TERRANCE SEREDA. Labs taught in Introductory University Chemistry (Chem 200) at Red Deer College

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1 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 1 of 10 Expt # Title of Experiment (brief details of experiment) 1 Buret and Scale Operation 2 3 Hydrates (Determination of the number of waters of hydration of an inorganic compound.) Determination of the Ideal Gas Constant (Determination of the gas constant, R, using the equation PV = nrt.) 4 Determination of the Molarity of an HCL solution / Determination of % Fe in unknown sample. (Determination of Fe using an oxidation reduction reaction between Fe (II) and permanganate.) Complex Nickel Compounds (Preparation of a coordination compound.) Complex Nickel Compounds (Determination of the formula for the coordination compound.) Redox Reactions: Oxidation states of Vanadium (Reduction of vanadium (V) to vanadium (IV) with sulfite, titrate with permanganate.) (Reduction of vanadium (V) to vanadium (II) with Zn, stabilize to vanadium (IV), titrate with permanganate.) Enthalpy of Reaction (Molar enthalpy of neutralization for HCl/NaOH, HOAc/NaOH; Molar enthalpy of formation of Mg using Mg/HCl, MgO/HCl.) Rates of Chemical Reactions: The kinetics of oxidation of Iodide by peroxydisulfate (Effect of concentration and temperature on rate, determine the overall order of the reaction, calculate the rate constant.) Chemical Equilibrium, (iodine-triiodide equilibrium constant) (Determination of the equilibrium constant for I -1 + I 2 I 3-1 in water.) Titration Curves (HCL titrated with NaOH, HOAc titrated with NaOH; determine equivalence point, pka.) Titration Curves (NH 3 titrated with HCL, H 3 PO 4 titrated with NaOH; determine equivalence point, pka) 17 Qualitative Analysis of Anions 18 / 19 Qualitative Analysis of Cations 20 Electrolysis (Determination of the Faraday constant and atomic weight of Cu)

2 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 2 of 10 Expt #1: Buret and Scale Operation - recording of volume and mass as an assessment of the proper recording of data Expt #2: Hydrates - repeated heating of a hydrate and mass determination of the anhydrous form - determination of the percentage mass of water - determination of anion and cation of hydrate - determination of the empirical formula of the hydrate Expt #3: Determination of the Ideal Gas Constant - generate N 2 gas from reaction of sulfamic acid and sodium nitrite - collect gas over water - calculate pressure of N 2 gas, accounting for vapour pressure of water and pressure due to the height of the water in the buret: where P nitrogen is derived from: P atm = P nitrogen + P water vapour + P hydrostatic (i.e., water column) - compensate for the volume difference due to the differential temperature P 1 V 1 /T 1 = P 2 V 2 /T 2 - calculate gas constant R from PV = nrt Expt #4: Determination of the Molarity of an HCl Solution - preparation of a primary standard, oxalic acid - standardization of a NaOH solution - titrate unknown HCl solution with NaOH, using an indicator to determine endpoint - calculation of Molarity of unknown HCl solution - calculation of Relative Standard Deviation Expt #5: Determination of % Fe using a Redox Titation - general redox equation: Fe +2 + MnO -1 4 Mn +2 + Fe +3 - standardization of a potassium permanganate solution using ferrous ammonium sulfate (FAS, Ammonium iron (II) sulfate), Fe(NH 4 ) 2 (SO 4 ) 2, as the primary standard - titrate potassium permanganate against known amount of FAS - calculate molarity of permanganate based on balanced equation: 5Fe +2 + MnO H +1 Mn Fe H 2 O - titrate Fe sample with permanganate solution - determine moles of Fe, determine mass of Fe (moles Fe x 55.8 g/mole Fe) - determine mass % Fe: mass Fe/mass of sample Expt #6: Complex Nickel Compounds - two complex compounds were prepared: Ni(NH 3 ) 4 (NO 2 ) 2 and Ni(NH 3 ) x (Cl) x

