A great video concept approach to each Big Idea may be found on BOZEMANSCIENCE at

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1 Big Idea 6: Chemical Equilibrium Problem Set: / 25 pts DUE: Fri. 4/10 Big Idea 6: AP Chemistry is a rigorous course that requires students to use their analytical and problem solving skills as they apply into the 6 Big Idea Divisions the College Board has designated. As such, according to the College Board, in order to receive a 5 on the AP Test a student should have the following understandings and abilities that I will break down further under Key Understandings and Abilities. Chemical equilibrium - is a dynamic, reversible state in which rates of opposing processes are equal. Solubility - Articulates when ranking of solubility does and does not follow the ranking of Ksp. Equilibrium and Thermodynamics - Articulates and cites inferences about the relation between the magnitude of K and the thermodynamic notion of ΔG, indicating favorability. Articulates and cites inferences about the conditions (ΔG and RT) under which K is close to 1. Identifies and explains free energy in complex systems, such as biological systems. Relates at the particulate level about the enthalpic and entropic effects accompanying dissolution of a salt. A great video concept approach to each Big Idea may be found on BOZEMANSCIENCE at Big Idea Sheet: Recall, your Big Idea Sheet will be graded based on the rubric you were provided when your notebook was setup. If you have misplaced it you may of course find a PDF copy on the website at When writing your Big Idea, be sure to refer to the rubric and check it again before you submit is. Big Idea Sheets are due at the end of each Big Idea Unit and may be submitted electronically or on paper. NO EXCEPTIONS OR LATE WORK. Rough drafts are welcome and will be handed/sent back with comments upon if arrangements are made with your teacher before the due date for use of a rough draft. Big IDEA 6 SHEET DUE: Fri. 4/10 Learning Objectives that should be covered on Big Idea Sheet = Key Understandings: Equilibrium - In many classes of reactions, it is important to consider both the forward and reverse reaction. LO 1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. Equilibrium Quotient - The current state of a system undergoing a reversible reaction can be characterized by the extent to which reactants have been converted to products. The relative quantities of reaction components are quantitatively described by the reaction quotient, Q. LO 2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K. Equilibrium Quotient vs. Equilibrium constant - When a system is at equilibrium, all macroscopic variables, such as concentrations, partial pressure, and temperature, do not change over time. Equilibrium results from an equality between the rates of the forward and reverse reactions, at which point Q = K. LO 3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as LeChatelier s principle, to infer the relative rate of the forward and reverse reactions. LO 4 The student can, given a set of initial conditions (concentrations or partial pressure) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached. LO 5 The student can, given data (tabular, graphical, etc ) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K. LO 6 The student can, given a set of initial condition (concentration or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction. LO 7 The student is able, for a reversible reaction that has a large or small K, to determine which chemical species will have very large versus very small concentrations at equilibrium. Le Chatelier s Principle - Systems at equilibrium are responsive to external perturbations, with the response leading to a change in the composition of the system. LO 8 The student is able to use LeChatelier s principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium. and Abilities: LO 9 The student is able to use LeChatelier s principle to design a set of conditions that will optimize a desired outcome, such as product yield. Le Chatelier s Principle and Equilibrium Quotient - A disturbance to a system at equilibrium causes Q to differ from K, thereby taking the system out of the original equilibrium state. The system responds by bringing Q back into agreement with K, thereby establishing a new equilibrium state. LO 10 The student is able to connect LeChatelier s principle to the comparison of Q to K by explaining the effects of the stress on Q and K. Acid-Base Equilibrium - Chemical equilibrium reasoning can be used to describe the proton transfer reactions of acid-base chemistry. LO 11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. LO 12 The student can reason about the distinction between strong and weak acid solutions with similar values of ph, including the percent ionization of the acids, the concentrations needed to achieve the same ph, and the amount of base needed to reach the equivalence point in a titration. LO 13 The student can interpret titration data for monoprotic or polyprotic acids involving titration of a weak or strong acid by a strong base (or a weak or strong base by a strong acid) to determine the concentration of the titrant and the pka for a weak acid, or the pkb for a weak base. LO 14 The student can, based on the dependence of Kw on temperature, reason that neutrality requires [H + ] = [OH - ] as opposed to requiring ph = 7, including especially the applications to biological systems. LO 15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the ph (and concentrations of all chemical species) in the resulting solution. LO 16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the ph and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations. LO 17 The students can, given an arbitrary mixture of weak and strong acids and bases (including polyprotic systems), determine which species will react strongly with one another (i.e., with K > 1) and what species will be present in large concentrations at equilibrium. Exclusion Statement: Numerical computation of the concentration of each species present in the titration curve for polyprotic acids is beyond the scope of this course and the AP Exam.

2 ph - The ph is an important characteristic of aqueous solutions that can be controlled with buffers. Comparing ph to pka allows one to determine the protonation state of a molecule with a labile proton. LO 18 The student can design a buffer solution with a target ph and buffer capacity by selecting appropriate conjugate acid-base pair and estimating the concentrations needed to achieve the desired capacity. LO 19 The student can relate the predominant form of a chemical species involving a labile proton (i.e., protonated/deprotonated form of a weak acid) to the ph of a solution and the pka associated with the labile proton. LO 20 The student can identify a solution as being a buffer solution and explain the buffer mechanism in terms of the reactions that would occur on addition of acid or base. Exclusion Statement: Computing the change in ph resulting from the addition of an acid or base to a buffer is beyond the scope of this course and the AP Exam. Exclusion Statement: The production of the Henderson-Hasselbalch equation by algebraic manipulation of the relevant equilibrium constant expression is beyond the scope of this course and the AP Exam. Le Chatelier s Principle and Spontaniety - External sources of energy can be used to drive change in cases where the Gibbs free energy change is positive. BI 5 LO 15 The student can use LeChatelier s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. BI 5 LO 16 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. College Board References and Resources: I have pasted the formulas/sections of the table of equations and constants that are pertinent to the Big Idea we are currently studying. Meaning that any mathematical procedures, constants, formulas, etc. not noted here, and used for the Big Idea, cannot be referenced during your AP examination. Solubility Product - The solubility of a substance can be understood in terms of chemical equilibrium. LO 21 The student can predict the solubility of a salt, or rank the solubility of salts, given the relevant Ksp values. LO 22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values. LO 23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, ph) that influence the solubility. LO 24 The student can analyze the enthalpic and entropic changes associated with the dissolution of a salt, using particulate level interactions and representations. Exclusion Statement: Computations of solubility as a function of ph are beyond the scope of this course and the AP Exam. Exclusion Statement: Computations of solubility in such solutions are beyond the scope of this course and the AP Exam. K and ΔG - When the difference in Gibbs free energy between reactants and products (ΔG ) is much larger than the thermal energy (RT), the equilibrium constant is either very small (for ΔG > 0) or very large (for ΔG < 0). When ΔG is comparable to the thermal energy (RT), the equilibrium constant is near 1. LO 25 The student is able to express the equilibrium constant in terms of ΔG and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process. Instructions: For the following two reactions, ON A SEPARATE SHEET OF PAPER, a) write the Keq expression in terms of concentration, Kc. b) given the equilibrium concentrations, state whether each equilibrium is product-favored, reactant-favored, or fairly even ([products] [reactants]). c) calculate the value of Kc. NW=NC 1. N2g) + 3 H2(g) 2 NH3(g) 2. HF(aq) H + (aq) + F - (aq) At equilibrium: [N2] = 1.50 M [H2] = 2.00 M [NH3]= 0.01 M At equilibrium: [HF] = 0.55 M [H + ] = M [F - ]= M Instructions: ON A SEPARATE SHEET OF PAPER, answer the following set of questions related to General Equilibria. NW=NC 4. Write equilibrium expressions for each of the following reactions: a) CaCO3(s) CaO(s) + CO2(g) b) Ni(s) + 4CO(g) Ni(CO)4(g) c) 5CO(g) + I2O5(s) I2(g) + 5CO2(g) d) Ca(HCO3)2(aq) CaCO3(s) + H2O(l) + CO2(g) e) AgCl(s) Ag + (aq) + Cl - (aq) 5. Write the equilibrium expression in terms of partial pressures (Kp) for each of the following reactions. (a) 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(g) K p = 1 x atm (b) N 2 (g) + O 2 (g) 2NO(g) K p = 5 x atm (c) 2HF(g) H 2 (g) + F 2 (g) K p = 1 x atm (d) 2NOCl(g) 2NO(g) + Cl 2 (g) K p = 4.7 x 10-4 atm (e) Rate the reactions in order of their increasing tendency to proceed toward completion: 6. Consider an equilibrium that occurs in two steps: H2S(aq) H + (aq) + HS - (aq) HS - (aq) H + (aq) + S 2- (aq) (a) Write the overall reaction. (b) How do the Kc s for the two steps (Kc1 & Kc2) relate to the Kc of the overall reaction (Kc)?

