lecture notes, week 2
lecture notes, week 2 chem 152
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This 10 page Class Notes was uploaded by Shelby Logsdon on Wednesday January 21, 2015. The Class Notes belongs to chem 152 at University of Washington taught by munira khalil in Winter2015. Since its upload, it has received 144 views. For similar materials see general chemistry in Chemistry at University of Washington.
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Date Created: 01/21/15
Chem 152 C 1515 Covedng Applications of Aqueous Equilibria ch8 Thermodynamics ch 910 Electrochemistry ch 11 Quantum Mechanics and Atomic Theory ch 12 Exam 1 Fri jan 30 Exam 2 Mon Mar 2 Final Mon Mar 16 230420 Practice exams will be posted exams cumulative Aleks code on class website Acid Base Equilibria strong acids ionize completely in water 0 the conjugate base of a strong acid is a very weak base weak acids do not ionize completely in water 0 conjugate base of a weak acid is strong enough to ionize water Hydronium and Hydroxide acids with a ka gt 1 strong acids will react completely to give H30 and the conjugate base bases with a kbgt 1 strong bases will react completely to give OH and the conjugate acid acids weaker than water have conjugate bases that are stronger than OH acids stronger than H30 have conjugate bases that are weaker than H2O conjugate base of a weak acid since HF is a weak acid its conjugate base is strong enough to hydrolyze water F H20 HF OH What is the pH of a 350 M solution of NaF Use Kw to nd Kb Kb14x10quot11 Common ion solutions Equilibrium concentrations of a weak acid will be affected by the presence of signi cant quantities of its conjugate base and vice versa What is the pH of a solution that is 450 M HF acid and 350 M NaF HF H20j F H30 450 0 350 0 x x x 450x 350x x Buffer solutions A buffer solution contains signi cant amounts of a weak acid and its conjugate base or a weak base and its conjugate acid Buffers are able to absorb a certain quantity of OH or H ions without signi cant pH change This works because 0 A weak acid HA is a better proton source than water so OH is neutralized and replaced with A o The conjugate base A want an extra proton more than water so H is neutralized and replaced with AH Only works when OH or H is much less than the concentration of HA and A Find the pH of a solution that is 5 M HF and NaF with Ka72x10quot4 HF H20 F H30 5 5 0 x x x 5x 5x x X 72x10quot4 M pH og72x10quot4 314 what happens to the pH if we add 010 mol of NaOH to 100 L of the 5 M HFNaF solution The OH added converts HF to F HF OH F H20 Moles of HF acid present initially 0 5M 100L5 moles OH then converts HF to F 0 5010 moles490 moles HF o 5010 moles 510 moles F Solve w new concentrations in ice table X692x10quot4 M pH316 Buffer is only effective to a certain amount of OH or H30 buffer capacity Ex What if we add 010 mol of HCI to 100 L of the 500 M HFNaF solution HF H20 D F H30 HCI H20 l H30 Cl H30 F l HF H20 Initial HF 500 mol HF Initial F 500 mol F 500010 490 mol F 500010 500 mol HF Use new concentrations in ice table HF H20 l H30 F 510M 0 490 M x x x 510x x 490x Ka 490510x X 749x10quot4 M pH313 Buffer only works if concentration of strong acid or base is much less than that of the weak acid or base Henderson hasselbalch equation pH pKa logHAA Buffer Capacity buffer capacity the amount of strong acid or base a buffer can absorb without a signi cant change in pH the buffer capacity depends on the amount of weak acid and conjugate base that is present the pH of a buffer that is 5 M in HF and NaF is 314 what is the pH of a solution that is 050 M in both species use HH equation to solve the pH of both buffers is the same but which has a larger buffer capacity 5 M bc of larger concentration buffers are most effective with the concentrations of weak acid and conjugate base are close to the same acidbase titration titration is a method of quantitative chemical analysis that is used to determine the unknown concentration of a species in solution ex Consider the titration of 5000 mL of 200 M HCI with 100 M NaOH initial pH of HC og200M 698 pH after 100 mL of NaOH mol OH 001 equivalence point number of moles of titrating solution moles of titrated solution pH at equivalence point for strong acidbase titration is 7 Weak acid strong base titration HA NaOH l NaA H20 Add base to neutralize acid like in a strong acidbase titration added base will neutralize the weak acid to produce