3 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 3 of 10 Expt #7: Complex Nickel Compounds - determination of formula for coordination compound Ni(NH 3 ) x (Cl) x - determination of %Cl using the Volhard titration: precipitate Cl -1 using excess Ag +, titrate excess Ag + with thiocyanate (forms AgSCN which is insoluble), indicator for end point is FeSCN +2, utilizing a small amount of Fe (as ferric nitrate) - determination of %Ni using EDTA (1Ni:1EDTA) titration using murexide indicator - %NH 3 determined by difference from total Ni and Cl - 3 possibilities: [Ni(NH 3 ) 6 ]Cl 2 ; [Ni(NH 3 ) 6 Cl]Cl; [Ni(NH 3 ) 6 Cl 2 ] Expt #9: Redox Reactions: Oxidation states of Vanadium Oxidation States of vanadium Oxidation Ionic form Name Color VO 3 metavanadate +1 pervanadyl + 5 VO 2 (dioxovanadium) VO 4 orthovanadate none yellow none + 4 VO +2 vanadyl blue + 3 V +3 vanadium III Green + 2 V +2 vanadium II violet - preparation of metavanadate from ammonium metavanadate dissolve NH 4 VO 3 in NaOH - preparation of the starting solution, pervanadyl add sulfuric acid to the metavanadate (solution turns yellow) VO H + +1 VO 2 + 2H 2 O - preparation of vanadyl (using the reducing agent sulfite, in the form SO 2 ) add sodium sulfite to the pervanadyl (solution turns blue) 2H VO 2 + SO 3 H 2 O + 2VO +2 + SO 4 - titrate vanadyl with permanganate 5VO +2 + MnO H 2 O 5VO H + + Mn +2 Expt #10: Redox Reactions: Oxidation states of Vanadium - reduction of pervanadyl with zinc (solution turns violet) 8H VO 2 + 3Zn 4H 2 O + 2V Zn +2 - stabilize by adding pervanadyl (convert to stable form, because V +2 and V +3 is air oxidized; VO +2 - not air oxidized) V VO 2 + 2H + 3VO +2 + H 2 O - titrate vanadyl with permanganate

4 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 4 of 10 Expt #11: Enthalpy of Reaction - enthalpy of neutralization for reactions of HCl/NaOH and HOAc/NaOH - monitor temperature for neutralization reaction - calculate heat of neutralization: q = msδt = (ml x g/ml)(j/ C mole)( C) - enthalpy change = ΔH = q - calculate molar enthalpy = ΔH/n - enthalpy of formation of MgO - determine heat of reaction for Mg + HCl, determine molar enthalpy change - determine heat of reaction for MgO + HCl, determine molar enthalpy change - molar enthalpy change for Mg +2 + H 2 O MgO + 2H +, inverse sign - using Hess law Mg + 2H + Mg +2 + H 2 ΔH 1 =? Mg +2 + H 2 O MgO + 2H + ΔH 2 =? H 2 + ½ O 2 H 2 O ΔH 2 = 85 KJ/mole Mg + ½ O 2 MgO ΔH f =? Expt #12: Rates of Chemical Reactions - determination of the effect of the concentration on the rate of a reaction - monitor rate of reaction between peroxdisulfate (S 2 O 8 ) and iodide (I - ): (i) S 2 O 8 + 2I - 2SO 4 + I 2 - follow rate of iodine formation by a secondary reaction, where [thiosulfate, S 2 O 3 ] is small and constant, (ii) is fast compared to (i), starch as indicator: (ii) 2S 2 O 3 + I 2 S 4 O 6 + 2I - - monitor time of reaction for varying concentration of [S 2 O 8 ] & [I - ] - determine rate of reaction: dividing iodine formed by reaction time Δ[I 2 ]/ Δt - determine the order of each reactant, n and m, using two rate equations where one concentration is held constant Rate 1 / Rate 2 = (k[constant-s 2 O 8 ] m [variable-i - ] n ) / (k[constant-s 2 O 8 ] m [variable-i - ] n ) Rate 1 / Rate 2 = ([variable-i - ] n ) / ([variable-i - ] n ) - determine the overall order of the reaction, sum n and m - calculate the rate constant for each of the experimental runs - determination of the effect of temperature on the rate of a reaction - monitor rate of reaction between peroxdisulfate (S 2 O 8 ) and iodide (I - ) by monitoring rate at different temperatures - monitor time of reaction for varying temperatures: 1, 10, 20, 40 C - calculate the rate constant, k, for each temperature - plot log k versus inverse temperature (1/T) - determine the activation energy E a