3 Instructions: ON A SEPARATE SHEET OF PAPER, answer the following set of questions related to RICE tables and Equilbira Problems. NW=NC 7. For the reaction: N2(g) + 3H2(g) 2NH3(g) The initial [N2] = 0.32 M and the initial [H2] = 0.66 M. At a certain temperature and pressure the equilibrium [H2] is found to be 0.30 M. What is Keq under these circumstances? 8. Suppose that 2.00 moles of HCl in a 1.00L glass flask slowly decomposes into H2 and Cl2. When equilibrium is reached, the concentrations of H2 and Cl2 are both M. What is the Keq? 9. Consider the equilibrium: 2N2O(g) + O2(g) 4NO(g) 3.00 moles of NO(g) are introduced into a 1.00-Liter evacuated flask. When the system comes to equilibrium, 1.00 mole of N2O(g) has formed. Determine the equilibrium concentrations of each substance. Calculate the Kc for the reaction based on these data. 10. At some temperature, Keq = 33 for the reaction H2 + I2 2HI. If initially, [H2] =.0600 M and [I2] =.0300 M, what are all three equilibrium concentrations? 11. Graphite (solid carbon) and carbon dioxide are kept at constant pressure at 1000 K until the following reaction reaches equilibrium. C(s) + CO2(g) 2CO(g) If Keq = 0.021, calculate the equilibrium concentration of CO if the concentration of CO2 was initially M. Instructions: Clearly circle the response that best satisfies the prompt giving only one answer to each question. DO NOT USE A CALCULATOR. The College Board does not allow the use of a calculator for Section I (the MC part of the exam, so it is time to start practicing without one. You may use ONLY a periodic table and the equations. While you should practice working as fast as possible, it is more important at this point in the course that you practice without a calculator, even if it slows you down. Look for the easy math common factors and rough estimation do not do long division to try to get exact values. Remember it is MC; use the answers to guide you. 12. When the substances in the equation below are at equilibrium at pressure P and temperature T, the equilibrium can be shifted to favor the products by: MgO(s) + H2(g) Mg(s) + H2O(g) ΔH = 14 kj (a) increasing the pressure in the reaction vessel while keeping the temperature constant (b) increasing the pressure by adding an inert gas such as argon (c) decreasing the temperature (d) allowing some hydrogen gas to escape at constant P and T (e) adding a catalyst A(g) + B(g) 2C(g) When 0.60 mole of A and 0.75 mole of B are placed in an evacuated 1.00 L flask, the reaction represented above occurs. After the reactants and the product reach equilibrium and the initial temperature is restored, the flask is found to contain 0.30 mole of product (C) Based on these results, the equilibrium constant, Kc for the reaction is: (a) 0.60 (b) 0.90 (c) 1.7 (d) 3.4 (e) What is the balanced net ionic equation that occurs when an excess of ammonia gas is bubbled through a solution saturated with silver chloride? (a) Ag + + Cl + NH4 + + OH NH4Cl + AgOH (b) Ag + + 2H2O + 2NH3 2NH4 + + Ag(OH)2 (c) Ag + + H2O + NH3 NH4 + + AgOH (d) AgCl + 2NH3 Cl + Ag(NH3)2 + (e) 2Ag + + H2O + 2NH3 2NH4 + + Ag2O 15. What is the molar solubility in water of PbI2? (The Ksp for PbI2 is 3.2 x 10-8 ) (a) 3.2 x 10-8 M (b) 8.0 x 10-8 M (c) (1.6)½ x 10-8 M (d) (1.6)⅓ x 10-8 M (e) 2 x 10-3 M 16. Which of the following systems would NOT experience a change in the number of moles of the substances present at equilibrium when the volume of the system is changed at constant temperature? (a) SO(g) + NO(g) SO2(g) + ½N2(g) (b) O2(g) + 2H2(g) 2H2O(g) (c) N2(g) + 2O2(g) 2NO2(g) (d) N2O4(g) 2NO2(g) (e) CH4(g) + 2O2(g) CO=(g) + 2H2O(g) 17. For the reaction 2W(g) 2X(g) + Y(g), the equilibrium constant, Kp, is 8 x 10 3 at 298 K. A mixture of the three gases at 298 is placed in a rigid metal cylinder, and the initial pressures are Px = 1 atm, Py = 0.8 atm, and Pw = 2 atm. At the instant of mixing, which of the following is true? (a) More product will form. (b) More reactant will form. (c) ΔS = 0 (d) ΔGº = 0 (e) ΔGº > 0 CaCO3(s) CaO(s) + CO2(g) 18. After the equilibrium represented above is established, some pure CO2(g) is added to the reaction vessel at constant temperature After equilibrium is reestablished, all of the following will happen EXCEPT (a) Kp for the reaction will increase (b) Kc for the reaction will remain the same (c) The concentration of CaCO3 in the reaction vessel remains constant. (d) The concentration of CaO in the reaction vessel remains constant. (e) ΔG will become zero. N2(g) + 3 H2(g) 2 NH3(g) ΔH = 92 KJ 19. Which of the following changes alone would cause a decrease in the value of Keq for the reaction represented above? (a) Removing NH3 (b) Increasing the temperature (c) Adding an inert gas such as argon (d) Adding more hydrogen gas at constant temperature (e) Adding a catalyst 2SO2(g) + O2(g) 2SO3(g) ΔH = 198 kj 20. Consider the equilibrium above. Which of the following changes will increase the concentration of SO2(g)? (a) Increasing the concentration of O2(g) (b) Increasing the pressure in the reactant vessel at constant temperature (c) Increasing the mass of SO3 present (d) Adding a catalyst (e) Decreasing the temperature 21. Consider the reaction system, NiO(s) + H2(g) Ni(s) + H2O(g) The equilibrium constant expression is: (a) Keq = [(Ni)(H2O)] / [(NiO)(H2)] (b) Keq = (Ni) / [(NiO)(H2)] (c) Keq = 1 / [(NiO)(H2)] (d) Keq = 1 / (H2) (e) Keq = (H2O) / (H2) 22. Given the equilibrium, 2SO2(g) + O2(g) 2SO3(g) If this equilibrium is established by beginning with equal number of moles of SO2 and O2 in an empty 1.0 L bulb, then the following must be true at equilibrium: (a) [SO2] = [SO3] (b) 2[SO2] = 2[SO3] (c) [SO2] = [O2] (d) [SO2] < [O2] (e) [SO2] > [O2] 23. Consider: N2O4(g) 2NO2(g) At 25 C, in a 1.0 L container 0.11 mole of N2O4 reacts to form 0.10 mol of N2O4 and 0.02 mole of NO2 at equilibrium. At 90 C, 0.11 mole of N2O4 results in mole of N2O4 and 0.12 mole of NO2 at equilibrium. From these data we can conclude (a) Equilibrium constants, in general, are always larger at higher temperatures. (b) N2O4 molecules react by a first order rate law. (c) the reaction is endothermic. (d) NO2 molecules react twice as fast as N2O4 molecules at 90 C (e) the equilibrium constant for the reaction above decreases with an increase in temperature. Questions refer to the following: At a given temperature, mole NO, mol Cl2 and mol ClNO were placed in a 10.0 Liter container. The following equilibrium is established: 2ClNO(g) 2NO(g) + Cl2(g) At equilibrium, mol of ClNO was present 24. The number of moles of Cl2 present at equilibrium is (a) (b) (c) (d) (e) 0.110

