b. c. d. e. 25. 0.33 mol 1.33 mol 1.50 mol 2.00 mol 6.00 mol H2 + F2 à 2 HF In the reaction represented above, what mass of HF is produced by the reaction of 3.0 × 1023 molecules of H2 with excess F2? (Assume the reaction goes to completion). 41. a. b. c. d. e. 1.0 g 4.0 g 10. g 20. g 40. g … LiHCO3(aq) + … H2SO4(aq) à … Li2SO4(aq) + H2O(l) + … CO2(g) When the equation above is balanced and the coefficients are reduced to lowest whole number terms, what is the coefficient of H2O(l)? a. b. c. d. e. 1 2 3 4 5 44. When a 3.22 g sample of an unknown hydrate of sodium sulfate, Na2SO4⋅xH2O(s), is heated, H2O (molar mass 18 g) is driven off. The mass of the anhydrous Na2SO4(s) (molar mass 142 g) that remains is 1.42 g. The value of x in the hydrate is a. b. c. d. e. 0.013 1.8 6.0 10. 20. 32 53. 63. What is the empirical formula of an oxide of chromium that is 48 percent oxygen by mass? a. b. c. d. e. CrO CrO2 CrO3 Cr2O Cr2O3 2 MnO4-­‐(aq) + 5 C2O42-­‐(aq) + 16 H+(aq) à 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l) Permanganate and oxalate ions react in an acidified solution according to the balanced equation above. How many moles of CO2(g) are produced when 20. mL of acidified 0.20 M KMnO4 solution is added to 50. mL of 0.10 M Na2C2O4 solution? 69. 74. a. b. c. d. e. 0.0040 mol 0.0050 mol 0.0090 mol 0.010 mol 0.020 mol Which of the following is NOT an accepted name for the formula given? a. b. c. d. e. CH3OH .. methanol CuO .. copper (I) oxide FeCl3 .. iron (III) chloride H2SO4 .. sulfuric acid SrCO3 .. strontium carbonate A student prepares a solution by dissolving 60.00 g of glucose (molar mass 180.2 g mol-­‐1) in enough distilled water to make 250.0 mL of solution. The molarity of the solution should be reported as a. b. c. d. e. 12.01 M 12.0 M 1.332 M 1.33 M 1.3 M 33 Big Idea 4: Rates of chemical reactions are determined by details of the molecular collisions Concepts List – KINETICS 1. 2. 3. 4. 5. 6. 7. 8. Rate definition Factors affecting rate a. b. c. d. e. [C] ΔT catalysis surface area nature of reactants – distinguish between homo-­‐ and heterogenous i. solids ii. Liquids iii. gases iv. Ions (solutions) Collision theory – orientation and energy Mechanism – relationship between ΔT, ΔS, ΔH – catalysis Orders Rate Law – differential versus integrated a. determined by i. experimental comparison (20% or less ) ii. graphing ( 80% or more ) b. zero, first, second – determining % remaining and/or % reacted ex. Ln (x2/x1) = kt Rate constants with units (units change with reaction order) a. unsuccessful versus effective collisions b. orientation and energy Vocabulary Reactants vs. products vs. catalyst vs. reaction intermediate Mechanisms are consistent if: -­‐ steps add up to balanced equation -­‐ slow step of mechanism will define the mechanistic rate law and rate law expression -­‐ no reaction intermediates in final rate law expression for comparison with the experimental rate law expression 34 Free Response Questions 2009B Q2 2. S2O32-­‐(aq) à SO32-­‐(aq) + S(s) in acidic solution A student performed an experiment to investigate the decomposition of sodium thiosulfate, Na2S2O3, in acidic solution, as represented by the equation above. In each trial the student mixed a different concentration of sodium thiosulfate with hydrochloric acid at constant temperature and determined the rate of disappearance of S2O32-­‐(aq). Data from five trials are given below in the table on the left and are plotted in the graph on the right. a. Identify the independent variable in the experiment. The initial concentration of S2O32−(aq) One point is earned for the correct answer. b. Determine the order of the reaction with respect to S2O32-­‐. Justify your answer by using the information above. Using trials 1 and 2 Rate2 / Rate1 = k2 [S2O32-­‐(aq)]2m / k2 [S2O32-­‐(aq)]1m 0.030 M s-­‐1 / 0.020 M s-­‐1 = [0.075]m / [0.050]m 1.5 = (1.5)m, so m = 1 and the reaction is first order with respect to S2O32-­‐. Note: Other correct justifications are acceptable. One point is earned for the correct order. One point is earned for a correct justification. 35 c. Determine the value of the rate constant, k, for the reaction. Include units in your answer. Show how you arrived at your answer. Rate = k [S2O32-­‐] => k = rate / [S2O32-­‐] Using the data from trial 1, k = 0.020 M s-­‐1 / 0.050 M = 0.40 s-­‐1 OR the rate constant is equal to the slope for the line k = (0.052 – 0.020) M s-­‐1 / (0.13 – 0.05) M = 0.32 M s-­‐1 / 0.08 M = 0.4 s-­‐1 One point is earned for the correct value. One point is earned for the correct units. d. In another trial the student mixed 0.10 M Na2S2O3 with hydrochloric acid. Calculate the amount of time it would take for the concentration of S2O32-­‐ to drop to 0.020 M. ln[A]t − ln[A]0 = − kt ln ([A]t / [A]0) = -­‐kt ln ([S2O32-­‐]t / [S2O32-­‐]0) = -­‐kt ln (0.020 / 0.10) = (-­‐0.40 s-­‐1) t => t = -­‐1.61 / -­‐.40 s-­‐1 = 4.0 s e. One point is earned for the correct setup. One point is earned for the correct answer with units. On the graph above, sketch the line that shows the results that would be expected if the student repeated the five trials at a temperature lower than that during the first set of trials. The line drawn should start on the y-­‐axis at a lower point than the line already plotted and should have a less steep slope. One point is earned for an acceptable line. 2013 PR Q4 4. 2 NO2(g) + F2(g) à 2NO2F(g) It is proposed that the reaction represented above proceeds via the mechanism represented by the two elementary steps shown below. Step I: NO2 + F2 à NO2F + F slow Step II: NO2 + F  NO2F fast reversible a. Step I of the proposed mechanism involves the collision between NO2 and F2 molecules. This step is slow even though such collisions occur very frequently in a mixture of NO2(g) and F2(g). Consider a specific collision between a molecule of NO2 and a molecule of F2. i. One factor that affects whether the collision will result in a reaction is the magnitude of the collision energy. Explain 36 Successful molecular collisions must have sufficient energy in order to result in reaction. Only collisions with sufficient energy to overcome the activation energy barrier, Ea, will be able to reach the transition state and begin to break the F-­‐F bond. 1 point is earned for a correct explanation that makes reference to the activation energy of the reaction. ii. Identify and explain one other factor that affects whether the collision will result in a reaction. For a collision to be successful, the molecules must have the correct orientation. Only collisions with the correct orientation will be able to begin to form an N-­‐F bond and begin to break an F-­‐F bond as the transition state is approached (that is, the molecules must contact each other at very specific locations on their surfaces for the transition state to be accessible.) 1 point is earned for identifying the relative orientation of the colliding molecules. 1 point is earned for an explanation that makes reference to specific parts (atoms or bonds) of the reacting molecules. b. Consider the following potential rate laws for the reaction. Circle the rate law below that is consistent with the mechanism proposed above. Explain the reasoning behind your choice in terms of the details of the elementary steps of the mechanism. Rate = k [NO2]2[F2] rate = k [NO2] [F2] The rate law that is consistent with the mechanism is the one on the right above (rate = k [NO2] [F2]). Step 1 is the slower step and the rate determining step in the mechanism. Since step 1 is an elementary reaction, its rate law is given by the stoichiometry of the reacting molecules, ratestep1 = k1 [NO2] [F2]. 1 point is earned for identifying the correct rate law with a proper explanation. The explanation must correlate the overall rate law with the rate law derived from the stoichiometry of the slow step of the reaction. Note: a statement relating the coefficients of the reactants in Step 1 to the exponents in the rate law indicates a correct understanding. 2012B #5 5. 2 H2O2(aq) à 2 H2O(l) + O2(g) 37 The decomposition of hydrogen peroxide to form water and oxygen gas is represented by the equation above. A proposed mechanism for the reaction, which involves the free radicals, HO• and HOO•, is represented by the three equations below. H2O2 à 2OH• H2O2 + OH• à H2O + HOO• fast HOO• + OH• à H2O + O2 fast a. Write the rate law consistent with the proposed mechanism above. Rate = k [H2O2] 1 point is earned for the correct rate law. b. The rate of the decomposition reaction was studied in an experiment, and the resulting data were plotted in the graph below. slow Using the graph, determine the time, in hours, needed for the concentration of H2O2 to change from i. 1.50 M to 0.75 M 38 Time = 0.71 – 0.21 = 0.50 ± 0.05 hours One point is earned for the correct time. ii. 0.80 M to 0.40 M Time = 1.17 – 0.66 = 0.51 ± 0.05 hours One point is earned for the correct time. c. The experimental data are consistent with the proposed mechanism. Explain The data show that the time remains constant when the concentration of the reactant decreases by one half, indicating that the reaction is first order. The rate law for the proposed mechanism has [H2O2] to the first power, indicating a first order reaction. 1 point is earned for a correct explanation. An electrochemical cell based on the decomposition of H2O2 can be constructed based on the half-­‐reactions in the table below. Standard Reduction Potential, Eo 0.88 V O2 + 2 H2O + 2 e-­‐ à H2O2 + 2 OH-­‐ -­‐ 0.15 V d. Calculate the value of the standard cell potential, Eo, for the cell. Combining the two reactions, with the second reaction reversed gives 2 H2O2 à 2 H2O + O2 1 point is earned for the correct value of Eo. e. Indicate whether ΔGo for the decomposition reaction is greater than 0, less than 0, or equal to 0. Justify your answer. ΔGo is less than 0. ΔGo = -­‐ n F Eo. Because Eo is positive, ΔGo must be negative. OR Eo is positive, thus the reaction is spontaneous. For spontaneous reactions, ΔGo is always negative. 1 point is earned for the correct answer with justification. f. The decomposition of H2O2(aq) is slow at 298 K, but a suitable catalyst greatly increases the rate of the decomposition reaction. Half-­‐reaction H2O2 + 2 e-­‐ à 2 OH-­‐ 39 Eo = 1.03 V i. Draw a circle around each of the quantities below that has a different value for the catalyzed reaction than for the uncatalyzed reaction. Keq ΔGo ΔHo Ea Only Ea should be circled. 1 point is earned for circling Ea and none of the other quantities. ii. For any quantity that you circled above, indicate whether its value is greater or less for the catalyzed reaction than for the uncatalyzed reaction. Explain why. Ea,the activation energy, is less for the catalyzed reaction than for the uncatalyzed reaction. A catalyst provided a new reaction mechanism that requires a lower minimum required potential energy for the reaction to occur. Thus the activation energy is lowered. Note: reference to other circled quantities is ignored. 1 point is earned for a correct answer with reference to an altered mechanism. 2010 B Q6 (2nd half) The gas phase decomposition of nitrous oxide has the following two step mechanism. Step 1 N2O à N2 + O Step 2 O + N2O à N2 + O2 f. Write the balanced equation for the overall reaction 2 N2O → 2 N2 + O2 One point is earned for the correct balanced equation. g. Is the oxygen atom, O, a catalyst for the reaction or is it an intermediate? Explain. The O atom is an intermediate because it is formed and then consumed during the course of the reaction. (Had it been a catalyst, it would have been present both at the beginning and the end of the reaction.) h. One point is earned for the correct choice with explanation. Identify the slower step in the mechanism if the rate law for the reaction was determined to be rate = k [N2O]. Justify your answer Step 1 is slower because N2O appears in Step 1 as the single reactant, which is consistent with the given rate law. One point is earned for the correct choice with justification. 40 2011 A Q6 (2 half) nd