5 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 5 of 10 Expt #13: Chemical Equilibrium (iodine-triiodide equilibrium constant) - determination of the equilibrium constant for the iodine-triiodide in water I -1 + I 2 I -1 3 ; chemical equilibrium - prepare aqueous solution of iodine triiodide - extract aqueous layer with CCl 4 and separate the two phases - titrate CCl 4 layer with thiosulfate, which contains I titrate aqueous layer with thiosulfate = total iodine in form of I 2 and I 3 - determine I 2 in aqueous layer knowing K for I 2 distribution between aqueous and CCl 4 layer; K = (I 2 org)/ (I 2 aqueous) = titration of aqueous layer gives I 2 aqueous and I -1 3 aqueous = Total - Total - I 2 aqueous = I -1 3 aqueous - amount of I -1 at equilibrium must be equal to initial [I -1 ] [I -1 3 aqueous], because for every I -1 3 formed, one I -1 was consumed - calculate K for chemical equilibrium: K aq = [I -1 3 ]/[ I 2 ] [I -1 ] Expt #15: Titration Curves (strong base/strong acid; strong base/weak acid) - titrate NaOH with HCl (unknown concentration), plot ph versus volume NaOH determine equivalence point determine concentration of unknown [HCl], using (M NaOH )(V NaOH ) = (M HCl )(V HCl ) - titrate NaOH with HOAc (unknown concentration), plot ph versus volume NaOH determine equivalence point determine concentration of unknown [HOAc], (M NaOH )(V NaOH ) = (M HOAc )(V HOAc ) determine the pka for HOAc (ph at which acid is ½ titrated) Expt #16 Titration Curves(weak base/strong acid; polyprotic acid) - titrate NH 3 with HCl, plot ph versus volume HCl determine equivalence point determine the pka for NH titrate H 3 PO 4 with NaOH, plot ph versus volume NaOH determine the pk a1 and pk a2 Expt #17: Qualitative Analysis of Anions - qualitative analysis of anions based on the formation of precipitates of, e.g., Ag +, Ba +2 and brown ring test for nitrate (add reagents H 2 SO 4 & FeSO 4, that ultimately form an addition compound FeNOSO 4 ) and addition of ammonia (for formation of ammonia complex).

6 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 6 of 10 Expt #18 &19: Qualitative Analysis of Cations - qualitative analysis performed for a series of cations from 5 different groups - unknowns contained 4 different cations - general determination of cations present based on solubility of corresponding chlorides, sulfides, e.g. AgCl, CuS - use of temperature to effect separation due to differing solubility products - additionally, precipitation used to isolate ions from mixture, to aid in subsequent identification of remaining cations - identification based on differential solubility of hydroxides - identification based on formation of ammonia complexes - Additional identification based on the color of aqueous solution subjected to a flame test. Expt #20: Electrolysis - set up electrolytic cell where Cu is the anode (+) and nichrome is the cathode (-) - anode: Cu Cu e - - cathode: 2H + + 2e - H 2 - perform electrolysis and record data on vol H 2 collected, current (I), weight Cu lost, time (sec) - calculate #moles of H 2 produced, using PV = nrt, where P hydrogen is derived from: P atm = P hydrogen + P water vapour + P water - calculate the number of moles of e - consumed: moles of e - = 2(#moles of H 2 produced) - from the current, calculate the #Coulombs of charge transferred: 1amp = 1 Coulomb/sec Coulomb = amps sec - calculate the Faraday constant (F) = #Coulombs/ moles of e - - calculate the atomic weight {atomic mass} for Cu using the definition: For redox reactions involving metals, the gram equivalent weight (GEW) of a metal is the weight of metal which reacts with or releases one mole of electrons. Therefore, the GEW is the mass of metal per mole of electrons transferred. GEW = #grams Cu/ moles of e - = gm Cu/ mole of e - Atomic weight = (gm Cu/ mole of e - )(2 mole of e - /mole Cu) or e.g. GEW = gm/2.5 x 10-3 moles e - = 32.1 gm/ mole of e - Atomic weight = (32.1 gm Cu/ mole of e - )(2 mole of e - /mole Cu) = 64.2 GEW = #grams Cu/ moles of e - = gm Cu/ equivalent Atomic weight = (gm Cu/equivalent)(2 equivalents Cu/mole Cu)