4 25. The equilibrium constant, Kc, is closest to: (a) 10-6 (b) 10-3 (c) 1 (d) 10 3 (e) At 985 C, the equilibrium constant for the reaction, H2(g) + CO2(g) H2O(g) + CO(g) is 1.5. What is the equilibrium constant for the reaction below? H2O(g) + CO(g) H2(g) + CO2(g) (a) -1.5 (b) 0.75 (c) 3.0 (d) 0.67 (e) There is no way of knowing what it would be 27. What is the relationship between Kp and Kc for the reaction, 2ICl(g) I2(g) + Cl2(g) (a) Kp = Kc (RT) 1 (b) Kp = Kc (RT) (c) Kp = Kc (RT) 2 (d) Kp = Kc (e) Kp = Kc (2RT) 28. What is the relationship between Kp and Kc for the reaction, 3A(g) 2B(g) + C(s) (a) Kp = Kc (RT) 1 (b) Kp = Kc (RT) (c) Kp = Kc (RT) 2 (d) Kp = Kc (e) Kp = Kc (2RT) 29. For the reaction system, N2(g) + 3 H2(g) 2NH3(g) + heat the conditions that would favor maximum conversion of the reactants to products would be (a) high temperature and high pressure (b) high temperature, pressure unimportant (c) high temperature and low pressure (d) low temperature and high pressure (e) low temperature and low pressure 30. Solid HgO, liquid Hg, and gaseous O2 are placed in a glass bulb and are allowed to reach equilibrium at a given temperature. 2HgO(s) 2Hg(L) + O2(g) ΔH = kcal The mass of HgO in the bulb could be increased by (a) adding more Hg. (b) removing some O2. (c) reducing the volume of the bulb. (d) increasing the temperature. (e) removing some Hg. 42. For the equilibrium system: H2O(g) + CO(g) H2(g) + CO2(g) ΔH = -42 kj/mol Kc equals 0.62 at 1260 K. If 0.10 mole each of H2O, CO, H2 and CO2 (each at 1260 K) were placed in a 1.0-Liter flask at 1260 K, when the system came to equilibrium. The temperature would. The mass of CO would. (a) decrease; increase (b) decrease; decrease (c) remain constant; increase (d) increase; decrease (e) increase; increase 43. Consider the reaction SO2(g) + ½O2(g) SO3(g) Kc = 49 at 1000 K What is the value of Kc for the reaction below? 2SO3(g) 2SO2(g) + O2(g) (a) 1 / 49 (b) 1 / 7 (c) 1 / (49) 2 (d) 7 (e) (49) For which reaction is Kc equal to Kp? (a) H2(g) + S(s) H2S(g) (b) 2H2O(g) 2H2(g) + O2(g) (c) 3H2(g) + N2(g) 2NH3(g) (d) H2(g) + Br2(L) 2HBr(g) (e) 2NO2(g) N2O4(g) heat + N2O4(g) 2NO2(g) Questions 46-48, consider an equilibrium system based on the reaction above This equilibrium mixture is contained in a piston. 46. Which occurs when the volume of the system is increased at constant temperature? #molecules total # of molecules of NO2 of all gases Kp (a) increases increases remains same (b) increases decreases remains same (c) increases remains same remains same (d) remains same decreases decreases (e) remains same increases decreases 47. Which occurs when a catalyst is added? partial pressure total pressure of N2O4 of all gases Kp: (a) decreases increases remains same (b) increases decreases remains same (c) stays the same stays the same remains same (d) increases decreases decreases (e) increases increases decreases 48. Which occurs when the temperature is increased at constant volume? #molecules total # of molecules of N2O4 of all gases Kp: (a) increases decreases remains same (b) decreases increases remains same (c) increases decreases increases (d) decreases decreases increases (e) decreases increases increases 49. 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g) + energy Which of the following changes to the system at equilibrium shown above would cause the concentration of H2O to increase? (a) The volume of the system was decreased at constant temperature. (b) The temperature of the system was increased at constant volume (c) NH3 was removed from the system. (d) N2 was removed from the system. (e) O2 was removed from the system. 4H2(g) + CS2(g) CH4(g) + 2H2S(g) A mixture of 2.50 mol H2(g), 1.50 mol CS2(g), 1.5 mol CH4(g) and 2.00 mol H2S(g) is placed in a 5.0 L rigid reaction vessel, and the system reaches equilibrium according to the equation above. When the equilibrium is achieved, the concentration of CH4(g) has become 0.25 mol L 1. Questions refer to this equilibrium system. 50. Changes in concentration occur as this system approaches equilibrium. Which expression gives the best comparison of the changes in those concentrations shown in the ratio to the right? (a) (c) (e) (b) (d) Δ[H2S] Δ[CS2] 51. What is the change in the number of moles of H2S(g) present as the system moves from its original state to the equilibrium described? (a) (b) 1.25 (c) (d) (e) What is the number of moles of CS2(g) at equilibrium? (a) 0.25 (b) 0.35 (c) 0.75 (d) 1.25 (e) What is the concentration in moles per liter of H2(g) at equilibrium? (a) 0.50 (b) 0.70 (c) 1.0 (d) 2.0 (e) Which correctly describes the values for ΔG, the free energy change, and Q, the reaction quotient, when the mixture was prepared. (a) ΔG = 0, Q = Keq (b) ΔG > 0, Q < Keq (c) ΔG > 0, Q > Keq (d) ΔG < 0, Q > Keq (e) ΔG < 0, Q < Keq N2(g) + 3H2(g) 2NH3(g) + 92 kj Questions refer to this equilibrium system based on the reversible reaction given above. 55. When the temperature of such an equilibrium system is increased at constant volume, which property is least affected? (a) density (b) pressure (c) conc. NH3(g) (d) conc. of N2(g) (e) average kinetic energy 56. How are the rates of the opposing reactions and the value of the equilibrium constant, Kp, affected when heat is added to this system? (a) The reaction rates remain unchanged and Kp decreases. (b) The forward reaction will be favored and Kp decreases. (c) The forward reaction will be favored and Kp increases. (d) The reverse reaction will be favored and Kp decreases. (e) The reverse reaction will be favored and Kp increases. 57. Which observation confirms the fact that equilibrium has been reached in such a system confined in a closed, rigid container? (a) The density remains constant. (b) The odor of ammonia can first be detected. (c) The pressure is decreasing at a constant rate (d) The partial pressure of hydrogen remains constant. (e) The mass of the system has decreased to a constant value. 58. Which occurs when such an equilibrium system is subjected to a stress by the addition of H2(g) and the system proceeds to a new equilibrium? i. Heat will be released. ii. Some of the added H2(g) will be consumed. iii. Some of the N2(g) present originally will be consumed. (a) ii only (b) iii only (c) i and iii only (d) ii and iii only (e) i, ii, and iii