In a second experiment, which is performed at a much higher temperature, a sample of ethanol gas and a copper catalyst are placed in a rigid, empty 1.0 L flask. The temperature of the flask is held constant, and the initial concentration of the ethanol gas is 0.0100 M. The ethanol begins to decompose according to the chemical reaction represented below. CH3 CH2 OH 𝑔
Cu
CH3 CHO 𝑔 + H2 (𝑔) The concentration of ethanol gas over time is used to create the three graphs below. c. Given that the reaction order is zero, one, or two, use the information in the graphs to respond to the following. i. Determine the order of the reaction with respect to ethanol. Justify your answer. The order of the reaction is zero. The plot on the left is a straight line, indicating that the rate of decrease in [ethanol] is constant as [ethanol] changes. Therefore the rate of reaction does not depend on [ethanol]. 1 point is earned for the correct choice with a valid justification. ii. Write the rate law for the reaction. rate = k iii. Determine the rate constant for the reaction, including units. 1 point is earned for the correct rate law. Rate = k = -­‐ Δ [ethanol] / Δt = -­‐ (0.0020 – 0.0100) mol L-­‐1 / 2000 s = 4.0 x 10-­‐6 M s-­‐1 d. 1 point is earned for the correct setup. 1 point is earned for the correct units. The pressure in the flask at the beginning of the experiment is 0.40 atm. If the ethanol completely decomposes, what is the final pressure in the flask? 41 The final pressure is 0.80 atm (twice the original pressure because the products represent twice as many moles of gas as the reactant). 1 point is earned for the correct final pressure. 2012 Q3 (2nd half) An experiment is carried out to measure the rate of the reaction, which is first order. A 4.70 x 10-­‐3 mol sample of CH3CH2NH2 is placed in a previously evacuated 2.00 L container at 773 K. After 20.0 minutes, the concentration of the CH3CH2NH2 is found to be 3.60 x 10-­‐4 mol/L. d. Calculate the rate constant for the reaction at 773 K. Include units with your answer. ln [A]t – ln [A]0 = -­‐ k t ln (3.60 x 10-­‐4 mol/L) – ln (4.70 x 10-­‐3 / 2) = -­‐ k (20.0 min) -­‐7.929 – (-­‐6.053) = -­‐ k (20.0 min) => k = 9.38 x 10-­‐2 min-­‐1 e. 1 point is earned for the initial concentration of CH3CH2NH2 . 1 point is earned for the correct setup of the first order integrated rate law equation. 1 point is earned for the calculated result with units. Calculate the initial rate, in M min-­‐1, of the reaction at 773 K. initial rate = k [CH3CH2NH2] = (9.38 x 10-­‐2 min-­‐1) (4.70 x 10-­‐3 mol / 2 L) = 2.20 x 10-­‐4 M min-­‐
f. 1 point is earned for the calculated result. !
If is plotted versus time for this reaction, would the plot result in a straight 1
[!H! !!! !!! ]
line or would it result in a curve? Explain your reasoning. The plot would produce a curve; had the reaction been second order the plot would have been a straight line. A plot of ln[CH3CH2NH2] vs. t would have yielded a straight line. 1 point is earned for the correct choice with explanation. Multiple Choice Questions C2 H4 g + H2
32. platinium
C2H4(g) is reduced by H2(g) in the presence of a solid platinum catalyst, as represented by the equation above. Factors that could affect the rate of the reaction include which of the following? I. II. Changes in the partial pressure of H2(g) Changes in the particle size of the platinum catalyst 42 C2 H6 (g) III. Changes in the temperature of the reaction system a. III only b. I and II only c. I and III only d. e. I, II, and III 43 II and III only Experiment 1 2 3 51. 61. 62. Initial [X] (mol L-­‐1) Initial [Y] (mol L-­‐1) Initial Rate of Formation of Z (mol L-­‐1 s-­‐1) 0.10 0.20 0.20 0.30 0.60 0.30 4.0 ×10-­‐4 1.6 ×10-­‐3 4.0 ×10-­‐4 The data in the table above were obtained for the reaction X + Y à Z. Which of the following is the rate law for the reaction? a. Rate = k[X]2 b. Rate = k[Y]2 c. Rate = k[X][Y] d. Rate = k[X]2[Y] e. Rate = k[X][Y]2 If the oxygen isotope 20O has a half-­‐life of 15 seconds, what fraction of a sample of pure 20O remains after 1.0 minute? a. 1/2 b. 1/4 c. 7/30 d. 1/8 e. 1/16 X à products Pure substance X decomposes according to the equation above. Which of the following graphs indicates that the rate of decomposition is second order in X ? D is correct 44 67. The role of a catalyst in a chemical reaction is to a. b. c. d. e. decrease the amount of reactants that must be used lower the activation energy for the reaction supply the activation energy required for the reaction to proceed increase the amounts of products formed at equilibrium increase the entropy change for the reaction 45 Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the direction of changes in matter Concepts – THERMODYNAMICS 1. ∆H0 rxn = ∑ ∆ Hf 0Products -­‐ ∑∆ Hf0 Reactants = ∑ Bond Energy Reactants -­‐ ∑ Bond energy Products ∆Hrxn -­‐ exothermic ∆Hrxn + endothermic 2. ∆S0 rxn = ∑ Sf0 Products -­‐ ∑ Sf0 Reactants ∆S0rxn -­‐ ordered ∆S0rxn + disordered 3. ∆G0 rxn = ∆H0rxn -­‐ T ∆S 0rxn ∆G0rxn -­‐ thermodynamically favored ∆G0rxn + thermodynamically unfavored 4. ∆G0rxn = -­‐ RT ln Q Q = Keq free energy and equilibrium 5. ∆G0 rxn = -­‐ nF E0 free energy and electrochemistry F = 96,500 coulombs / mole electrons Faraday’s constant 6. Phase diagrams 7. ∆H rxn = q = m ( c ) ( ∆T ) Free Response Questions 2013 PR Q2 2. A student is given the task of determining the enthalpy of reaction for the reaction between HCl(aq) and NaOH(aq). The following materials are available. 1.00 M HCl(aq) 1.00 M NaOH(aq) distilled water 2.00 M HCl(aq) 2.00 M NaOH(aq) goggles gloves lab coat Insulated cups with covers 46 Thermometer (± 0.1 oC) stirring rod The student may select from the glassware listed in the table below. Glassware items Precision 250 mL Erlenmeyer flasks ± 25 mL 100 mL beakers ± 10 mL ± 0.1 mL 100 mL graduated cylinders a. b. The student selects two 100 mL beakers, uses them to measure 50 mL each of 1.00 M HCl(aq) solution and 1.00 M NaOH(aq) solution, and measures an initial temperature of 24.5 oC for each solution. Then the student pours the two solutions into an insulated cup, stirs the mixture, covers the cup, and records a maximum temperature of 29.9 oC. i. Is the experimental design sufficient to determine the enthalpy of reaction to a precision of two significant figures? Justify your answer. No. The use of the beakers to measure 50 ± 10 mL of solutions limits the precision of the volume measurements and of the calculations to ±20% or 1 significant figure. 1 point is earned for the correct answer with the correct explanation. ii. List two specific changes to the experiment that will allow the student to determine the enthalpy of reaction to a precision of three significant figures. Explain. Use graduated cylinders to measure the volumes of acid and base allowing a volume precision of ±0.1 mL or 3 significant figures for a volume of 50.0 mL AND Use the 2.00 M HCl and 2.00 M NaOH solutions (instead of 1.00 M) to get a larger ΔT, thereby improving the relative precision is ΔT to ±1% 1 point is earned for the change of glassware to graduated cylinders with a proper explanation. 1 point is earned for using the 2.00 M solutions for improved relative precision in temperature. Note: doubling the volumes will not increase ΔT or significantly improve volume precision. A second student is given two solutions, 75.0 mL of 1.00 M HCl and 75.0 mL of 1.00 M NaOH, each at 25.0 oC. The student pours the solutions into an insulated cup, stirs the mixture, covers the cup, and records the maximum temperature of the mixture. i. The student calculates the amount of heat evolved in the experiment to be 4.1 kJ. Calculate the student’s experimental value for the enthalpy of reaction, in kJ/molrxn. 47 c. 75 mL (1.00 mol HCl (or NaOH)) / 1000 mL = 0.0750 mol HCl or NaOH ΔH = -­‐4.1 kJ / 0.0750 mol of reactants = -­‐55 kJ / molrxn 1 point is earned for the correct calculation of moles of reactants. 1 point is earned for the correct substitution and answer. ii. The student assumes that the thermometer and the calorimeter do not absorb energy during the reaction. Does this assumption result in a calculated value of the enthalpy of reaction that is higher than, lower than, or the same as it would have been had the heat capacities of the thermometer and calorimeter been taken into account? Justify your answer. The calculated value of the enthalpy of reaction will be lower (smaller or less negative) than it would have been had the thermometer and calorimeter been taken into account. The thermometer and calorimeter will absorb some of the heat of reaction. This lost heat is ignored in the original calculation of ΔHrxn making it smaller in magnitude (less negative). OR The actual heat capacity of the system is the sum of the heat capacities of the water, thermometer, and calorimeter. The assumed heat capacity of the system (water only) is less than the actual value, resulting in a lower (less negative) calculated value of ΔHrxn. 1 point is earned for the correct prediction. 1 point is earned for an acceptable justification. iii. One assumption in interpreting the results of the experiment is that the reaction between HCl(aq) and NaOH(aq) goes to completion. Justify the validity of this assumption in terms of the equilibrium constant for the reaction. H+ + OH-­‐ à H2O, the reaction between HCl(aq) and NaOH(aq) is the reverse of H2O à H+ + OH-­‐; the autoionization of water (for which K = Kw = 1 x 10-­‐14). Thus the value of K for the neutralization reaction is the reciprocal of Kw, of 1.0 x 1014, a very large number. Thus the neutralization reaction goes virtually to completion. 