7 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 7 of 10 APPENDIX (1) Experiment #9/10: 1. 2H + + SO 3 H 2 O + SO 2 {reducing agent} {sulfite in acid is degraded into SO 2, which can be air oxidized to sulfate, SO 4 } 2. 2H 2 O + SO 2 SO 4 + 4H + + 2e - ref: Chem lab manual 3. 1e VO 2 + 2H + VO +2 + H 2 O Overall: 2H + + 2VO SO 3 H 2 O + 2VO +2 + SO 4 (2) Experiment #9/10,12: SO 2 sulfur dioxide, {IV} SO 3 sulfite, {sulfate (IV)} SO 4 sulfate, {sulfate (VI)} S 2 O 3 thiosulfate S 2 O 8 peroxydisulfate, {VI)} S 4 O 6 tetrathionate (3) Experiment #17: - Ag + : AgCl, AgBr, Ag 2 SO 4, AgI - Ba +2 : BaSO 4 - CO 3 {carbonates} in HCl CO 2 gas - nitrates - brown ring test - ammonia for cation complexes (4) Experiment #18/19: Group I: ppt when Cl - added PbCl 2, Hg 2 Cl 2, AgCl Group II: ppt when thioacetamide added Bi 2 S 3, CuS, CdS Group III: insoluble in strong NaOH Fe(OH) 3, Mn(OH) 2, Co(OH) 2, Ni(OH) 2 Group IV: soluble in strong NaOH Al(OH) 4, Cr(OH) 4, Zn(OH) 4 Group V: did not react to this point Ca, NH 4 (5) Electrochemistry basics: 1. VANO Voltaic cell/anode/negative/oxidation 2. Voltaic Cell: voltage of cell E cell = E cathode E anode If E cell is positive, then the cell is spontaneous. For a spontaneous cell, the anode must always be a smaller number, i.e., a smaller positive, negative or more negative redox potential. Metal(s) M(aq) M(aq) Metal(s); { Anode Cathode } convention { Anode ( ) Cathode (+) } 3. Example: voltaic cell Anode (-), Zn Zn e - Cathode (+), Cu e - Cu Cu e - Cu E º = V Zn e - Zn E º = V {since V is neg, must be at anode} 4. Oxidation always occurs at the anode for both voltaic cells or electrolysis. 5. SRAN: Strong ReducingAgent is more Negative redox potential 6. Multiplying a half reaction by a number does not change the Redox Potential.