5 Instructions: Clearly circle the response that best satisfies the prompt giving only one answer to each question. DO NOT USE A CALCULATOR. The College Board does not allow the use of a calculator for Section I (the MC part of the exam, so it is time to start practicing without one. You may use ONLY a periodic table and the equations. While you should practice working as fast as possible, it is more important at this point in the course that you practice without a calculator, even if it slows you down. Look for the easy math common factors and rough estimation do not do long division to try to get exact values. Remember it is MC; use the answers to guide you. 59. Temperature is often given with Ksp values because: (a) the solubility of solids always increases with increasing temperature (b) the solubility of solids varies with temperature changes. (c) solubility changes with temperature but Ksp values do not. (d) Ksp varies with temperature even though concentrations do not. (e) the number of ions varies with the kind of salt that is dissolving. 60. The Ksp expression for silver phosphate is (a) Ksp = [Ag + ][PO4 3 ] (b) Ksp=[Ag + ] 2 [PO4 3 ] (c) Ksp = [Ag + ] 3 [PO4 3 ] (d) Ksp=[Ag + ][PO4 3 ] 3 (e) Ksp = [Ag + ][PO4 3 ] [Ag3PO4] 61. The solubility of nickel(ii) hydroxide, Ksp = , is about (a) (b) (c) (e) (d) 62. What is the molar solubility in water of PbI2? (The Ksp for PbI2 is ) (a) M (b) M (c) (1.6 )½ 10 8 M (d) (1.6) ⅓ 10 8 M (e) M 63. Determine the solubility of lead(ii) fluoride, Ksp = , in a M lead(ii) nitrate solution. (a) M (b) M (c) M (d) M (e) M 64. The solubility of silver sulfide is M. Determine the Ksp of this salt. (a) (b) [ ] [ ] (c) [ ] 2 [ ] (d) [ ] [ ] (e) [ ] 2 [ ] 65. Equal volumes of M KCl and M AgNO3 are mixed the Ksp for silver chloride is As these two solutions are combined, (a) a precipitate of AgCl forms. (b) there is no precipitate formed (c) NaCl will precipitate (d) AgNO3 will precipitate (e) the [Na + ] will become M 66. The solubility of metal(ii) sulfide, Ksp = , is about (a) (b) 10 4 (c) (d) (e) Calculate the ph of a saturated metal hydroxide, XOH, whose Ksp = (a) 4.0 (b) 6.0 (c) 7.0 (d) 8.0 (e) Calculate the Ksp of a saturated metal hydroxide, X(OH)2, solution whose ph = (a) (b) (c) (d) (e) HI(g) + Cl2(g) 2HCl(g) + I2(g) + energy A gaseous reaction occurs and comes to equilibrium as shown above. Which of the following changes to the system will serve to increase the number of moles of I2 present at equilibrium? (a) Increasing the volume at constant temperature (b) Decreasing the volume at constant temperature (c) Adding a mole of inert gas at constant volume (d) Increasing the temperature at constant volume (e) Decreasing the temperature at constant volume 70. In the reaction shown below, 0.50 mole of Br2 and 0.50 moles of I2 are placed in an evacuated 1.00 liter vessel and allowed to reach equilibrium. What is the value of Kc if the vessel contains 0.84 moles of IBr at equilibrium? I2(g) + Br2(g) 2IBr(g) (a) 2.0 (b) 8.8 (c) 11.0 (d) 110 (e) Which is the correct set up to determine Kc for the reaction below? 4CuO(s) + CH4(g) CO2(g) + 4Cu(s) + 2H2O(g) (a) [CO2 ][Cu] 4 [H2O] 2 (b) [CuO] 4 [CH4] _ [CuO] 4 [CH4 ] [CO2 ][Cu] 4 [H2O] 2 (c) [CO2 ] [H2O] 2 (d) [CO2][H2O] 2 _ [CH4 ] [CuO] 4 [CH4 ] (e) [CO2 ] 2 [Cu] 4 [H2O] 2 [CuO] 4 [CH4 ] 72. An evacuated 1.00-liter vessel is injected with moles of sulfur trioxide gas, SO3. The vessel is then heated to a high temperature where the SO3 partially decomposes to form the products SO2 and O2 in the reaction shown below: 2SO3(g) 2SO2(g) + O2(g) The temperature is kept constant, and the amount of SO3 in the vessel at equilibrium is mol. What is the value of Kc at this temperature? (a) (b) (c) (d) (e) A sample of solid potassium nitrate is placed in water. The solid potassium nitrate comes to equilibrium with its dissolved ions by the endothermic process shown below. KNO3(s) + energy K + (aq) + NO3 (aq) Which of the following changes to the system would increase the concentration of K + ions at equilibrium? (a) The volume of the solution is increased (b) The volume of the solution is decreased (c) Additional solid KNO3 is added to the solution. (d) The temperature of the solution is increased (e) The temperature of the solution is decreased 74. A 3.00-liter reaction vessel is filled with carbon monoxide gas, CO, and chlorine gas, Cl2. The mixture is heated to 670 K and allowed to reach equilibrium according to the balanced equation shown below: CO(g) + Cl2(g) COCl2(g) At equilibrium, the mixture contains moles CO, moles Cl2, and 1.11 moles COCl2. What is Kc at this temperature? (a) (b) (c) 12.0 (d) 411 (e) The reaction below is allowed to come to equilibrium at room temperature. At equilibrium, the partial pressure of H2O is 296 mm Hg, Cl2O is 15 mm Hg, and HOCl is 20 mm Hg. What is the value of Kp at this temperature? H2O(g) + Cl2O(g) 2HOCl(g) (a) 222 (b) 11 (c) (d) (e) For the reaction shown below, Kp = at 400 K. 2NH3(g) N2(g) + 3H2(g) What is the value of Kc at this temperature? (a) (b) (c) (d) (e) At 373 K, the reaction shown below has an equilibrium constant, Kc = COCl2(g) CO(g) + Cl2(g) After placing a mixture of gases in the reaction vessel, the concentrations were measured to be [COCl2] = M, [CO] = M, and [Cl2] = M. Which statement below accurately describes the reaction? (a) The reaction is at equilibrium (b) The reaction is not at equilibrium, and it is proceeding to the left (c) The reaction is not at equilibrium, and it is proceeding to the right (d) The reaction quotient is equal to Kc (e) The reaction quotient is less than Kc

6 78. In the reaction 3W + X 2Y + Z, all substances are gases. The reaction is initiated by adding equal number of moles of W and of X. When equilibrium is reached, (a) [Y] = [Z] (b) [X] = [Y] (c) [W] = [X] (d) [X] > [W] (e) [W] + [X] = [Y] + [Z] 79. Consider the reactions below, both at the same temperature: Rx I X Y K = 1 x 108 Rx II Z Y K = 1 x 105 (a) I is 3 times faster than II. (b) I is 1000 times faster than II. (c) II is 3 times faster than I. (d) II is 1000 times faster than I. (e) The size of K and the time required to reach equilibrium are not directly related 80. The reaction 3H2(g) + N2(g) 2NH3(g) has an enthalpy of change of 92 kj. Increasing the temperature of this equilibrium system causes (a) an increase in [NH3] (b) an increase in [N2] (c) a decrease in [H2] (d) an increase in K (e) a decrease in pressure at constant volume 81. Consider N2(g) + O2(g) 2NO(g). The reaction was initiated by adding 15.0 moles of NO to a 1.0-L flask. At equilibrium, 3.0 moles of oxygen are present in the 1.0-L flask. The value of K must be (a) 0.33 (b) 3.0 (c) 5.0 (d) 9.0 (e) At a certain temperature, the synthesis of ammonia gas from nitrogen and hydrogen gases, shown below, has a value for K of If [H2] = [N2] = 0.10 M and [NH3] = 0.20 M N2 + 3H2 2NH3 (a) the reaction would shift toward the ammonia (b) the reaction would shift toward the N2 and the H2 (c) the system is at equilibrium, therefore no shifting will occur (d) the reaction will shift toward a new equilibrium position, but the direction cannot be determined from these data (e) the equilibrium may shift but it is not possible to calculate Q without knowing the temperature 82. The equilibrium constant, K, may be used to determine the K of other reactions. I. When the equation is reversed the reciprocal of the original K becomes the value of K for the new equation. II. When the equation is doubled the square root of the original K becomes the value for the new equation. III. When the equation for a reaction is tripled the equilibrium expression for the new equation is triple the original K. Of the above three statements, those which are always true are (a) I only (b) II only (c) III only (d) I and III (e) I, II, and III 83. The equilibrium P4(g) + 6Cl2(g) 4PCl3(L) is established at 10 C. The equilibrium constant expression is (a) _[PCl3 ]_ (b) [PCl3] 4 _ [P4][Cl2 ] [P4 ][Cl2] 6 (c) [P4 ] [Cl2] 6 (d)_ 1 [PCl3 ] 4 [PCl3 ] 4 (e)_ 1 [P4 ][Cl2] Ammonium hydrogen sulfide will decompose into ammonia gas and hydrogen sulfide gas when heated. Consider the equilibrium system NH4HS(s) NH3(g) + H2S(g) which is developed from mole of NH4HS in a 100-L cylinder. At equilibrium the total pressure is found to be atm. Kp will be equal to (a) (b) (c) (d) 4.00 (e) a value impossible to calculate from only these data 85. At a certain temperature, it has been determined that K = 8.0 for the reaction below. If we also determine that the equilibrium mixture contains 0.80 mole of H2O(g), mole of CO2(g), and mole of CO(g), in an 8.0-L flask, what must be the number of moles of H2(g) at equilibrium? H2O(g) + CO(g) CO2(g) + H2(g) (a) mole (b) 0.80 mole (c) 6.4 moles (d) 8.0 moles (e) 64 moles 86. Which of the following systems at equilibrium are not affected by a change in pressure caused by changing the volume at constant temperature? (a) H2(g) + Cl2(g) 2HCl(g) (b) H2(g) + I2(s) 2HI(g) (c) N2(g) + 3H2(g) 2NH3(g) (d) 2NH3(g) N2(g) + 3H2(g) (e) 3O2(g) 2O3(g) 87. After the equilibrium represented below is established, some pure O2(g) is injected into the reaction vessel at constant temperature After equilibrium is reestablished, which of the following has a lower value compared to its value at the original equilibrium? 2CO2(g) 2CO(g) + O2(g) (a) Keq for the reaction (b) The total pressure in the reaction vessel (c) The amount of CO2(g) in the reaction vessel (d) The amount of O2(g) in the reaction vessel (e) The amount of CO(g) in the reaction vessel 88. Which of the following changes alone would cause a decrease in the value of Keq for the reaction represented below? 2SO2(g) + O2(g) 2SO3(g) ΔH 0 = -197 kj (a) Decreasing the temperature (b) Increasing the temperature (c) Decreasing the volume of the reaction vessel (d) Increasing the volume of the reaction vessel (e) Adding a catalyst 89. In which of the following systems would the number of moles of the substances present at equilibrium NOT be shifted by a change in the volume of the system at a constant temperature? (a) 2SO2(g) + O2(g) 2SO3(g) (b) N2(g) + 3H2(g) 2NH3(g) (c) NO2(g) + SO2(g) SO3(g) + NO(g) (d) N2O4(g) 2NO2(g) (e) CO(g) + 3H2(g) CH4(g) + H2O(g) 90. For: N2(g) + 3H2(g) 2NH3(g) + energy Some N2 and H2 are mixed in a container at 200 C, and the system reaches equilibrium according to the equation above Which of the following causes an increase in the number of moles of NH3 present at equilibrium? I. Decreasing the volume of the container II. Raising the temperature III. Adding a mole of Ar gas at constant volume (a) I only (b) II only (c) I and III only (d) II and III only (e) I, II, and III

7 Instructions: Write the generic equation that symbolized the dissociation of acetic acid an ICE box. Acetic acid, Ka = Ka = [ ][ ] [ ] 91. Calculate the [H+] for a 0.10 M acetic acid solution. Then calculate % ionization 1.8 x 10 5 = [ ][ ] I [ ] x = [ H + ] = C E % ionization = ph = 92. Calculate the [H + ] for a M acetic acid solution. Then calculate % ionization I C E 1.8 x 10 5 = [ ][ ] [ ] x = [ H + ] = % ionization = ph = 93. Calculate the [H + ] for a M acetic acid solution. Then calculate % ionization I C E 1.8 x 10 5 = [ ][ ] [ ] x = [ H + ] = % ionization = ph = 94. Calculate the [H + ] for a M acetic acid solution. Then calculate % ionization I C E 1.8 x 10 5 = [ ][ ] [ ] x = [ H + ] = % ionization = ph = 95. What do these calculations tell us about percent ionization as it relates to the concentration of a weak acid? Instructions: ON A SEPARATE SHEET OF PAPER, answer the following set of questions related to Buffers. NW=NC 96. Calculate the ph of the solution that results from adding 5 drops (~0.25 ml) of 1.0 M HCl to 50 ml of water. 97. Calculate the ph of the solution that results from adding 5 drops (~0.25 ml) of 1.0 M NaOH to 50 ml of water. 98. Write a reaction to represent the dissociation of aqueous acetic acid, HC2H3O2. Write the equilibrium expression. a. Calculate the ph a 125 ml of vinegar, which is 0.80 M acetic acid, Ka = b. Calculate the ph of 50.0 ml of 0.80 M acetic acid to which 1.64 g of sodium acetate (MM 82 g/mol) has been added. (Assume no volume change occurs when the solid is added.) c. Calculate the ph of 50.0 ml of 0.80 M acetic acid to which 3.28 g of sodium acetate has been added. (Assume no volume change.) d. Calculate the ph of the resulting solution when 1.0 ml of 1.0 M HCl is added to the solution described in part c. e. Calculate the ph of the resulting solution when 2.0 ml of 1.0 M NaOH is added to the solution described in part c. 99. Write a reaction to represent the ionization of hypochlorous acid. Write the equilibrium expression. a. Calculate the ph of 0.55 M hypochlorous acid. (Ka = ) b. Calculate the ph of ml of 0.55 M hypochlorous acid to which 3.49 g of potassium hypochlorite (MM g/mol) has been added. (Assume no volume change occurs when the solid is added.) c. Calculate the mass of potassium hypochlorite that must be added to ml of 0.55 M hypochlorous acid to produce a solution with a ph of 7.00 (Assume no volume change occurs when the solid is added.) 