1 point is earned for the correct justification. A third student calculates a value for the enthalpy of reaction that is significantly higher than the accepted value. i. Identify a specific error in procedure made by the student that will result in a calculated value for the enthalpy of reaction that is higher than the accepted 48 value. (Vague statements like “human error” or “incorrect calculations” will not earn credit). The student read the thermometer incorrectly in such a way to result in a calculated value of ΔT that was too high (either read Ti too low or read Tf too high). OR The student mistakenly used 2.00 M acid and 2.00 M base, thinking they were both 1.00 M. 1 point is earned for an acceptable procedural error that results in a higher calculated value. ii. Explain how the error that you identified in part c i) leads to a calculated value for the enthalpy of reaction that is higher than the accepted value. The calculation of the molar enthalpy of reaction may be expressed as Molar ΔHrxn = -­‐ (masssoln x c x ΔT) / nrxn If there is a measurement error that results in a ΔT that is too high, the magnitude (i.e. the absolute value) of the calculated molar enthalpy of reactions will be too high. 1 point is earned for an explanation that is consistent with the stated procedural error. 2012 B Q 3 3. CaSO4•2H2O(s)  CaSO4(s) + 2 H2O(g) The hydrate CaSO4•2H2O(s) can be heated to form the anhydrous salt, CaSO4(s), as shown by the reaction represented above. a. Using the data in the table below, calculate the value of ΔGo, in kJ / molrxn, for the reaction at 298 K. Substance ΔGfo at 298 K (kJ / mol) CaSO4•2H2O(s) -­‐1795.70 CaSO4(s) -­‐1320.30 H2O(g) -­‐228.59 ΔGo = Σ ΔGfo products – Σ ΔGfo reactants 49 ` are optional. = -­‐1320.30 + (2 x -­‐228.59) – (-­‐1795.70 kJ mol-­‐1) = 18.22 kJ mol-­‐1 1 point is earned for the mole factor for water. 1 point is earned for the answer (units b. Given that the value of ΔHo for the reaction at 298 K is +105 kJ / molrxn, calculate the value of ΔSo for the reaction at 298 K. Include units with your answer. ΔGo = ΔHo – T ΔSo ΔSo = (ΔHo – ΔGo) / T = (105 – 18.22) kJ mol-­‐1 / 298 K = 0.29 kJ mol-­‐1 K-­‐1 or 290 J mol-­‐1 K-­‐1 1 point is earned for the correct substitution of ΔGo, ΔHo, and T. 1 point is earned for the correct answer with correct units. 2009B Q5 5. Answer the following questions about nitrogen, hydrogen, and ammonia. a. In the boxes below, draw the complete Lewis electron-­‐dot diagrams for N2 and NH3. :NN: and nitrogen in the middle, 3 single N-­‐H bonds and a lone pair on N The correct structures are described above. Two points are earned for the correct Lewis electron dot diagrams (1 point each). b. Calculate the standard free energy change, ΔGo, that occurs when 12.0 g of H2(g) reacts with excess N2(g) at 298 K according to the reaction represented below. N2(g) + 3 H2(g)  2 NH3(g) ΔG298o = -­‐34 kJ mol-­‐1 12.0 g H2 x (1 mol H2 / 2.0 g H2) x (1 mol reaction / 3 mol H2) x (-­‐34 kJ / 1 mol reaction) = -­‐
68 kJ 1 point is earned for the correct stoichiometry. 1 point is earned for the correct answer. 50 c. Given that ΔH298o for the reaction is -­‐92.2 kJ mol-­‐1, which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain. ΔH298o = Σ(bond energy of the reactions) – Σ(bond energy of the products) Based on the equation above, for ΔH298o to be negative, the total bond energy of the products must be larger than the total bond energy of the reactants. OR more energy is released as products bonds are formed than is absorbed as reactant bonds are broken. One point is earned for the correct answer with the correct equation and explanation. d. The value of the standard entropy change, ΔS298o, for the reaction is -­‐199 J mol-­‐1 K-­‐1. Explain why the value of ΔS298o is negative. e. All of the reactants and products in the reaction are in the gas phase, so the sign of the entropy change will depend on the number of moles of particles in the reactants and products. There are more moles of reactants (four) compared with moles of products (two), so there is a greater number of microstates in the reactants than in the products. Therefore the entropy decreases as the reaction proceeds (fewer possible microstates), and the sign of the entropy change is negative. One point is earned for the correct explanation. Assume that ΔHo and ΔSo for the reaction are independent of temperature. i. Explain why there is a temperature above 298 K at which the algebraic sign of the value of ΔGo changes. ΔG° = ΔH° − TΔS° As the temperature increases |TΔS°| will at some point exceed |ΔH°|. Because both ΔH° and ΔS° are negative, the sign of ΔG° will then change from negative to positive. One point is earned for the correct explanation. ii. Theoretically, the best yields of ammonia should be achieved at low temperatures and high pressures. Explain. Low temperatures: The reaction is exothermic. By Le Chatelier’s principle, decreasing the temperature drives the reaction to the right to produce more heat energy, and thus more ammonia is produced. High pressures: For this reaction, higher pressure is achieved by decreasing the volume of the container. As pressure increases, the reaction equilibrium shifts in the direction that reduces the total number of particles (by Le Chatelier’s 51 principle). In this case, the product has fewer moles of particles than the reactants; thus product would be favored. Higher pressure therefore results in an increase in the amount of ammonia. One point is earned for explaining ncreased yield at low temperatures. One point is earned for explaining increased yield at high pressures. Multiple Choice Questions 2013 29. Which of the following processes involves the greatest increase in entropy? a. SO3(g) + H2(g) à SO2(g) + H2O(g) b. N2(g) + 3 H2(g) à 2 NH3(g) c. Ag+(aq) + Cl-­‐(aq) à AgCl(s) d. C2H2(g) + 2 H2(g) à C2H6(g) e. MgSO3(s) à MgO(s) + SO2(g) 2 NH3(g) → 3 H2(g) + N2(g) ΔHo298 = 92 kJ/molrxn 37. According to the information above, what is the standard enthalpy of formation, ΔHfo, for NH3(g) at 298 K ? a. b. c. d. e. −92 kJ/mol −46 kJ/mol 46 kJ/mol 92 kJ/mol 184 kJ/mol 52 46. In an insulated cup of negligible heat capacity, 50. g of water at 40.°C is mixed with 30. g of water at 20.°C. The final temperature of the mixture is closest to a. 22°C b. 27°C c. 30.°C d. 33°C e. 38°C 2 H2(g) + O2(g) à 2 H2O(g) 52. For the reaction represented above at 25°C, what are the signs of ΔH°, ΔS°, and ΔG° ? a. b. c. d. e. ΔHo Δ So ΔGo + + + -­‐ -­‐ + + -­‐ -­‐ -­‐ + -­‐ -­‐ -­‐ + 2012 31. Under which of the following conditions can an endothermic reaction be thermodynamically favorable? a. b. c. d. e. 53. ΔG is positive ΔS is negative T ΔS > ΔH TΔS = 0 There are no conditions under which an endothermic reaction can be thermodynamically favorable. 4 NH3(g) + 3 O2(g) à 2 N2(g) + 6 H2O(g) If the standard molar heats of formation of ammonia, NH3(g), and gaseous water, H2O(g), are -­‐
46 kJ / mol and -­‐242 kJ / mol, respectively, what is the value of ΔH298o for the reaction represented above? a. b. c. d. e. -­‐190 kJ / molrxn -­‐290 kJ / molrxn -­‐580 kJ / molrxn -­‐1270 kJ / molrxn -­‐1640 kJ / molrxn 53 54. When a magnesium wire is dipped into a solution of lead (II) nitrate, a black deposit forms on the wire. Which of the following can be concluded from this observation? a. b. c. d. e. The standard reduction potential, Eo, for Pb2+(aq) is greater than that for Mg2+(aq). Mg(s) is less easily oxidized than Pb(s) An external source of potential must have been supplied. The magnesium wire will be the cathode of a Mg / Pb cell. Pb(s) can spontaneously displace Mg2+(aq) from solution. 54 Big Idea 6: Any bond or intermolecular attraction that can be formed can be broken. These two processes are in a dynamic competition, sensitive to initial conditions and external perturbations. Concept List – EQUILIBRIUM All Problems are equilibrium problems because All problems involve stoichiometry: soluble salts, strong acids, strong bases Some problems involve equilibrium: “insoluble” salts, weak acids, weak bases For chemical reactions – Keq, Kc, and Kp are the important quantities For physical changes – Ka, Kb, Ksp, Kionize, and Kdissocation are the important quantities Important points 1. Law of mass action aA + bB + …  rR +sS + xxx Kc = [R]r [S]s … / [A]a [B]b … 2. Kc for molarity for ions and gases 3. Kp with atm, or mmHg for gases 4. Orientation of collisions 5. Shifting equilibrium – Le Chatlier’s Principle a. b. c. d. e. f. g. 6. Relationship / connection between these Kp = Kc (RT)Δn solid liquid catalyst inert gas added temperature changes (increasing T favors endothermic processes) only factors in equation constant will affect Keq eg. CaCO3(s)  CaO(s) + CO2(g) pressure / volume changes Important vocabulary Driving force Favors (reactants or products) Shifts (in LeChatelier arguments) 55 7. K > 1 products favored 8. 9. K < 1 reactants favored Excluded: solids, pure liquids, water (in aqueous solution) Typical question: Given Kc and the starting concentration of reactants, find the concentration (or pH !) of products at equilibrium. Example: Kc of acetic acid = 1.754 × 10-­‐5. Find the pH of a 0.100 M solution of acetic acid. 10. 11. 12. 13. Equilibrium constant for a reverse reaction = 1 / K of the value of the forward reaction. When using Hess’s Law: Koverall = K1 × K2 If out of equilibrium: Calculate the reaction quotient (Q) in a similar fashion to the way an equilibrium constant would be found. If: Q < K forward reaction occurs to reach equilibrium Q > K reverse reaction occurs to reach equilibrium Problem solving: Learn when to make an approximation (needed for multiple choice and free response questions!). 5% rules usually works when value of K is 10-­‐2 or smaller than value of known concentrations. Example: A  B + C K = 3.0 × 10-­‐6 If [A] = 5.0 M; find [C] at equilibrium If greater than 5 % use the quadratic equation: ax2 + bx + c = 0 − b ± b 2 − 4ac
x=
2a
AP Chemistry Concept List – ACID – BASE
pH = -­‐ log [H+] pOH = -­‐ log [OH-­‐] Kw = [H+] [OH-­‐] = 1×10-­‐14 at 25 oC If you know one quantity, you know the other three 56 pH
[H+]
pOH
-
[OH ]
Definitions Acid Base +
Donates H Donates OH-­‐ Donates protons Accepts protons -­‐ {anions?} Conjugate Acid – Base Pairs 1. HCl + H2O → H3O+ + Cl-­‐ 2. NH3 + H2O  NH4+ + OH-­‐ 3. HSO4-­‐ + H2O  H3O+ + SO42-­‐ 4. CO32-­‐ + H3O+  HCO3-­‐ + H2O A. Ka Weak Acid HCN  H+ + CN-­‐ B. Ka =
What is the pH of a 0.5 M HCN solution? Kb Kb =
C. [H + ] [CN − ]
= 6.2 × 10 −10 [HCN]
Weak base NH3 + H2O  NH4+ + OH-­‐ [ NH +4 ] [OH − ]