8 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 8 of 10 APPENDIX, continued: (6) Experiment #20: Electrolysis for Cu e - Cu E º = V 1/2O 2 + 2H e H 2 O E º = 1.23 V 1. EAPO Electrolytic cell/anode/positive/oxidation 2. { Cathode, reduction ( ) l (+) Anode, oxidation } 3. ** For Voltaic cell put most negative at the Anode, (using SRAN). 4. For an Electrolytic cell, do just the opposite, i.e., put most negative at the Cathode. 5. More Neg or (least Pos = copper = { Cathode, reduction ( ), Cu, 0.342V l (+) Anode, oxidation, H 2 O, 1.23V } 7. Cathode: Cu e - Cu Anode: H 2 O 1/2O 2 + 2H e 8. for Voltaic cell: E cell = E cathode - E anode = 0.342V V = -0.89V. Need 0.9 Volts to force the reaction to go. (7) Experiment #11: 1. Heat content = Enthalpy = H 2. HCl, NaOH (H) NaCl, H 2 O (H), exothermic, ΔH = (-) 3. Urea (solid), H 2 O (H) Urea (dissolved in water ) (H), endothermic, ΔH = (+) 4. C = heat capacity (heat needed to increase temp by 1 C º ) Specific heat capacity = (heat needed to increase temp of 1 gram by 1 C º ) 5. ΔG = ΔH TΔS; ΔG = (-), the reaction is favored 6. ΔS (-) = least disordered (e.g., ice) water gas ΔS (+) = most disordered (8) Oxidation Reduction Reactions: 1. SOAP: Strong Oxidizing Agent is more Positive redox potential SRAN: Strong ReducingAgent is more Negative redox potential 2. half reactions: V oxidizing agent.923v reducing agent V oxidizing agent 0.857V reducing agent 3..84V reducing agent 0.92V oxidizing agent V ** reducing agent 2.0V 0.56V -0.85V ** oxidizing agent

9 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 9 of 10 APPENDIX, continued: (9) Experiment #15: (acid / base chemistry) (10) Other 1. px = -logx ph = -log [H]; poh = -log [OH] 2. ph + poh = K a K b = Kw = 10-14, for a conjugate acid/base pair 3. ph = pka + log([a - ]/[HA]) 4. pka = ph at the point where it is ½ titrated 5. ph of salt solutions: (i) salf of a strong acid/strong base = ph7 {NaCl} (ii) salt of strong base/weak acid = ph >7 {Na-Acetate} (iii) salt of weak base/strong acid = ph <7 {NH 4 Cl } (iv) salt of weak base/weak acid = ph determined by competing equilibrium H 3 PO 4 + OH - - H 2 PO 4 HPO 4 + H + - H 2 PO 4 HPO 4 + H + K a = 6.2 x 10-8 H 2 PO H 2 O H 3 PO 4 + OH - K b ( H 2 PO 4 ) = K w / K a ( H 3 PO 4 ) = / 7.5 x 10-3 K b ( H 2 PO 4 ) = 1.3 x Since K a >> K b, a solution of Na H 2 PO 4 would be acidic. 1. log function: log a (a x ) = x; antilog function: a log a (x) = x log a (x) = (1/ln a)(ln x); (2.303) log 10 (x) = ln x 2. NO 2 = nitrite NO 3 = nitrate CO 2 3 = carbonate ClO = hypochlorite ClO 2 = chlorite ClO 3 = chlorate ClO 4 = perchlorate MnO 2 4 = permanganate CrO 2 4 = chromate 3. Lewis Structure: 1. Find total # electrons for all atoms = valence e + account for charge. Account for charge -1 for each (+) and +1 for each ( ). 2. Find # e for individual atoms = 2(H atoms) + 8(for all other atoms). 3. # bonding e is (2) (1). 4. # bonds = (3)/2. 5. # unshared e = (total e ) (bonding e ) = (1) (3). 4. Formal Charge = group # - [# bonds - # unshared e ]

10 Teaching Dossier for Chem 200, T. Sereda, 2011 Page 10 of 10 APPENDIX, continued: 6. Balancing Equations 1. In Acid: Split into the two half reactions. Assign oxidation numbers. Add e to accounr for oxidation state. Balance the e by multpying by appropriate factor. Balance the atoms that are oxidized or reduced. Balance oxygen atoms by adding H 2 O. Balance hydrogen atoms by adding H + based on number of H 2 O. 2. In Base: Using OH Using H + Balance in acid. Add x OH to both sides based on number of H +. The x OH will equal the number H 2 O. Eliminate water from one side. 7. Nernst equation calculate cell voltage based on concentration of cell components. 1. Separate into two half cells: Anode (0.01 Molar): M e M V Cathode (0.5 Molar): 2N e 2N 0.799V Anode: E a = / 2 (log 1/[0.01]) = V Cathode: E c = / 2 (log 1/[0.5] 2 ) = 0.78V E cell = E c - E a = 0.78 (-0.461) = 1.24V