ANSWERS AND EXPLANATIONS TO PRACTICE EXAM II FREE RESPONSE QUESTIONS 1. Mandatory Calculation - Equilibrium 10 points total Ag 2 S (s) W 2 Ag (aq) S 2- (aq) a) Write the K sp and calculate its value. (solids do not appear in K expressions) K sp = [Ag ] 2 [S 2- ] Calculate the value: Given the value for S 2- as 3.42 x 10-17 M find the value of Ag. (2:1 ratio) Ag 2 S W 2 Ag S 2-2(3.42 x 10-17 ) 3.42 x 10-17 Substitute and solve. K sp = (6.84 x 10-17 ) 2 (3.42 x 10-17 ) = 1.60 x 10-49 (b) Calculate solubility in M for Ag 2 S in a 0.00250 M K 2 S solution. Find the silver ion concentration using the K sp value calculated above. ( ) ( ) Reaction Ag 2 S W 2 Ag S 2- Initial --- 0.00250 M Change --- 2x x Equilibrium --- 2x.00250 x K 1.60 x10 S 2 Ag -47 = 6.40 x10 M Ag = 2x -49 2 sp Ag = = 2-2.50 x10-3 -24 1pt Ag =8.00 x10 M Ratio of Ag2S to Ag is 2:1 so... -24 1molAg2S -24 1pt 8.00 x10 M = 4.00 x10 M 1molAg (c) Will a precipitate form? Compare Q to K. 2 AgNO 3 K 2 S Ag 2 S 2 KNO 3
First, calculate the number of moles in solution by multiplying volume and concentration for each component: -3 9.00 x10 mol.275l =.02475mol Ag L -2 7.00 x10 mol 2-.500 L =.035molS L Calculate new molarities: Remember to divide by total volume..02475mol Ag = =.0330M.750L 2-.035mol S = =.0467 M.750L Calculate Q: Q = [Ag ] 2 [S 2- ] Q = (.0330) 2 (.0467) Q = 5.09 x 10-5 No precipitate will form. Q > K (d) (i) Write a balanced equation. Cancel all common terms and end with: AgCl (s) W Ag (aq) Cl - (aq) Ag (aq) NH 3 (aq) W Ag(NH 3 ) (aq) Ag(NH 3 ) (aq) NH 3 (aq) W Ag(NH 3 ) 2 (aq) AgCl (s) 2 NH 3 (aq) W Ag(NH 3 ) 2 (aq) Cl - (aq) (ii) Calculate the K overall. Multiply all K values together. K = (1.0 x 10-10 ) (2.1 x 10-3 ) (8.2 x 10 3 ) K = 1.72 x 10-9 (iii) Write the equilibrium expression for the overall reaction. - Ag(NH 3) 2 Cl Keq = 2 NH [ ] 3
2. Choice Problem---Electrochemistry 9 points total (2pt) (a) Label on the diagram: anode ---Zn ; cathode---ni; electron flow---from anode to cathode 1 point for labeling both electrodes correctly and 1 point for showing proper electron flow. Ni Zn 1.0 M Ni(NO 3 ) 2 1.0 M Cu(NO 3 ) 2 (b) Write the half-cell reactions occurring at each electrode. Zn Zn 2 2 e - Ni 2 2 e - Ni (c) Calculate the voltage of the cell. Remember to change the sign on the voltage for zinc since it is oxidized. Eº cell = Eº ox Eº red Eº cell =.76 (-.25) =.51 volts (d) Calculate Gº for the reaction. (1 volt = J/coul) Gº = -nfe º Gº = - (2 mole e - ) (96,500 coul/mol e-) (.51 Joule/coul) = - 98,000 J (e) Calculate K for the reaction. Gº = - RT lnk o ΔG ln K = - RT -98,000 J =- (8.31J mol/k) 298K = 1.5x10 17 ( )
(2pt) (f) Calculate the voltage of the cell when Ni 2 changes to 0.250 M. o RT E cell =Ecell - lnq nf 1 point for correct calculation of Q and 1 point for set up and answer. ( 8.31J/mol K)( 298K) ( 1.00) E cell =.51- ln 2mole - 96,500coulomb/mole -.250 E cell ( )( ) ( ) =.49 volts remember that J/coul = volts ( ) This answer makes logical sense. When the concentration of a reactant is decreased, the system will shift to relieve the stress. 3. Choice Calculation Stoichiometry and Colligative Properties 9 points total (a) Determine the empirical formula from the data given. The empirical formula is the simplest formula for a compound. The first step in solving this problem is to convert everything into moles. To find the number of grams of oxygen: 100 (31.7 5.30) = 63.0 1 mol C 31.70 g C = 2.639 mol C 12.01 g C 1 mol H 5.30 g H = 5.2475 mol H 1.01 g H 1 mol O 63.0 g O = 3.9375 mol O 16.00 g O The second step is to calculate the mole ratio. This will give subscripts. The moles of carbon are the smallest so divide everything by the moles of carbon. 2.639 mol C = 1 2.639 mol 5.2475mol H = 2 2.639 mol 3.9375mol O = 1.49 2.639 mol The ratio does not end with whole numbers. To keep the same ratio of 1:2:1.49; multiply all atoms by 2.