= 1.8 × 10 −5 [ NH3 ]
What is the pH of a 0.5 M NH2OH solution? Ksp Insoluble Salts MgF2(s)  Mg2+ + 2F-­‐ 57 Theory Arrhenius Bronsted – Lowry D. Ksp = [Mg2+] [F-­‐]2 = 6.6 × 10-­‐9 pH = pKa + log
What is the solubility of MgF2 in molarity? Buffers – a weak acid/base and its soluble salt (conjugate base or acid) mixture [base]
[acid]
What is the pH of a 0.5 M HC2H3O2 in 2 M NaC2H3O2 solution? Ka = 1.8 × 10-­‐5 E. Salts of Weak Acids and Weak Bases What is the pH of a 1 M NaC2H3O2 solution? Titrations and Endpoints At endpoint: acid moles = base moles or [H+] = [OH-­‐] Strong acid – strong base endpoint pH = 7 Strong acid – weak base endpoint pH < 7 Weak acid – strong base endpoint pH > 7 The last two are important because of conjugate acid and base pairs 11. Acid strength – know the 6 strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4 (removal of the first H+ only) a) binary acids – acid strength increased with increasing size and electronegativity of the “other element”. (NOTE: Size predominates over electronegativity in determining acid strength.) Example: H2Te > H2O and HF > NH3 b) oxoacids – Acid strength increases with increasing: 1) electronegativity 2) number of bonded oxygen atoms 3) oxidation state 58 of the “central atom”. However, need to show as electron withdrawing groups rather than trends (trends need to be explained as a result of chemical principles rather than solely as a trend). Example: HClO4 [O3Cl(OH)] is very acidic NaOH is very basic Acid strength also increases with DECREASING radii of the “central atom” Example: HOCl (bond between Cl and OH is covalent – acidic) HOI (bond between I and OH is ionic – basic) 12. HA + H2O  H3O+ + A-­‐ Acid Ionization Constant (Ka): Ka =
[A − ][H 3O + ]
[HA]
Example: HF + H2O  H3O+ + F-­‐ Ka =
[F − ][H 3O + ]
[HF]
13. What is the pH of 0.5 M HCN solution for which Ka = 6.2 × 10-­‐10? Base Ionization Constant (Kb): Kb =
[BH − ][OH − ]
[B]
Example: F-­‐ + H2O  HF + OH-­‐ Kb =
[HF][OH − ]
[F − ]
B + H2O  BH+ + OH-­‐ 14. 15. 16. What is the pH of a 0.5 M NH2OH solution for which Kb = 6.6 × 10-­‐9? Do equal number of Ka and Kb problems as they are equally likely! Ka × Kb = Kw = 10-­‐14 for conjugate acid/base pairs @ 25 oC! Percent ionization = [H+]equilibrium / [HA]initial × 100 Buffers: Similar concentrations of a weak acid and its conjugate base -­‐or-­‐ 59 Similar concentrations of a weak base and its conjugate acid If these concentrations are large in comparison to SMALL amounts of added acid or base, equilibrium will be shifted slightly and the pH change resisted. Consider: HA  H+ + A-­‐ Ka =
[A − ][H 3O + ]
[HA]
[H+] = Ka [HA] / [A-­‐] pH = pKa – log [HA] / [A-­‐] or pH = pKa + log [A-­‐] / [HA] (Henderson-­‐Hasselbach equation) B + H2O  HB+ + OH-­‐ [BH + ][OH − ]
Kb =
[B]
[OH-­‐] = Kb [B] / [HB+] or pOH = pKb + log [HB+]/[B] (Henderson-­‐Hasselbach equation) What is the pH of a solution which is 0.5 M HC2H3O2 in 2 M NaC2H3O2 for which Ka = 1.8 × 10-­‐5? 17. 18. Polyprotic Acids: H3PO4, H2SO4, H2C2O4, etc. Equivalence Point – the point at which stoichiometric amounts of reactants have reacted. NOTE: This only occurs at pH = 7 for the reaction of a strong acid with a strong base. The equivalence point will occur ABOVE pH = 7 (more basic) for a weak acid / strong base titration. (the conjugate base of the weak acid will react with water.) The equivalence point will occur BELOW pH = 7 for a weak base / strong acid titration (the conjugate acid of the weak base with react with water). 19. 20. Indicators – select bases on the pH at the equivalence point. Titration curves: a) Weak acid / strong base HA + OH-­‐  A-­‐ + H2O 60 NOTE: Graph should have “pH” as the vertical axis and “added base” as the horizontal axis. The graph should be in an “S” shape. The middle of the lower part of the “S” indicates the point of maximum buffering where [HA] / [A-­‐] = 1. The middle of the “S” is the equivalence point (above pH = 7) and [HA] = 0. The top part of the “S” levels off at the pH of the base solution. b) Weak base / strong acid B + H3O+  BH+ + H2O NOTE: Graph should have “pH” as the vertical axis and “added base” as the horizontal axis. The graph should be in a “backwards S” shape. The middle of the upper part of the “backwards S” indicates the point of maximum buffering where [B] / [HB+] = 1. The middle of the “backwards S” is the equivalence point (below pH = 7) and [B] = 0. The bottom part of the “backwards S” levels off at the pH of the acid solution. c) Weak diprotic acid / strong base H2A + OH-­‐  HA-­‐ + H2O HA-­‐ + OH-­‐  A2-­‐ + H2O NOTE: Graph should have “pH” as the vertical axis and “added base” as the horizontal axis. The graph should be in a “double S” shape. The middle of the lower part of the “first S” indicates the point of maximum buffering of the first buffering zone where [H2A] / [HA-­‐] = 1. The middle of the “first S” is the first equivalence point where [H2A] = 0. The top of the “first S” (i.e. the lower part of the “second S”) indicates the point of maximum buffering of the second buffering zone where [HA-­‐] / [A2-­‐] = 1. The middle of the “second S” is the second equivalence point where [HA-­‐
] = 0. The top part of the “second S” levels off at the pH of the base solution. 21. Solubility Product (Ksp) Example 1: Co(OH)2(s)  Co2+ + 2OH-­‐ Ksp = [Co2+][OH-­‐]2 (don’t forget – molar concentration of OH-­‐ is twice the solubility) Example 2: Solubility of Ag2SO4 is 0.016 mol L-­‐1 (5.0 g L-­‐1). Find the Ksp of Ag2SO4. (Answer: Ksp = 1.5 × 10-­‐5) 22. Ion product (Qi) – equivalent to the “reaction quotient” Ksp > Qi all ions in solution; more solid will dissolve Qi = Ksp equilibrium – solution is saturated Ksp < Qi precipitation will occur until Qi = Ksp 61 Free Response Questions 2013 PR Q3 3. SO2Cl2(g)  SO2(g) + Cl2(g) A 4.32 g sample of liquid SO2Cl2 is placed in a rigid, evacuated 1.50 L reaction vessel. As the container is heated to 400. K, the sample vaporizes completely and starts to decompose according to the equation above. The decomposition reaction is endothermic. a. If no decomposition occurred, what would be the pressure, in atm, of the SO2Cl2(g) in the vessel at 400. K? Assuming no decomposition, Moles (SO2Cl2) = m / M = 4.32 g / 134.96 g mol-­‐1 = 0.0320 mol P (SO2Cl2) = n RT / V = (0.0320 mol) (0.0821 L atm mol-­‐1 K-­‐1) (400 K) / (1.50 L) = 0.701 atm 1 point is earned for the correct calculation of moles of SO2Cl2 (may be implicit). b. When the system has reached equilibrium at 400. K, the total pressure in the container is 1.26 atm. Calculate the partial pressures, in atm of SO2Cl2(g), SO2(g), and Cl2(g) in the container at 400. K. Pressures at equilibrium at 400 K: SO2Cl2(g) à SO2(g) + Cl2(g) Total 0.701 –x x x 0.701 + x Ptotal = 0.701 + x = 1.26 atm x = P(SO2) = P(Cl2) = 0.56 atm P(SO2Cl2) = 0.701 – x = 0.14 atm 1 point is earned for the correct setup. 1 point is earned for the correct calculation of pressures. c. For the decomposition reaction at 400. K, i. write the equilibrium constant expression for Kp for the reaction, and Kp = P(SO2) P(Cl2) / P(SO2Cl2) 1 point is earned for the correct Kp expression. Note: the pressure subscripts must be specific (i.e. SO2, Cl2, and SO2Cl2 – NOT A, B, C, D, …) 62 ii. calculate the value of the equilibrium constant, Kp. KP = (0.56) (0.56) / 0.14 = 2.2 1 point is earned for the correct calculation of KP that is consistent with the KP expression stated in part c ii and with the partial pressures calculated in part b. d. The temperature of the equilibrium mixture is increased to 425 K. Will the value of Kp increase, decrease, or remain the same? Justify your prediction. At a higher temperature, KP will increase. According to LeChatelier’s principle, raising the temperature of an endothermic reaction at equilibrium adds a thermal stress that increases the value of KP and produces more products. 1 point is earned for the correct prediction. 1 point is earned for a proper justification in terms of LeChatelier’s principle. e. In another experiment, the original partial pressures of SO2Cl2(g), SO2(g), and Cl2(g) are 1.0 atm each at 400. K. Predict whether the amount of SO2Cl2(g) in the container will increase, decrease, or remain the same. Justify your prediction. The amount of SO2Cl2 in the container will decrease. Initially QP = 1.0 < 2.2 = KP, thus the reaction will consume SO2Cl2 as it proceeds in the forward direction to reestablish equilibrium. 1 point is earned for the correct prediction. 1 point is earned for an acceptable justification. Note: the justification must consider the relative values of QP and KP. 2010B #1 1. The compound butane, C4H10, occurs in two isomeric forms, n-­‐butane and isobutane (2-­‐methyl propane). Both compounds exist as gases at 25 oC and 1.0 atm. a. b. Draw the structural formula of each of the isomers (include all atoms). Clearly label each structure. Two points are earned for two correct structures with correct labels. (Note: 1 point can be earned for either two correct structures that are mislabeled or one correct structure with or without correct label.) OR 1 point can be earned for two skeletal structures (hydrogen atoms not shown) with proper labels. On the basis of molecular structure, identify the isomer than has the higher boiling point. Justify your answer. 