100. Write a reaction to represent the hydrolysis of methylamine. (CH3NH2, Kb = ) Write the equilibrium expression. Then write the reaction of the hydrolysis that occurs when methylamide chloride (CH3NH3Cl). Write the equilibrium expression and calculate the Ka for this reaction. a. Calculate the ph of 75.0 ml of M aqueous methylamine. b. Calculate the ph of 0.25 g of methylamide chloride (MM g/mol) dissolved to make 75.0 ml of solution. c. Calculate the ph of 75.0 ml of M aqueous methylamine to which 0.25 g of methylamide chloride has been added. (Assume no volume change occurs when the solid is added.) d. Calculate the ph of methylamine / methylamide chloride buffer that contains equal number of moles of both the weak base and its conjugate weak acid. e. Calculate the ph of the solution that results when 1.5 ml of 1.0 M HCl is added to the solution described in part c Write a reaction to represent the dissociation of aqueous ascorbic acid, HC6H7O6. Write the equilibrium expression. a. Calculate the ph a 225 ml of an aqueous solution of 0.50 M ascorbic acid, Ka = 8.0 x b. Calculate the ph of ml of 0.50 M ascorbic acid to which 1.1 g of potassium ascorbate, KC6H7O6 (MM g/mol) has been added. (Assume no volume change occurs when the solid is added.) c. Calculate the mass of potassium ascorbate that must be added to 0.50 M ascorbic acid when making a solution with a volume of ml to generate a ph of 4.10 d. Calculate the ph of the resulting solution when 1.0 ml of 1.0 M HCl is added to the solution described in part c. e. Calculate the ph of the resulting solution when 1.0 ml of 1.0 M NaOH is added to the solution described in part c. Instructions: ON A SEPARATE SHEET OF PAPER, do the following Acid Base Titration (SB + SA) question considering the changes (ph, [H + ], [OH ] ) when titrating 50.0 ml of M nitric acid with M sodium hydroxide. NW=NC 102. Is nitric acid a strong or weak acid? Write a reaction that represents its ionization in water Is sodium hydroxide a strong or weak base? Write a reaction that represents its ionization in water Determine the initial ph of the 50.0 ml of M nitric acid solution Determine the initial ph of the M sodium hydroxide solution What volume of sodium hydroxide would you need to add to the 50.0 ml of M nitric acid if you wanted to completely neutralize this acid? 107. What would be the ph of this resulting solution from the previous question? 108. Calculate the ph of the resulting solution when 5.0 ml of the M sodium hydroxide is added to 50.0 ml of M nitric acid Calculate the ph of the resulting solution when 40.0 ml of the M sodium hydroxide is added to 50.0 ml of M nitric acid Calculate the ph of the resulting solution when 75.0 ml of the M sodium hydroxide is added to 50.0 ml of M nitric acid Calculate the ph of the resulting solution when 90.0 ml of the M sodium hydroxide is added to 50.0 ml of M nitric acid.

8 112. Make a sketch of the titration curve using the volumes and ph values calculated for the titration of 30 ml of 0.00 M sodium hydroxide with M hydrochloric acid right. As you construct this graph, you will get a much more accurate sense of the titration curve if you make two more calculations and place two more points on the graph Calculate the ph of the resulting solution when 23.0 ml of the M hydrochloric acid is added to 30.0 ml of M potassium hydroxide Calculate the ph of the resulting solution when 25.0 ml of the M hydrochloric acid is added to 30.0 ml of M potassium hydroxide. Instructions: ON A SEPARATE SHEET OF PAPER, do the following Acid Base Titration (WA + SB) question considering the changes (ph, [H + ], [OH ] ). Consider a M solution of ammonia, NH3 to which you plan to add M hydroiodic acid, HI. The Kb of ammonia is NW=NC Instructions: ON A SEPARATE SHEET OF PAPER, do the following Acid Base Titration (WA + SB) question considering the changes (ph, [H + ], [OH ] ) when titrating 50.0 ml of 0.20 M solution of lactic acid, HCH5O3 with 0.40 M potassium hydroxide. The Ka of lactic acid = NW=NC 115. Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HC3H5O What volume of 0.40 M KOH would be needed to neutralize this 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3? (This should be the first step you would take to solve any titration problem, even if you weren t asked to do it.) 117. Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with 10.0 ml of 0.40 M KOH? 118. Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 12.5 ml of 0.40 M KOH? 119. Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 20.0 ml of 0.40 M KOH? 120. Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 25.0 ml of 0.40 M KOH? 121. Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 30.0 ml of 0.40 M KOH? 122. Make a sketch of the titration curve using the volumes and ph values calculated for the titration of 50 ml of 0.20 M lactic acid with 0.40 M KOH. Placing a point on the graph for problems and # s 123 & 124 below. As you construct this graph, you will get a much more accurate sense of the curve if you make two more calculations and place two more points on the graph Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 24.0 ml of 0.40 M KOH? 124. Calculate the ph of 50.0 ml of a 0.20 M solution of lactic acid, HCH5O3 after it has been titrated with a total of 26.0 ml of 0.40 M KOH? 125. Calculate the ph of 20.0 ml of a M solution of ammonia, NH What volume of M HI would be needed to neutralize this 20.0 ml of a M solution of ammonia, NH3? 127. Calculate the ph of 20.0 ml of M ammonia, NH3 after it has been titrated with a total of 5.0 ml of M hydroiodic acid, HI? 128. Calculate the ph of 20.0 ml of M ammonia, NH3 after it has been titrated with a total of 15.0 ml of M hydroiodic acid, HI? 129. Calculate the ph of 20.0 ml of M ammonia, NH3 after it has been titrated with a total of 20.0 ml of M hydroiodic acid, HI? 130. Calculate the ph of 20.0 ml of M ammonia, NH3 after it has been titrated with a total of 30.0 ml of M hydroiodic acid, HI? 131. Calculate the ph of 20.0 ml of M ammonia, NH3 after it has been titrated with a total of 50.0 ml of M hydroiodic acid, HI? Make a sketch of the titration curve using the volumes and ph values calculated for the titration of 20 ml of M ammonia with M hydroiodic acid. As you construct this graph, you will get a much more accurate sense of the curve if you make two more calculations and place two more points on the graph Calculate the ph of 20.0 ml of M ammonia, NH3 after it has been titrated with a total of 25.0 ml of M hydroiodic acid, HI? 133. Calculate the ph of 20.0 ml of M ammonia, NH3 after it has been titrated with a total of 35.0 ml of M hydroiodic acid, HI?