The empirical formula is: C 2 H 4 O 3 (b) (i) Calculate the molar mass from the freezing point data given. for calculating molality Molar mass = grams/ mole The problem gives the number of grams but moles must be found. Begin with the freezing point formula. T f = K f m i i = 1 since the compound is a nonelectrolyte Calculate the temperature change: Normal freezing temperature: 5.51 ºC Freezing point with solute added: 1.46 ºC T f = 5.51 1.46 = 4.05 ºC Solve the equation for molality: o ΔT f 4.05 C m = m = o =.791 m K f 5.12 C/m for calculating molar mass Find the number of moles from the molality equation. m = mol solute kg solvent mol solute = m kg solvent =.791 mol/kg.250 kg =.198 moles solute Solve for molar mass: molar mass = 30.00 g solute = 155 g/mol.198 mol solute (ii) Find the molecular formula: molec.wt. 155 Molecular formula = = =2 emp.wt. 76 2(C 2 H 4 O 3 ) = C 4 H 8 O 6 (iii) Calculate the mole fraction of the solute. First, calculate the moles of each component. 1 mol C4H8O 6 30.00 g C4H8O 6 =.1935 mol 155 g 1 mol C6H6 250.0 g C6H 6 = 3.205 mol 78 g
Calculate the mole fraction for solute. moles solute Χ solute = total moles (c).1935 = =.0569.1935 3.205 Calculate the osmotic pressure. Π = MRT First, calculate the molarity. Find liters from the density. M D= V M 280 g V = = = 243mL =.243L D 1.15g/mL.1935molC4H8O6 M = =.796M.243L Substitute and solve for osmotic pressure. Π = (.796 mol/l) (.0821 L atm/mol K) (298 K) = 19.5 atm Determine the vapor pressure. Vapor pressure is lowered for a nonvolatile solute according to Raoult s law. P solvent = (X solute ) (P solvent ) Calculate the change in pressure by substituting in to Raoult s Law. P solvent = (.0569) (95 mm Hg) = 5.4 mm Hg Calculate the new vapor pressure. 95 mm Hg 5.4 mm Hg = 90 mm Hg
4. Mandatory -- Net Ionic Equations choose 5 of 8 (3 points for each equation for a total of 15 points) For each reaction: 1 point is awarded for correct reactants and 2 points for correct products. All substances that ionize (strong acids, strong bases and soluble salts) must be written in ionic form. All spectator ions must be cancelled. Do not take time to balance. a) A solution of ammonium thiocyanate is added to a solution of iron (III) nitrate. Fe 3 SCN - Fe(SCN) 2 This is a classic on the AP exam. The formation of this complex ion is brick-red in color. b) Carbon dioxide gas is bubbled through a concentrated solution of sodium hydroxide. CO 2 OH - CO 3 2- H 2 O Carbon dioxide is a nonmetallic oxide and sodium hydroxide is a strong base. This is a neutralization. c) Solutions of manganese(ii) chloride and sodium sulfide are mixed. Mn 2 S 2- MnS Sulfides are not very soluble. d) Solid potassium chlorate is heated in the presence of a catalyst. KClO 3 KCl O 2 Solids are written together. e) Phosphorus pentachloride powder is sprinkled over distilled water. PCl 5 H 2 O H 3 PO 4 H Cl - A combination of acids will be formed. Be sure to keep the oxidation state of P the same on both sides of the equation. HCl is a strong acid and thus, ionizes. f) A solution of potassium dichromate is added to an acidified solution of iron(ii) bromide.