63 The isomer n-­‐butane has the higher boiling point. London (dispersion) forces are greater among molecules of n-­‐butane than they are among molecules of isobutane because molecules of n-­‐butane, with its linear structure, can approach one another more closely and can form a greater number of induced temporary dipoles than molecules of isobutane, with its more compact structure, can form. One point is earned for the correct choice of isomer with justification. The two isomers exist in equilibrium as represented by the equation below. n-­‐butane(g) isobutane(g) Kc = 2.5 at 25 oC Suppose that a 0.010 mol sample of pure n-­‐butane is placed in an evacuated 1.0 L rigid container at 25 oC. c. Write the expression for the equilibrium constant, Kc, for the reaction. Kc = [isobutene] / [n-­‐butane] One point is earned for the correct equation. d. Calculate the initial pressure in the container when the n-­‐butane is first introduced (before the reaction starts). P = n R T / V = (0.010 mol) (0.0821 L atm mol-­‐1 K-­‐1) (298 K) / 1.0 L = 0.24 atm 1 point is earned for the correct substitution and numerical answer. e. The n-­‐butane reacts until equilibrium has been established at 25 oC. i. Calculate the total pressure in the container at equilibrium. Justify your answer. The total pressure in the container remains the same, 0.24 atm. As the reaction proceeds, the number of molecules in the container remains constant; one molecule of isobutane is produced for each molecule of n-­‐butane consumed. ii. One point is earned for the correct answer with justification. Calculate the molar concentration of each species at equilibrium. Kc = [isobutane] / [n-­‐butane] = x / (0.010 –x) = 2.5 x = 2.5 (0.010 – x) = 0.025 – 2.5 x 3.5 x = 0.025 à x = 0.071 M isobutane (0.0100 M – 0.0071 M) = 0.003 M n-­‐butane 64 1 point is earned for the correct setup. One point is earned for both correct numerical answers. iii. If the volume of the system is reduced to half of its original volume, what will the new concentration of n-­‐butane after equilibrium has been reestablished at 25 oC? Justify your answer. Halving the volume of the container at equilibrium doubles the pressure of both isobutane and n-­‐butane, which has no effect on the equilibrium because the stoichiometry of the reaction is one mole of product produced for each mole of reactant consumed. Since the number of moles of each isomer is unchanged but the volume is reduced by half, concentrations of both isomers are doubled and the concentration of n-­‐butane will be 2 × 0.003 M = 0.006 M. One point is earned for the correct answer with justification. Suppose that in another experiment a 0.010 mol sample of pure isobutane is placed in an evacuated 1.0 L rigid container and allowed to come to equilibrium at 25 oC. e. Calculate the molar concentration of each species after equilibrium has been established. f. g. The concentrations of isobutane and n-­‐butane would be the same as they were calculated in part (e)(ii), 0.0071 M and 0.003 M, respectively. One point is earned for correct numerical answers or a correct statement regarding their equivalence to values obtained in part (e)(ii). 2011 B Q 6 c. A small amount of liquid ethyl methanoate (boiling point 54 oC) was placed in a rigid closed 2.0 L container containing argon gas at an initial pressure of 1.00 atm and a temperature of 20 oC. The pressure in the container was monitored for 70. seconds after the ethyl methanoate was added, and the data in the graph below were obtained. It was observed that some liquid ethyl methanoate remained in the flask after 70.0 seconds. (Assume that the volume of the remaining liquid is negligible compared to the total volume of the container.) 65 i. Explain why the pressure in the flask increased during the first 60. seconds. Some of the liquid ethyl methanoate is going into the gas (vapor) phase. 1 point is earned for the correct explanation. ii. Explain, in terms of processes occurring at the molecular level, why the pressure in the flask remained constant after 60. seconds. At the equilibrium vapor pressure, the rate of molecules passing from the liquid to the gas phase (vaporizing) equals the rate of gas phase molecules passing into the liquid phase (condensing). 1 point is earned for the correct explanation. iii. What is the value of the partial pressure of ethyl methanoate vapor in the container at 60. seconds? 1.25 atm − 1.00 atm = 0.25 atm 1 point is earned for the correct answer
iv. After 80. seconds, additional liquid ethyl methanoate is added to the container at 20 oC. Does the partial pressure of the ethyl methanoate vapor in the container increase, decrease, or stay the same? Explain. (Assume that the volume of the additional liquid ethyl methanoate in the container is negligible compared to the total volume of the container.) 66 The partial pressure of the vapor stays the same because the equilibrium vapor pressure for 20°C has already been reached. Because the temperature remains constant, the vapor pressure would remain unchanged. 1 point is earned for the correct answer with an explanation. 2009 B Q1 1. A pure 14.85 g sample of the weak base ethylamime, C2H5NH2, is dissolved in enough water to make 500. mL of solution. a. Calculate the molar concentration of the C2H5NH2 in the solution. n (C2H5NH2) = 14.85 g C2H5NH2 x (1 mol C2H5NH2 / 45.09 g C2N5NH2) = .3293 mol C2H5NH2 M (C2H5NH2) = 0.3293 mol C2H5NH2 / 0.500 L = 0.659 M 1 point is earned for the correct number of moles. 1 point is earned for the correct concentration. The aqueous ethylamine reacts with water according to the equation below. C2H5NH2(aq) + H2O(l)  C2H5NH3+(aq) + OH-­‐(aq) b. Write the equilibrium constant expression for the reaction between C2H5NH2(aq) and water. Kb = [C2H5NH3+] [OH-­‐] / [C2H5NH2] 1 point is earned for the correct expression. c. Of C2H5NH2(aq) and C2H5NH3+(aq), which is present in the solution at the higher concentration at equilibrium? Justify your answer. C2H5NH2 is present in the solution at the higher concentration at equilibrium. Ethylamine is a weak base, and thus it has a small Kb value. Therefore only partial dissociation of C2H5NH2 occurs in water, and [C2H5NH3+] is thus less than [C2H5NH2]. d. One point is earned for the correct answer with justification. A different solution is made by mixing 500. mL of 0.500 M C2H5NH2 with 500. mL of 0.200 M HCl. Assume that volumes are additive. The pH of the resulting solution is found to be 10.93. i. Calculate the concentration of OH-­‐(aq) in the solution. 67 pH = −log[H+] => [H+] = 10−10.93 = 1.17 × 10−11 [OH−] = Kw / [H+] = (1 x 10-­‐14) / (1.17 x 10-­‐11) = 8.5 x 10-­‐4 M OR pOH = 14 − pH = 14 − 10.93 = 3.07 pOH = −log[OH−] [OH−] = 10−3.07 = 8.5 × 10−4 M ii. equation. One point is earned for the correct concentration. Write the net ionic equation that represents the reaction that occurs when the C2H5NH2 solution is mixed with the HCl solution. C2H5NH2 + H3O+ → C2H5NH3+ + H2O iii. Calculate the molar concentration of the C2H5NH3+(aq) that is formed in the reaction. Moles of C2H5NH2 = 0.500 L x 0.500 mol / 1 L = 0.250 mol Moles of H3O+ = 0.500 L x 0.200 mol / 1.00 L = 0.100 mol initial value change final value iv. [C2H5NH2] 0.250 −0.100 0.150 One point is earned for the correct [H3O+] 0.100 −0.100 ~ 0 [C2H5NH3+] ~ 0 +0.100 0.100 [C2H5NH3+] = 0.100 mol C2H5NH3 / 1.0 L = 0.100 M One point is earned for the correct number of moles of C2H5NH2 and H3O+. One point is earned for the correct concentration. Calculate the value of Kb for C2H5NH2. [C2H5NH2] = 0.150 mol C2H5NH2 / 1.00 L = 0.150 M Kb =[C2H5NH3+] [OH-­‐] / [C2H5NH2] = (0.100) (8.5 x 10-­‐4) / (0.150) = 5.67 x 10-­‐4 One point is earned for the correct calculation of the molarity of C2H5NH2 after neutralization. One point is earned for the correct value. 2010 Q1 1. Several reactions are carried out using AgBr, a cream-­‐colored silver salt for which the value of the solubility-­‐product constant, Ksp, is 5.0 × 10-­‐13 at 298 K 68 a. Write the expression for the solubility-­‐product constant, Ksp, of AgBr. Ksp = [Ag+][Br−] One point is earned for the correct expression (ion charges must be present; parentheses instead of square brackets not accepted). b. × 10−7 M Calculate the value of [Ag+] in 50.0 mL of a saturated solution of AgBr at 298 K. Let x = equilibrium concentration of Ag+ (and of Br−). Then Ksp = 5.0 × 10−13 = x2 ⇒ x = 7.1 One point is earned for the correct value with supporting work (units not necessary). c. d. A 50.0 mL sample of distilled water is added to the solution described in part b, which is in a beaker with some solid AgBr at the bottom. The solution is stirred and equilibrium is reestablished. Some solid AgBr remains in the beaker. Is the value of [Ag+] greater than, less than, or equal to the value you calculated in part b? Justify your answer. The value of [Ag+] after addition of distilled water is equal to the value in part (b). The concentration of ions in solution in equilibrium with a solid does not depend on the volume of the solution. One point is earned for the correct answer with justification. Calculate the minimum volume of distilled water, in liters, necessary to completely dissolve a 5.0 g sample of AgBr(s) at 298 K. (The molar mass of AgBr is 188 g mol-­‐1). 5.0 g AgBr x (1 mol AgBr / 188 g AgBr) = 0.0266 mol AgBr 0.0266 mol / V = 7.1 x 10-­‐7 mol L-­‐1 à V = 3.7 x 104 L One point is earned for the calculation of moles of dissolved AgBr. One point is earned for the correct answer for the volume of water. e. A student mixes 10.0 mL of 1.5 × 10-­‐4 M AgNO3 with 2.0 mL of 5.0 × 10-­‐4 M NaBr and stirs the resulting mixture. What will the student observe? Justify you answer with calculations. [Ag+] = (10.0 mL) (1.5 x 10-­‐4 M) / (12.0 mL) = 1.3 x 10-­‐4 M [Br-­‐] = (2.0 mL) (5.0 x 10-­‐4 M) / (12.0 mL) = 8.3 x 10-­‐5 M Q = [Ag+] [Br-­‐] = (1.3 x 10-­‐4 M) (8.3 x 10-­‐5 M) = 1.1 x 10-­‐8 1.1 x 10-­‐8 > 5.0 x 10-­‐13, therefore a precipitate will form. 69 One point is earned for calculation of concentration of ions. One point is earned for calculation of Q and conclusion based on comparison between Q and Ksp. One point is earned for indicating the precipitation of AgBr. f. The color of another salt of silver, AgI(s) is yellow. A student adds a solution of NaI to a test tube containing a small amount of solid, cream-­‐colored AgBr. After stirring the contents of the test tube, the student observes that the solid in the test tube changes color from cream to yellow. i. Write the chemical equation for the reaction that occurred in the test tube. AgBr(s) + I−(aq) → AgI(s) + Br−(aq) OR AgBr(s) + NaI(aq) → AgI(s) + NaBr(aq) One point is earned for the correct equation. ii. Which salt has the greater value of Ksp: AgBr or AgI? Justify your answer. AgBr has the greater value of Ksp. The precipitate will consist of the less soluble salt when both I−(aq) and Br−(aq) are present. Because the color of the precipitate in the test tube turns yellow, it must be AgI(s) that precipitates; therefore Ksp for AgBr must be greater than Ksp for AgI. OR Keq for the displacement reaction is of Ksp of AgBr / Ksp of AgI Because yellow AgI forms, Keq > 1; therefore Ksp of AgBr > Ksp of AgI. One point is earned for the correct choice with justification. 2011 B Q1 1. Answer the following questions about the solubility and reactions of the ionic compounds M(OH)2 and MCO3, where M represents an unidentified metal. a. Identify the charge of the M ion in the ionic compounds above. 2+ 1 point is earned for the correct charge. b. At 25 oC, a saturated solution of M(OH)2 has a pH of 9.15. i. Calculate the molar concentration of OH-­‐(aq) in the saturated solution. pOH = 14 – pH; pOH = 14 − 9.15 = 4.85; [OH−] = 10−4.85 = 1.4 × 10−5 M 1 point is earned for the correct concentration. 70 ii. Write the solubility product expression for M(OH)2. Ksp = [M2+] [OH−]2 1 point is earned for the correct expression. iii. Calculate the value of the solubility product constant, Ksp, for M(OH)2 at 25 oC. [M2+] = (1/2)[OH−] = (1/2)(1.4 × 10−5 M) = 7.0 × 10−6 M Ksp = [M2+] [OH-­‐]2 = (7.0 x 10-­‐6) (1.4 x 10-­‐5)2 = 1.4 x 10-­‐15 1 point is earned for the correct relationship between [M2+] and [OH−]. 