9 Instructions: Sketch Titration Curves given the following information. SHOW ALL WORK ON A SEPARATE SHEET OF PAPER. NW=NC 134. Sketch the curve that represents the titration of 40.0 ml of M HCl (in the flask) with M NaOH. Choose a few key points to calculate; starting ph, volume to reach equivalence point, ph at 50 ml) 137. Sketch the curve that represents the titration of 40.0 ml of M nitrous acid, HNO2 (in the flask) with M NaOH. Ka of HNO2 = (Calculate a few key points; starting ph, ph and volume at equivalence point, halfway, after 50 ml) 135. On the same graph, sketch the curve that represents the titration of 40.0 ml of M HCl (in the flask) with M Ba(OH)2 (Choose a few key points to calculate; starting ph, volume to reach equivalence point, ph at 50 ml) 138. Sketch the curve that represents the titration of 40.0 ml of M hydrocyanic acid, HCN with M NaOH. Ka of HOCl = (Calculate a few key points; starting ph, ph and volume at equivalence point, halfway, after 50 ml) 136. Sketch the curve that represents the titration of 20.0 ml of M NaOH (in the flask) with M HNO3 (Choose just a few key points to calculate; starting ph, volume to reach equivalence point, ph at 50 ml) 139. Sketch the curve that represents the titration of 30.0 ml of M ammonia, NH3 (in the flask) with M HCl. Kb of NH3 = (Calculate a few key points; starting ph, ph and volume at equivalence point, halfway, after 50 ml) 140. Sketch the curve that represents the titration of 30.0 ml of M pyridine, C5H5N (in the flask) with M HNO3. Kb of C5H5N = (Calculate a few key points; starting ph, ph and volume at equivalence point, halfway, after 50 ml)

10 Instructions: ON A SEPARATE SHEET OF PAPER answer the following set of questions related to putting all of this together; Acid-Base Equilbria, Salts, and Buffers. NW=NC 141. Calculate the ph of a solution made by pouring 5.0 ml of 0.20 M HCl into 100. ml of water Calculate the ph of a solution made by pouring 10.0 ml of 0.20 M NaOH into 100. ml of water Calculate the ph of 100. ml of 0.20 M nitrous acid, HNO2 (Ka = ) Calculate the ph of 100. ml of 0.20 M ammonia, NH3 (Kb = ) Calculate the ph of a solution made by dissolving 1.38 g of NaNO2 (MM = 69.0 g/mol) into enough water to produce 100. ml of solution Calculate the resulting ph after adding 10.0 ml of 0.20 M NaOH into the solution from problem # Calculate the resulting ph after adding 20.0 ml of 0.20 M HCl into the solution from problem # Calculate the ph of a solution produced by dissolving 1.38 g of NaNO2 into 100. ml of 0.20 M nitrous acid, HNO2. (Assume that there is no volume change.) 149. Calculate the resulting ph after adding 10.0 ml of 0.20 M NaOH into the solution from problem # Calculate the resulting ph after adding 20.0 ml of 0.20 M HCl into the solution from problem # Calculate the resulting ph after adding 110. ml of 0.20 M HCl into the solution from problem # Calculate the resulting ph after adding 120. ml of 0.20 M NaOH into the solution from problem # Calculate the resulting ph after adding 100. ml of 0.20 M HCl into the solution from problem # Calculate the resulting ph after adding 100. ml of 0.20 M NaOH into the solution from problem # Calculate the resulting ph after adding ml of 0.22 M NaOH into the solution from problem # Calculate the resulting ph after adding ml of 0.22 M HCl into the solution from problem # Calculate the resulting ph after adding 10.0 ml of 0.20 M HCl into the solution from problem # Calculate the resulting ph after adding 20.0 ml of 0.20 M NaOH into the solution from problem # 144. Ka or Kb as appropriate for the listed substance for Problems 159 and 160 HCN Ka = HF Ka = HC2H3O2 Ka = Al 3+ Ka = NH3 Kb = (C2H5)2NH Kb = K s for H3PO4 Ka1 = Ka2 = Ka3 = For (a) (g) which of the following salts, when dissolved in water, at 25ºC, cause a change in ph? If there is a ph change, will the ph be above or below 7? Write a hydrolysis reaction that illustrates the cause of the ph change. Calculate the ph if 2.0 g of the salt is dissolved in 100 ml of water. a. barium chloride, BaCl2 (208.2 g/mol) b. ammonium perchlorate, NH4ClO4 ( g/mol) c. sodium cyanide, NaCN (49.01 g/mol) d. ammonium nitrate, NH4NO3 ( g/mol) e. potassium acetate, KC2H3O2 ( g/mol) f. diethylamide bromide, (C2H5)2NH2Br ( g/mol) g. aluminum perchlorate, Al(ClO4)3 ( g/mol) 160. For (a) (n) will the ph be 7, above 7, or be below 7. a. equal volumes of 1 M HF and 1 M HCl b. equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2 c. equal volumes of 1 M HF and 0.5 M NaOH d. equal volumes of 1 M NH3 and 1 M NH4NO3 e. 50 ml of 1 M NH3 and 25 of 1 M HNO3 f. 50 ml of 1 M NaOH and 25 of 1 M HNO3 g. 100 ml of 1 M NH3 and 50 ml of 1 M NaOH h. equal volumes of 1 M NaBr and 1 M KCl i. equal volumes of 1 M KOH and 1 M HClO4 j. equal volumes of 1 M CsOH and 1 M HF k. equal volumes of 1 M HC2H3O2 and 2 M NaOH l. equal volumes of 1 M CsF and 2 M HCl m. equal volumes of 1 M H3PO4 and 1 M NaOH n. equal volumes of 1 M NaH2PO4 and 1 M Na2HPO4 Instructions: Clearly circle the response that best satisfies the prompt giving only one answer to each question. DO NOT USE A CALCULATOR. The College Board does not allow the use of a calculator for Section I (the MC part of the exam, so it is time to start practicing without one. You may use ONLY a periodic table and the equations. While you should practice working as fast as possible, it is more important at this point in the course that you practice without a calculator, even if it slows you down. Look for the easy math common factors and rough estimation do not do long division to try to get exact values. Remember it is MC; use the answers to guide you A mole sample of which compound below, when placed in water gives the lowest concentration of anions. (a) NaNO2 (b) HNO2 (c) Pb(NO3)2 (d) Al(NO3)3 (e) a and b would produce the same amount 162. The conjugate base of HSO4 is (a) OH (b) H2SO 4 (c) SO4 2 (d) HSO4 (e) H3SO What is the ph of an aqueous solution at 25.0 C in which [H + ] is M? (a) 2.60 (b) 2.60 (c) 3.40 (d) 3.40 (e) What is the [OH ] in a solution that has a ph of 5 (a) M (b) M (c) M (d) M (e) M 165. What is the concentration of hydronium ions, H3O +, in a solution at 25.0 C with poh = 4.282? (a) 4.28 M (b) 9.72 M (c) M (d) M (e) M 166. What is the ph of a M solution of barium hydroxide? (a) 4.0 (b) 4.3 (c) 5.0 (d) 9.7 (e) HZ is a weak acid. An aqueous solution of HZ is prepared by dissolving mol of HZ in sufficient water to yield 1.0 L of solution. The ph of the solution was 5.00 at 25.0 C. Calculate the Ka of HZ. (a) (b) (c) (d) (e) What is the concentration of the hydroxide ion in pure water at 25 C? (a) 0 M (b) M (c) M (d) 14 M (e) 7.00 M 169. In a basic solution,. (a) [H3O + ] = [OH ] (b) [H3O + ] > [OH ] (c) [H3O+] < [OH ] (d) [H3O + ] = 0 M (e) [OH ] > 7.00