2- Cr 2 O 7 H Fe 2 Cr 3 H 2 O Fe 3 Watch for hints such as acidified, this indicates a redox reaction. Be sure to have one element oxidized Fe here went from 2 to 3 and one element reduced---cr here went from 6 to 3. The water is necessary to balance all atoms. g) A small piece of potassium metal is added to distilled water. K H 2 O K OH - H 2 Active metals react with water to form a strong base and hydrogen gas. h) A concentrated solution of hydrochloric acid is added to solid sodium permanganate. H Cl - - MnO 4 Mn 2 Cl 2 H 2 O Permanganate in the presence of an acid reduces to Mn 2 (from its 7 state) and halides, Cl -, will oxidize to form the free halogen.
5. Mandatory Essay Kinetics---9 points total In an essay of this type, be sure to read the prompt and address whatever is asked. 2 NO 2 (g) ÿ 2 NO (g) O 2 (g) (a) Complete the graph. Label the reactant and each product. (3pt) 1point for [NO 2 ] decreasing and leveling off 1 point for [NO] and [O 2 ] increasing 1 point for [NO] having a steeper curve than [O 2 ] since the ratio is 2:1 0 50 100 150 200 250 300 350 400 time (sec) (b) Answer the questions using data given: (i) Write rate-law and justify. rate = k [NO 2 ] 2 When experimental data is given with concentration and time, the order is determined from graphical means. A straight line for the following will yield the orders listed: Graph Order ln [A] vs time first 1/[A] vs time second [A] vs time zero These are listed from most common to least common in frequency. (ii) How to find rate constant, k. The slope of the straight line in the second graph, 1/[NO 2 ] vs. time equals the rate constant and 1/ [NO 2 ] is the y-intercept. 1 1 =kt [ NO2] [ NO2] o y = mx b (iii) The units for a second order reaction: L/mol s (or M -1 s -1 ) (c)
(i) Write a balanced equation from mechanism. Cancel common terms from both sides. I. H 2 O 2 ÿ 2 OH II. H 2 O 2 OH ÿ H 2 O HO 2 III. HO 2 OH ÿ H 2 O O 2 Final equation: 2 H 2 O 2 ÿ 2 H 2 O O 2 (ii) Which is rate-determining step and why? Step 1 must be the rate determining step. The rate law is given to be rate = k [H 2 O 2 ] in the beginning of this question. If the second step was the slow step, the rate law would be rate = k[h 2 O 2 ] 2.