1 point is earned for the correct value. c. For the metal carbonate, MCO3, the value of the solubility product constant, Ksp, is 7.4 x 10-­‐14 at 25 oC. On the basis of this information and your results in part b), which compound, M(OH)2 or MCO3, has the greater molar solubility in water at 25 oC? Justify your answer with a calculation. For M(OH)2 : [M2+] and molar solubility = 7.0 × 10−6 M For MCO3: Ksp = 7.4 × 10−14 = [M2+][CO32−] [M2+] and molar solubility = 2.7 × 10−7 M Because 7.0 × 10−6 M > 2.7 × 10−7 M, M(OH)2 has the greater molar solubility. 1 point is earned for the molar solubility of MCO3. 1 point is earned for an answer consistent with the calculated molar solubility. d. MCO3 decomposes at high temperatures, as shown by the reaction represented below. MCO3(s)  MO(s) + CO2(g) A sample of MCO3 is placed in a previously evacuated container, heated to 423 K, and allowed to come to equilibrium. Some solid MCO3 remains in the container. The value of KP for the reaction at 423 K is 0.0012. i. Write the equilibrium constant expression for KP of the reaction. KP = P(CO2) ii. Determine the pressure, in atm, of CO2(g) in the container at equilibrium at 423 K. P(CO2) = 0.0012 atm iii. Indicate whether the value of ΔGo for the reaction at 423 K is positive, negative, or zero. Justify your answer. 1 point is earned for the correct expression. 1 point is earned for the correct pressure. 71 ΔG° = −RT lnK; K = 0.0012 < 1, thus ln K is negative; therefore ΔG° is positive. 1 point is earned for the correct answer with justification. 2011 Q1 1. Each of three beakers contains 25.0 mL of a 0.100 M solution of HCl, NH3, or NH4Cl, as shown above. Each solution is at 25 oC. a. Determine the pH of the solution in beaker 1. Justify your answer. pH = -­‐log[H+] = -­‐log(0.100) = 1.000 b. In beaker 2, the reaction NH3(aq) + H2O(l)  NH4+(aq) + OH-­‐(aq) occurs. The value of Kb for NH3(aq) is 1.8 x 10-­‐5 at 25 oC. i. Write the Kb expression for the reaction of NH3(aq) with H2O(l). Kb = [NH4+] [OH-­‐] / [NH3] ii.) Calculate the [OH-­‐] in the solution in beaker 2. Let [OH−] = x, then Kb = (x)(x) (0.100 -­‐ x) 1 point is earned for the correct pH. 1 point is earned for the correct expression. Assume that x << 0.100 M, then 1.8 × 10−5 = x2 / 0.100 ⇒ x = [OH−] = 1.3 × 10-­‐3 M 1 point is earned for the correct setup. 1 point is earned for the correct answer. c.) In beaker 3, the reaction NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) occurs. i.) Calculate the value of Ka for NH4+ at 25 oC. Ka = Kw / Kb = 1.0 x 10-­‐14 / 1.8 x 10-­‐5 = 5.6 x 10-­‐10 1 point is earned for the correct answer. ii.) The contents of beaker 2 are poured into beaker 3 and the resulting solution is stirred. Assume that volumes are additive. Calculate the pH of the resulting solution. 72 In the resulting solution, [NH3] = [NH4+]; Ka = 5.6 × 10-­‐10 = [NH3][H3O+] / [NH4+] Thus [H3O+] = 5.6 × 10-­‐10; pH = -­‐log(5.6 × 10-­‐10) = 9.25 1 point is earned for noting that the solution is a buffer with [NH3] = [NH4+]. 1 point is earned for the correct pH. d.) The contents of beaker 1 are poured into the solution made in part c) ii). The resulting solution is stirred. Assume that volumes are additive. i.) Is the resulting solution an effective buffer? Justify your answer. The resulting solution is not an effective buffer. Virtually all of the NH3 in the solution formed in (c)(ii) will react with the H3O+ from solution 1: NH3 + H3O+ → NH4+ + H2O leaving mostly NH4+ in the final solution. Since only one member of the NH4+/NH3 conjugate acid-­‐base pair is left, the solution cannot buffer both base and acid. 1 point is earned for the correct response with an acceptable justification. ii.) Calculate the final [NH4+] in the resulting solution at 25 oC. moles = (volume)(molarity) moles H3O+ in sol. 1 = (0.0250)(0.100) = 0.00250 mol moles NH3 in sol. 2 = (0.0250)(0.100) = 0.00250 mol moles NH4+ in sol. 3 = (0.0250)(0.100) = 0.00250 mol When the solutions are mixed, the H3O+ and NH3 react to form NH4+, resulting in a total of 0.00500 mol NH4+. The final volume is the sum (25.0 + 25.0 + 25.0) = 75.0 mL. The final concentration of NH4+ = (0.00500 mol/0.0750 L) = 0.0667 M. 1 point is earned for the correct calculation of moles of NH4+. 1 point is earned for the correct calculation of the final volume and concentration. 2012 Q1 1. A 1.22 g sample of a pure monoprotic acid, HA, was dissolved in distilled water. The HA solution was then titrated with 0.250 M NaOH. The pH was measured throughout the titration, and the equivalence point was reached when 40.0 mL of the NaOH solution had been added. The data from the titration are recorded in the table below. Volume of 0.250 M NaOH pH of titrated Solution Added (mL) 73 0.00 ? 10.0 3.72 20.0 4.20 30.0 ? 40.0 8.62 50.0 12.40 a. Explain how the data in the table above provide evidence that HA is a weak acid rather than a strong acid. The pH at the equivalence point is above 7, which indicates that HA is a weak acid. 1 point is earned for the correct explanation. b. Write the balanced net-­‐ionic equation for the reaction that occurs when the solution of NaOH is added to the solution of HA. HA(aq) + OH-­‐(aq) à A-­‐(aq) + H2O(l) c. 1 point is earned for writing the net-­‐ionic equation balanced for mass and charge. Calculate the number of moles of HA that were titrated. At the equivalence point, the number of moles of base added equals the number of moles of acid initially present. 0.0400 L (0.250 mol NaOH / 1 L) (1 mol HA / 1 mol NaOH) = 0.100 mol HA d. 1 point is earned for the correct number of moles. Calculate the molar mass of HA. MM = mass of acid / moles of acid = 1.22 g / 0.0100 mol = 122 g mol-­‐1 1 point is earned for the correct molar mass. The equation for the dissociation reaction of HA in water is shown below. HA(aq) + H2O(l)  H3O+(aq) + A-­‐(aq) e. Ka = 6.3 x 10-­‐5 Assume that the initial concentration of the HA solution (before any NaOH solution was added) is 0.200 M. Determine the pH of the initial HA solution. 74 Ka = [H3O+] [A-­‐] / [HA] 6.3 x 10-­‐5 = (x) (x) / (0.200 – x); assume x << 0.200 M x = [H3O+] = 3.5 x 10-­‐3 M pH = -­‐ log [H3O+] = -­‐ log (3.5 x 10-­‐3) = 2.45 f. 1 point is earned for the appropriate substitution into the Ka expression. 1 point is earned for the correct [H3O+]. 1 point is earned for the calculation of pH. Calculate the value of [H3O+] in the solution after 30.0 mL of NaOH is added and the total volume of the solution is 80.0 mL HA + OH-­‐à A-­‐ + mol before rxn: 0.0100 mol after rxn: 0.00250 0.00750 0.00000 0.00000 0.00750 H2O [HA] = 0.00250 mol / 0.0800 L = 3.13 x 10-­‐2 M [A-­‐] = 0.00750 mol / 0.0800 L = 9.38 x 10-­‐2 M Ka = [H3O+] [A-­‐] / [HA] 6.3 x 10-­‐5 = (x) (9.38 x 10-­‐2 + x) / (3.13 x 10-­‐2 – x) Assume that x << 9.38 x 10-­‐2 and 3.13 x 10-­‐2 , Then 6.3 x 10-­‐5 = (x) (9.38 x 10-­‐2) / (3.13 x 10-­‐2) X = [H3O+] 2.10 x 10-­‐5 M. 1 point is earned for the correct calculation of moles of A-­‐ and HA after the reaction. 1 point is earned for the appropriate substitution into the equilibrium expression. 1 point is earned for the correct calculation of [H3O+]. 2012B Q1 1. CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH-­‐(aq) Kb = 4.4 x 10-­‐4 Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH of the solution to be 11.77 a. Write the expression of the equilibrium constant, Kb, for methylamine. Kb = [CH3NH3+] [OH-­‐] / [CH3NH2] One point is earned for the correct expression. 75 b. Calculate the molar concentration of OH-­‐ in the 50.0 mL sample of the methylamine solution. pH = 11.77 => [H+] = 10-­‐11.77 = 1.7 x 10-­‐12 [OH-­‐] = Kw / [H+] = 1.0 x 10-­‐14 / 1.7 x 10-­‐12 = 5.9 x 10-­‐3 OR pOH = 14 – pH = 2.23 => [OH-­‐] = 10-­‐2.23 = 5.9 x 10-­‐3 1 point is earned for correct [OH-­‐] c. Calculate the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established. Kb = [CH3NH3+] [OH-­‐] / [CH3NH2] 4.4 x 10-­‐4 = (5.9 x 10-­‐3) (5.9 x 10-­‐3) / (x – 5.9 x 10-­‐3) x = 0.085 M One point is earned for [CH3NH3+] = [OH-­‐]. One point is earned for the correct initial molar concentration. Note: An approximated molar concentration does NOT earn the second point. The 50.0 mL sample of the methylamine solution is titrated with an HCl solution of unknown concentration. The equivalence point of the titration is reached after a volume of 36.0 mL of the HCl solution is added. The pH of the solution at the equivalence point is 5.98. d. Write the net-­‐ionic equation that represents the reaction that takes place during the titration. CH3NH2 + H3O+ à CH3NH3+ + H2O OR One point is earned for a correct equation. e. Calculate the concentration of the HCl solution used to titrate the methylamine. (0.085 mol / 1000. mL ) x 50.0 mL = 0.00425 mol CH3NH2 CH3NH2 + H+ à CH3NH3+ (0.00425 mol HCl / 36.0 mL) x (1000. mL / 1.000 L) = 0.12 M One point is earned for equal moles of acid and base. One point is earned for the correct concentration. 76 f. Using the axes provided, sketch the titration curve that results from the titration described above. On the graph, clearly label the equivalence point of the titration. One point is earned for a curve starting at a pH between 11 and 12. One point is earned for labeling the equivalence po9int at V = 36.0 mL HCl and a pH of 5.98. One point is earned for general shape of the curve for a weak acid / strong base titration. Multiple Choice Questions 2013 26. 30. For which of the following reaction mixtures at equilibrium would reducing the volume of the container at constant temperature cause the equilibrium to shift toward the products? a. SrCO3(s)  SrO(s) + CO2(g) b. C(s) + CO2(g)  2 CO(g) c. CO(g) + H2O(g)  CO2(g) + H2(g) d. PCl5(g)  PCl3(g) + Cl2(g) e. 2 NO2(g)  N2O4(g) The equilibrium constant for the gas phase reaction above is 95 at 25 oC. What is the value of the equilibrium constant for the following reaction at 25 oC? O2(g) + 4 NO2(g)  2 N2O5(g) 40. a. (95)2 b. 95 c. (95)1/2 d. 1/95 e. 1/(95)2 HSO4-­‐(aq) + CO32-­‐(aq)  SO42-­‐(aq) + HCO3-­‐(aq) The reaction represented above is observed to proceed spontaneously to the right in aqueous solution. In this system the strongest base is a. b. c. d. e. SO42-­‐(aq) CO32-­‐(aq) H2O(l ) HCO3-­‐(aq) HSO4-­‐(aq) 77 55. Which of the following statements is true for the equilibrium vapor pressure of a liquid in a closed system? a. b. c. d. e. It remains constant when the temperature increases. It decreases to half its original value if the volume of the gas phase is doubled. It increases to twice its original value if the volume of the liquid phase is doubled. It decreases to half its original value if the surface area of the liquid is reduced by one-­‐