11 170. Which one of the following statements regarding Kw is false? (a) pkw is at 25 C (b) The value of Kw is always (c) Kw changes with temperature (d) The value of Kw shows that water is a weak acid (e) Kw is known as the ion product of water Of the following substances, which of the following will form basic aqueous solutions? NH4Cl Cu(NO3)2 K2CO3 NaF (a) NH4Cl, Cu(NO3)2 (b) NH4Cl, K2CO3 (c) NaF only (d) K2CO3, NaF (e) K2CO3, NaF, Cu(NO3)2 (f) NH4Cl only (g) None of the above will be basic, they are all just salt water with a ph of Which of the following salts will have a ph of 7.0 for an aqueous 0.10 M solution at 25.0 C NaOCl KNO2 NH4Cl Ca(OCl)2 (a) NaOCl (b) KNO2 (c) NH4Cl (d) Ca(OCl)2 (e) None of them would have a ph of 7, they all change the ph 173. What is the poh of a M solution of NaNO2? HNO2, Ka = (a) 3.15 (b) 3.60 (c) 6.70 (d) (e) A 0.1-molar solution of acetic acid (CH3COOH) has a ph of about (a) 1 (b) 3 (c) 7 (d) 10 (e) Ka, the acid dissociation constant, for an acid is at room temperature At this temperature, what is the approximate percent dissociation of the acid in a 1.0 M solution? (a) 0.03% (b) 0.09% (c) 3% (d) 5% (e) 9% 176. What is the ionization constant, Ka for a weak monoprotic acid if a 0.30-molar solution has a ph of 4.0? (a) (b) (c) (d) (e) Phenol, C6H5OH, has a Ka = What is the ph of a M solution of phenol? (a) 2 (b) 5 (c) 6 (d) 10 (e) Determine the OH concentration in 1.0 M aniline (C6H5NH2) solution. (Kb for aniline is ) (a) M (b) M (c) M (d) M (e) M 179. IO3 + HC2H3O2 HIO3 + C2H3O2 The above equation has an equilibrium constant that is less than 1. What are the relative strengths of the acids and bases? 180. H2C3H2O4 + 2H2O 2H3O + + C3H2O4 2 As shown above, malonic acid is a diprotic acid. The successive equilibrium constants are (Ka1) and (Ka2). What is the equilibrium constant for the above reaction? (a) (b) (c) (d) (e) H2PO4 + H2O H3O + + HPO4 2 Which species, in the above equilibrium, behave as bases? I. HPO4 2 II. H2PO4 III. H2O (a) I only (b) I and II (c) II and III (d) I and III (e) III only 182. HC3H5O2 + HCOO HCOOH + C3H5O 2 The equilibrium constant, K, for the above equilibrium is This value implies which of the following? (a) The concentration of HC3H5O2 and HCOO will always be equal (b) C3H5O2 is a stronger base than HCOO (c) HC3H5O2 is a stronger acid that HCOOH (d) HCOO- is a stronger base than C3H5O2 (e) The value of the equilibrium does not depend on the temperature 183. The calculation of [H + ] concentration and ph for weak acids is more complex than for strong acids due to (a) the incomplete ionization of weak acids. (b) the low Ka value for strong acids. (c) the more complex atomic structures of strong acids. (d) the low percent ionization of strong acids. (e) the inconsistent Kb value for strong acids The general reaction of an acid dissolving in water may be shown as HA + HOH H3O + + A One of the two conjugate acid base pair for this reaction is (a) HA and HOH. (b) HA and A (c) HOH and A (d) H3O + and A (e) HA and H3O Strong acids are those which (a) have an equilibrium lying far to the left. (b) yield a really weak ( pathetic in fact) conjugate base when reacting with water. (c) have a conjugate base which is a stronger base than water. (d) readily remove the H + ions from water. (e) are only slightly dissociated (ionized) at equilibrium When calculating the ph of a hydrofluoric acid solution (Ka = ) from its concentration, the contribution of water ionizing (Kw = ) is usually ignored because (a) hydrofluoric acid is such a weak acid (b) hydrofluoric acid can dissolve glass. (c) the ionization of water provides relatively few H + ions (d) the [OH ] for pure water is unknown. (e) the conjugate base of HF is such a strong base 187. The percent dissociation (percent ionization) for weak acids (a) is always the same for a given acid, no matter what the concentration. (b) usually increases as the acid becomes more concentrated (c) compares the amount of acid that has dissociated at equilibrium with the initial concentration of the acid (d) may only be used to express the dissociation of weak acids. (e) has no meaning for polyprotic acids The [OH ] of a certain aqueous solution is M. The ph of this same solution must be (a) (b) 5.00 (c) 7.00 (d) 9.00 (e) In many calculations for the ph of a weak acid from the concentration of the acid, an assumption is made that often takes the form [HA] 0 x = [HA] 0. This (a) is valid because x is very small compared to the initial concentration of the weak acid (b) is valid because the concentration of the acid changes by such large amounts. (c) is valid because the actual value of x cannot be known. (d) is valid because ph is not dependent upon the concentration of the weak acid (e) approximation is always shown to be valid and so need not be checked 190. HA is a weak acid which is 4.0% dissociated at 0.100M. Determine the Ka for this acid (a) (b) (c) (d) 1.6 (e) What is the ph of a 0.01-molar solution of NaOH? (a) 1 (b) 2 (c) 8 (d) 10 (e) What is the volume of 0.05-molar HCl that is required to neutralize 50 ml of a 0.10-molar Mg(OH)2 solution? (a) 100 ml (b) 200 ml (c) 300 ml (d) 400 ml (e) 500 ml 193. Which of the following best describes the ph of a 0.01-molar solution of HBrO? (Ka = ) (a) Less than or equal to 2 (b) Between 2 and 7 (c) 7 (d) Between 7 and 11 (e) Greater than or equal to 11