6. Mandatory Laboratory Question---Thermochemistry 9 points total (a) Equation for heat evolved. q = m C p T (2pt) (b) Use the graph to find initial starting temperature. 1 point for drawing line of best fit and crossing the y-axis. 1 point for estimating answer about 41.5ºC. 43 Temperature vs. Time after mixing 42.5 42 41.5 41 40.5 40 39.5 0 50 100 150 200 Temperature ( C) (3pt) (c) Use Hess s Law to find H for the following reaction: NH 3 HCl NH 4 Cl H =? 1 point for writing reactions; 1 point for answer; 1 point for proper significant figures. H (kj/mol) HCl NaOH NaCl H 2 O -55.0 NH 4 Cl NaOH NH 4 OH NaCl (NH 4 OH breaks down into NH 3 and H 2 O) Reaction is: NH 4 Cl NaOH NH 3 H 2 O NaCl -5.50 Keep the first equation as is: HCl NaOH NaCl H 2 O (-55.0) Reverse the second equation: NH 3 H 2 O NaCl NH 4 Cl NaOH (5.50) Cancel common terms: NH 3 HCl NH 4 Cl (-50.0) (d) Calculate % error. actual- theoretical % error = 100 theoretical -50.0-(-45.0) %error = 100 = 10% -50.0
(e) Calculate heat of reaction. First, find mass and difference in temperature. q = m C p T mass = 20.0 ml 30.0 ml = 50.0 ml 1.02g M = VD = 50.0 ml =51.0g ml T = final initial 25.3 21.4 = 3.9 ºC q = (51.0 g) (4.18 J/g ºC) (3.9 ºC) = -831 J The reaction is exothermic, the temperature rose, therefore, the heat is negative. Calculate heat with the correction: (3.9 ºC ) (11 J/ºC) = 43 J (lost to calorimeter) -831 J -43 J = 874 J 7. Choice Essay ---Chemical Principles 8 points total This question represents what is commonly referred to as a kitchen-sink question. Various chemical concepts are needed to accurately explain common phenomenon. (2pt) (2pt) (2pt) (2pt) (a) Water expands upon freezing. Water in the liquid state has particles that are strongly attracted to each other through hydrogen bonds. The water molecule has a bent structure with two unshared pairs of electrons. When the water molecules begin to organize to form a crystalline structure, they arrange themselves into a hexagonal pattern. At the center of the structure is an air pocket. This shape increases the volume of the ice and thus, decreases its density. (b) 7s 2 p 6 is a noble liquid. The electron configuration, s 2 p 6 is the ending configuration for elements in the last column on the periodic table. The element would most likely be in the liquid state because of its large size and great number of electrons. Weak attractive forces, London dispersion forces, become much stronger the more electrons present because there is greater chance for polarizability. (c) HF is not a strong acid but HCl, HBr and HI are strong acids. HF does not ionize 100% when in solution. In weak solutions, HF molecules hydrogen bond with water molecules and do not ionize much. In concentrated solutions, HF molecules dimerize with each other. The other hydrohalic acids essentially dissociate completely in aqueous solutions. (d) Silver chloride changes color when exposed to light. Silver chloride undergoes a photochemical reaction with light and decomposes according to the following equation and turns a very dark grayish-purple color: 2AgCl (s) hν 2Ag (s) Cl 2 (g)
8. Choice Essay ---Acid-Base Titration 8 points total (2pt) (a) Choose the conjugate pair that is best for preparing a buffer with ph = 6.80 HSO 3 H 2 SO 3 K a = 1.0 x 10 7 Bisulfite/sulfite ions would make the best buffer choice. Using the relationship, ph = pka /- 1, it is easy to see that the equation above would yield a ph of 6-8. Log base 10 is the power of ten in scientific notation. The log of 1.0 is 0. (b) List lab equipment needed to make the buffer. 1 point for any three; 2 points for any five volumetric flask pipet distilled water analytical balance ph meter stirring rod; magnetic stirrer (c) Sketch the curve for weak acid with NaOH.
14 2 18 36 40 (2pt) 1 point for beginning low around ph of 2-3 and ending high near 14 1 point for the general shape of the curve plateau; up; plateau (d) Explain how to use titration curve to determine the Ka of the acid. The Ka for a weak acid can be found by the following relationship: ph = pka at ½ way to equivalence point Given that the equivalence point occurs at 36.00 ml, draw a vertical line at 18.0 ml of NaOH until it intersects the titration curve and read the ph at this point. Take the antilog of the ph and the Ka value will be found.