half. It is independent of the volume of the vapor phase. X(g) + Y(g)  2 Z(g) 60. When 4.00 mol each of X(g) and Y(g) are placed in a 1.00 L vessel and allowed to react at constant temperature according to the equation above, 6.00 mol of Z(g) is produced. What is the value of the equilibrium constant, Kc ? a. 3 b. 6 c. 8 d. 16 e. 36 2013 AB HSO4-­‐(aq) + CO32-­‐(aq)  SO42-­‐(aq) + HCO3-­‐(aq) 40. 45. The reaction represented above is observed to proceed spontaneously to the right in aqueous solution. In this system the strongest base is a. SO42-­‐(aq) b. CO32-­‐(aq) c. H2O(l) d. HCO3-­‐(aq) e. HSO4-­‐(aq) A 0.10 M solution of a weak monoprotic acid has a pH equal to 4.0. The ionization constant, Ka, of the acid is a. b. c. d. e. 1 × 10-­‐3 1 × 10-­‐4 1 × 10-­‐7 1 × 10-­‐8 1 × 10-­‐9 78 56. An unknown acid is dissolved in 25 mL of water and titrated with 0.100 M NaOH. The results are shown in the titration curve above. Which of the following could be the unknown acid? a. b. c. d. e. Fluoroacetic acid, pKa = 2.6 Glycolic acid, pKa = 3.8 Propanoic acid, pKa = 4.9 Hypochlorous acid, pKa = 7.5 Boric acid, pKa = 9.3 79 Laboratory Concepts – LABORATORY QUESTIONS – contained in several Big Ideas and the Science Practices 1. Vocabulary words ppt (precipitate) Free Response Questions 2010 Q2 Thermo Lab 2. A student performs an experiment to determine the molar enthalpy of solution of urea, H2NCONH2. The student places 91.95 g of water at 25 oC into a coffee-­‐cup calorimeter and immerses a thermometer in the water. After 50 s, the student adds 5.13 g of solid urea, also at 25 oC, to the water and measures the temperature of the solution as the urea dissolves. A plot of the temperature data is shown in the graph below. a. Determine the change in temperature of the solution that results from the dissolution of the urea. ΔT = 21.8 − 25.0 = −3.2 Celsius degrees One point is earned for the correct temperature change.
b. According to the data, is the dissolution of urea in water an endothermic process or an exothermic process? Justify your answer. The process is endothermic. The decrease in temperature indicates that the process for the dissolution of urea in water requires energy. One point is earned for the correct choice with justification. 80 c. Assume that the specific heat capacity of the calorimeter is negligible and that the specific heat capacity of the solution of urea and water is 4.2 J g-­‐1 oC-­‐1 throughout the experiment. i. Calculate the heat of dissolution of the urea in joules. Assuming that no heat energy is lost from the calorimeter and given that the calorimeter has a negligible heat capacity, the sum of the heat of dissolution, qsoln and the change in heat energy of the urea-­‐water mixture must equal zero. qsoln + mcΔT = 0 ⇒ qsoln = − mcΔT msoln = 5.13 g + 91.95 g = 97.08 g qsoln = −(97.08 g)(4.2 J g −1°C−1)(−3.2°C) = 1.3 × 103 J One point is earned for the correct setup. One point is earned for the correct numerical result for the heat of dissolution. ii. Calculate the molar enthalpy of solution, ΔHsolno, of urea in kJ mol-­‐1. ΔHsolno = qsoln / mol solute Molar mass of urea = 60.00 g mol-­‐1 Moles of urea = 5.13 g urea (1 mol urea / 60.0 g urea) = 0.0855 mol ΔHsolno = 1.3 x 103 J / 0.0855 mol = 1.5 x 104 J mol-­‐1 = 15 kJ mol-­‐1 One point is earned for the calculation of moles of urea. One point is earned for the correct numerical result with correct algebraic sign. d. Using the information in the table below, calculate the value of the molar entropy of solution, ΔSsolno, of urea at 298 K. Include units with your answer. ΔHsolno of urea 14.0 kJ mol-­‐1 ΔGsolno of urea -­‐6.9 kJ mol-­‐1 Accepted value ΔG° = ΔH°−TΔS° => − 6.9 kJ mol−1 = 14.0 kJ mol−1− (298 K)(ΔS°) = 0.0701 kJ mol−1 K−1 = 70.1 J mol−1 K−1 One point is earned for the correct setup. One point is earned for the correct numerical result with correct units. e. The student repeats the experiment and this time obtains a result for ΔHsolno of urea that is 11 percent below the accepted value. Calculate the value of ΔHsolno that the student obtained in this second trial. 81 Error = (0.11)(14.0 kJ mol−1) = 1.54 kJ mol−1 14.0 kJ mol−1 − 1.54 kJ mol−1 = 12.5 kJ mol−1 One point is earned for the correct numerical result. f. The student performs a third trial of the experiment but this time adds urea that has been taken directly from a refrigerator at 5 oC. What effect, in any, would using the cold urea instead of urea at 25 oC have on the experimentally obtained value of ΔHsolno? Justify your answer. There would be an increase in the obtained value for ΔHsolno because the colder urea would have caused a larger negative temperature change. One point is earned for the correct prediction with justification. 2011B #5 5. A student is instructed to prepare 100.0 mL of 1.250 M NaOH from a stock solution of 5.000 M NaOH. The student follows the proper safety guidelines. a. Calculate the volume of 5.000 M NaOH needed to accurately prepare 100.0 mL of 1.250 M NaOH. M1V1 = M2V2 => V1 = M2V2 / M1 = (1.250 M)(100.0 mL) / 5.000 M = 25.00 mL 1 point is earned for the correct volume. b. Describe the steps in a procedure to prepare 100.0 mL of 1.250 M NaOH using 5.000 M NaOH and equipment selected from the list below. Balance 25 mL Erlenmeyer flask 100 mL graduated cylinder 100 mL volumetric flask 50 mL buret 100 mL Florence flask 25 mL pipet 100 mL beaker Eyedropper Drying oven Wash bottle of distilled H2O Crucible Pipet 25.00 mL of 5.000 M NaOH solution into the 100 mL volumetric flask. Fill the volumetric flask to the calibration line with distilled water; using an eyedropper for the last few drops is advised. Cap the volumetric flask and invert several times to ensure homogeneity. 1 point is earned for descriptions of any two of the three steps. An additional point is earned if all three steps are described. c. The student is given 50.0 mL of a 1.00 M solution of a weak, monoprotic acid, HA. The solution is titrated with the 1.250 M NaOH to the endpoint. (Assume that the endpoint is at the equivalence point.) 82 i. Explain why the solution is basic at the equivalence point of the titration. Include a chemical equation as part of your explanation. When a weak acid is titrated with a strong base, the reaction forms water and the A− ion. HA + OH− → A− + H2O The A− ion formed in the titration reacts with the solvent water to release OH− ions, making the solution basic at the equivalence point. A− + H2O  HA + OH− ii. 1 point is earned for either the correct equation or a clear statement that the conjugate base, A–, is a (weak) base. 1 point is earned for indicating that the solution is basic because of the formation of OH–. Identify the indicator in the table below that would be best for the titration. Justify your choice. Indicator pKa Methyl Red 5 Bromothymol blue 7 Phenolphthalein 9 Because the pH is basic at the equivalence point, it is best to use an indicator that changes color in basic solution. Therefore, phenolphthalein would be the best indicator for the titration. 1 point is earned for an answer consistent with the answer to part (c)(i) with justification. d. The student is given another 50.0 mL sample of 1.00 M HA, which the student adds to the solution that had been titrated to the endpoint in part c). The result is a solution with a pH of 5.0. i. What is the value of the acid dissociation constant, Ka, for the weak acid? Explain your reasoning. The resulting solution is at the half-­‐equivalence-­‐point, where [HA] = [A−] , thus pH = pKa = 5.0 ⇒ Ka = 1 × 10−5. 1 point is earned for showing that the system is at the half-­‐equivalence point. 1 point is earned for the correct value of Ka. ii. Explain why the addition of a few drops of 1.250 M NaOH to the resulting solution does not appreciably change its pH. The resulting solution is a buffer; therefore adding a few drops of acid or base does not appreciably change the pH. 1 point is earned for indicating that the solution is a buffer. 83 2012 Q6 6. In a laboratory experiment, Pb and an unknown metal Q were immersed in solutions containing aqueous ions of unknown metals Q and X. The following reactions summarize the observations. Observation 1: Pb(s) + X2+(aq) à Pb2+(aq) + X(s) Observation 2: Q(s) + X2+(aq) à no reaction Observation 3: Pb(s) + Q2+(aq) à Pb2+(aq) + Q(s) a. On the basis of the reactions indicated above, arrange the three metals, Pb, Q, and X, in order from least reactive to the most reactive on the lines provided below. ___________ , _______________, _______________ Least reactive metal Most reactive metal Q, X, Pb 2 points are earned for the correctly ordered relationship. (1 point earned for Q, Pb, X or X, Q, Pb) The diagram below shows an electrochemical cell that is constructed with a Pb electrode immersed in 100. mL of 1.0 M Pb(NO3)2(aq) and an electrode made of metal X immersed in 100. mL of 1.0 M X(NO3)2(aq). A salt bridge containing saturated aqueous KNO3 connects the anode compartment to the cathode compartment. The electrodes are connected to an external circuit containing a switch, which is open. When a voltmeter is connected to the circuit as shown, the reading on the voltmeter is 0.47 V. When the switch is closed, electrons flow through the switch from the Pb electrode toward the X electrode. b. Write the equation for the half reaction that occurs at the anode. 84 Pb(s)→ Pb2+(aq) + 2 e− c. 1 point is earned for the correct equation. The value of the standard potential for the cell, Eo, is 0.47 V. i. Determine the standard reduction potential for the half reaction that occurs at the cathode. Ecello =Ecathode o -­‐ Eanode o Ecathodeo = Ecello + Eanodeo Ecathodeo = 0.47 + (-­‐0.13) = 0.34 V ii. 1 point is earned for the calculated reduction potential with mathematical justification. Determine the identity of metal X. The metal is copper. 1 point is earned for identification of the metal. d. Describe what happens to the mass of each electrode as the cell operates. The mass of the Pb electrode decreases and the mass of the Cu electrode increases. 1 point is earned for both descriptions. e. During a laboratory session, students set up the electrochemical cell shown above. For each of the following three scenarios, choose the correct value of the cell voltage and justify your choice. i. A student bumps the cell setup, resulting in the salt bridge losing contact with the solution in the cathode compartment. Is V equal to 0.47 or is V equal to 0? Justify your choice. ii. V = 0 V. The transfer of ions through the salt bridge will stop. A charge imbalance between the half-­‐cells will prevent electrons from flowing through the wire. 1 point is earned for the correct choice with an appropriate explanation. A student spills a small amount of 0.5 M Na2SO4(aq) into the compartment with the Pb electrode, resulting in the formation of a precipitate. Is V less than 0.47 or is V greater than 0.47 ? Justify your choice. V > 0.47 V. The sulfate ion will react with the Pb2+ ion to form a precipitate. This results in a thermodynamically favored anode half-­‐cell reaction and hence a larger potential difference. The choice may also be justified using the Nernst equation. Ecell = Ecello – (R T / n F) ln ([Pb2+] / [Cu2+]) 85 Decreasing the [Pb2+] will increase the cell voltage. iii. 1 point is earned for the correct choice with an appropriate explanation. After the laboratory session is over, a student leaves the switch closed. The next day, the student opens the switch and reads the voltmeter. Is V less than 0.47 or is V equal to 0.47? Justify your choice. V < 0.47 V. Over time, [Pb2+] increases and [Cu2+] decreases, making both half-­‐
cell reactions less thermodynamically favorable. The choice may also be justified using the Nernst equation. Increasing [Pb2+] and decreasing [Cu2+] decreases the cell voltage. The choice may also be justified by stating that the voltage is zero as a result of the establishment of equilibrium. 1 point is earned for the correct choice with an appropriate explanation 2011 A 2 2. A student is assigned the task of determining the mass percent of silver in an alloy of copper and silver by dissolving a sample of the alloy in excess nitric acid and then precipitating the silver as AgCl. First the student prepares 50. mL of 6 M HNO3. a. The student is provided with a stock solution of 16 M HNO3, two 100 mL graduated cylinders that can be read to ± 1 mL, a 100 mL beaker that can be read to ± 10 mL, safety goggles, rubber gloves, a glass stirring rod, a dropper, and distilled H2O. i. Calculate the volume, in mL, of 16 M HNO3 that the student should use for preparing 50. mL of 6 M HNO3. moles before dilution = moles after dilution MiVi= MfVf (16 M)(Vi) = (6 M)(50. mL) Vi= 19 mL or 20 mL (to one significant figure) 1 point is earned for the correct volume. ii. Briefly list the steps of an appropriate and safe procedure for preparing the 50. mL of 6 M HNO3. Only materials selected from those provided to the student (listed above) may be used. Wear safety goggles and rubber gloves. Then measure 19 mL of 16M HNO3 using a 100 mL graduated cylinder. Measure 31 mL of distilled H2O using a 100 mL graduated cylinder. Transfer the water to a 100 mL beaker. Add the acid to the water with stirring. 86 iii. 1 point is earned for properly measuring the volume of 16M HNO3 and preparing a 6 M HNO3 acid solution. 1 point is earned for wearing protective gear and for adding acid to water. Explain why it is not necessary to use a volumetric flask (calibrated to 50.00 mL ± 0.05 mL) to perform the dilution. The graduated cylinders provide sufficient precision in volume measurement to provide two significant figures, making the use of the volumetric flask unnecessary. 1 point is earned for an acceptable explanation. iv. During the preparation of the solution, the student accidently spills about 1 mL of 16 M HNO3 on the bench top. The student finds three bottles containing liquids sitting near the spill: a bottle of distilled water, a bottle of 5 percent NaHCO3(aq), and a bottle of saturated NaCl(aq). Which of the liquids is best to use in cleaning up the spill? Justify your choice. NaHCO3(aq) should be used. The HCO3− ion will react as a base to neutralize the HNO3. 1 point is earned for the correct choice with explanation.
Then the student pours 25 mL of the 6 M HNO3 into a beaker and adds a 0.6489 g sample of the alloy. After the sample completely reacts with the acid, some saturated NaCl(aq) is added to the beaker, resulting in the formation of an AgCl precipitate. Additional NaCl(aq) is added until no more precipitate is observed to form. The precipitate is filtered, dried, and weighed to constant mass in a filter crucible. The data are shown in the table below. Mass of sample of copper -­‐ silver alloy 0.6489 g Mass of dry filter crucible 28.7210 g Mass of filter crucible and precipitate (1st weighing) 29.3587 g Mass of filter crucible and precipitate (2nd weighing) 29.2599 g Mass of filter crucible and precipitate (3rd weighing) 29.2598 g b. Calculate the number of moles of AgCl precipitate collected. mass of AgCl collected = (29.2598 – 28.7210) g = 0.5388 g / ((107.87 + 35.45) g mol−1) = 3.759 × 10−3 mol AgCl 1 point is earned for the correct mass of AgCl. 1 point is earned for the correct number of moles of AgCl given with the correct number of significant figures. 87 c. Calculate the mass percent of silver in the alloy of copper and silver. 3.759 × 10−3 mol Ag × (107.87 g Ag / 1 mol Ag) = 0.4055 g Ag 0.4055 g / 0.6489 g × 100% = 62.49% Ag 1 point is earned for the correct setup and the correct calculation of the mass of Ag. 1 point is earned for the correct percent of Ag. 1992 Q 7 7. Four bottles, each containing about 5 grams of finely powdered white substance, are found in a laboratory. Near the bottles are four labels specifying high purity and indicating that the substances are glucose (C6H12O6), sodium chloride (NaCl), aluminum oxide (Al2O3), and zinc sulfate (ZnSO4). Assume that these labels belong to the bottles and that each bottle contains a single substance. Describe the tests you would conduct to determine which label belongs to which bottle. Give the results you would expect for each test. General Principles -­‐ distinct and unambiguous identification of these unknowns relative to each other. By the structure of the question, if 3 out of 4 substances are unambiguously identified, the fourth may be identified by exclusion. If attempted, this must be stated to earn final 2 points. The series may take to form of a flow chart, excluding samples from further tests as they are identified. Or, the series may be a series of tests, each applied to each substance, the total results of all tests being used to identify the substance. Scoring 1) Each of the 4 substances correctly and unambiguously identified earns 2 points. 2) Tests which are usable and are misapplied, misinterpreted, etc. can earn at most 1 point. 3) Full credit for a substance cannot be earned unless the identification is both correct and unambiguous. 4) The use of physical tests only may be acceptible if: i) Unambiguous results can be stated based upon well-­‐known properties, (and easily separated). ii) Potentially ambiguous results can be resolved by reference to established values from sources (for example, solubility from CRC). iii) Unusual physical behaviors, different between compounds, but characteristics of the compound are explained. (For example; when using a colligative property to find FW, the dissociation factor must be cited.) iv) Properties must be distinguishable to within errors of reasonably performed experiments. Note: Tasting is explicitly excluded as a test technique. 88 1997 #9 An experiment is to be performed to determine the mass percent of sulfate in an unknown soluble sulfate salt. The equipment shown above is available for the experiment. A drying oven is also available. a. Briefly list the steps needed to carry out this experiment. 2 points Mix unknown and BaCl2(aq) as reactants. Collect precipitate / set up filtration b. What experimental data need to be collected to calculate the mass percent of sulfate in the unknown? 2 points Mass of unknown salt as reactant (sulfate="salt"=unknown salt, unless otherwise specified) Mass BaSO4 (must be specified) as dried precipitate/product Note: "Dried" must appear to earn all 4 points for (a) and (b) c. List the calculations necessary to determine the mass percent of sulfate in the unknowm. 2 points Mass BaSO4 à moles SO42¯ˉ à mass SO42¯ˉ (to be used in) à mass SO42¯ˉ / mass unknown Notes: A list alone is acceptable. Method, if correct, acceptable as list. Response must clearly distinguish between SO42¯ˉ, BaSO4, and unknown sulfate. Only one of two points earned if mass SO42¯ˉ incorrect but fraction for percent clearly indicates part (of original salt) / whole (of original salt). d. Would 0.20 M MgCl2 be an acceptable substitute for the BaCl2 solution provided for this experiment? Explain. 2 points MgCl2 is NOT an acceptable substitute for BaCl2. MgCl2 is too soluble. 89 Note: 1 point earned if response indicates MgCl2 is acceptable and reason given is that Mg2+ behaves like Ba2+ to form an insoluble SO42¯ˉ precipitate (response must previously specify BaSO4 as product) Multiple Choice 2013 33. 38. 48. For an experiment, a student needs 100.0 mL of 0.4220 M NaCl. If the student starts with NaCl(s) and distilled water, which of the following pieces of laboratory glassware should the student use to prepare the solution with the greatest accuracy? a. 25 mL volumetric pipet b. 100 mL Erlenmeyer flask c. 100 mL graduated cylinder d. 100 mL volumetric flask e. 1 L beaker The percentage of silver in a solid sample is determined gravimetrically by converting the silver to Ag+(aq) and precipitating it as silver chloride. Failure to do which of the following could cause errors in the analysis? I. Account for the mass of the weighing paper when determining the mass of the sample II. Measure the temperature during the precipitation reaction III. Wash the precipitate IV. Heat the AgCl precipitate to constant mass a. I only b. I and II c. I and IV d. II and III e. I, III, and IV Potassium hydrogen phthalate, KHP, is used as a primary standard for determining the concentration of a solution of NaOH by titration. If the KHP has not been dried before weighing, the calculated molarity of the NaOH would be a. higher than the actual value, since water is included in the apparent mass of KHP b. higher than the actual value, since the presence of water requires a larger volume of titrant c. lower than the actual value, since NaOH absorbs water d. unaffected, since KHP is a strong acid 90 e. unaffected, since water is routinely added before the titration NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) 65. A student is trying to determine the heat of reaction for the acid-­‐base neutralization reaction represented above. The student uses 0.50 M NaOH and 0.50 M HCl solutions. Which of the following situations, by itself, would most likely result in the LEAST error in the calculated value of the heat of reaction? a. The thermometer was incorrectly calibrated and read 0.5 Celsius degree too high during the procedure. b. The volume of the acid solution added to the calorimeter was actually 1.0 mL less than what was recorded. c. The calorimeter was poorly insulated, and some heat escaped to the atmosphere during the procedure. d. The actual molarity of the base solution was 0.53 M but was recorded as 0.50 M. e. The final temperature of the mixture was taken before the contents of the calorimeter had reached thermal equilibrium. 91