water and the conjugate base of the acid Equilibrium lies very far towards products such that we can consider added OH to have reacted completely with the weak acid Finally we can calculate new pH of the solution using equilibrium concepts and the new concentrations Ex calculate the pH during the titration of 2000 mL of 450 M HF acid Ka 72x10quot4 after adding volumes of 450 M NaOH a 000 ml b 500mL c1000 mLmidpoint d 2000mL eq pt A 000 mL of NaOH pH175 HF H20 H30 F Ka 72x10quot4 H30FHF xquot2450 450M 0 0 xquot2 324x10quot4 x 180x10quot2M x x x pH og180x10quot2 175 450x x x B 500mL NaOH HF OH H2O F moles OH 00230 mol lnitial moles HF 00900 mol moles HF remaining 00670 mol Moles F 00230mol total volume 250 mL F0920 M HF 268 M pH pKa log FHF pH 314 log0920268 pH 3140464 pH 268 C 1000 mL NaOH midpoint 00450 mol OH 00450 mol HF 00450 mol F pH pKa log FHF pH pKa pH 314 D 2000 mL NaOH equivalence point Species present Na F H20 Thermodynamics 0th law de nes temperature 1st law de nes energy 2nol law de nes entropy 3rCI law provides a numerical value to entropy Energy potential vs kinetic Potential energy the energy an object has by virtue of its placement in a eld of force like gravity PE mgh m mass g acceleration due to gravity 98 msquot2 h height Kinetic energy the energy an object has by virtue of its motion KE 12 mvquot2 m mass v velocity lnternal energy EEkEp Molecular kinetic energy 0 How do molecules move around Translation motion through space Rotation motion abou the center of mass Vibration motion directed through chemical bonds Molecular potential energy 0 Molecules are subject to more forces than just gravity Electrostatic intermolecular attractions and chemical bonds Thermodynamics Energy exchange work heat Types of energy 0 Potential 0 Kinetic all exist as work heat 0 Nuclear etc Need to differentiate between system and surroundings System that part of the universe you are interested in 0 Open matter and heat exchange 0 Closed only heat exchange no matter 0 Isolated no heat or matter exchange Surroundings the rest of the universe First law of thermodynamics E q w E change in internal energy q heat w work internal energy for a closed system is conserved internal energy is a state function and not a path function ie it depends only on its initial and nal conditions and not how you got them Work transfer of energy from system to surroundings and vice versa work work done on the system or transferred to they system energy of the system increases work work done by the system or transferred from the system to the surroundings energy of the system increases work forceextemal x displacement Heat a form of energy associated w temperature Governs random motion of molecules Energy that is owing into or out of system At thermal equilibrium heat stops owing between system and surroundings q heat is added to the system energy increases q heat is transferred from system to surroundings energy decreases Heat does not equal temperature 0th law of thermodynamics de nes temperature Exothermic process heat is released from the system Endothermic process heat is gained by system De ning PV work The expansion of gas against a constant external pressure is a common type of work in a chemical processes 39 Pext FextA I Iext Pext X A w Pext x V w Pext x V Ex 13 x 10quot8J of heat is applied to the air in a hot air balloon causing this volume of gas to in ate from 4x 10quot6 L to 45x 10quot6 L W Pext V Pext Vf39vi 1atm 45X10A6L 4X10A6L 1 atm 5X10A6 L R 8314Jmol K 1013J L atm 1000L 1 mquot3 w 5 x 10quot6 L atm 1013JL atm 51 x 10quot7J E q w 13x10A8J 51x10A7J 8 x 10quot7J Enthalpy H enthalpy H EPV E internal energy P pressure V volume H E PV E PV VP q w PV VP q PV PV VP q VP H Constant pressure process PO qp constant pressure H qp enthalpy heat Speci c heat capacity The amount of heat required to change the temperature by one degree for one mole of a particular system substance At constant volume CV qudT At constant pressure Cp dqpdT Units of C JKmol QVnCVT qpnCpT Thermodynamics of ideal gases All substances have some capacity to stroe heat energy this heat is stored at kinetic energy KE 32 RT KE depends on temperature dea gasses have 3 degrees of freedom xyz planes When heat is added under constant volume translational KE only is increased When heat is added under constant pressure system volume is increased due to increased translational KE System performs PV work System must absorb more heat constant pressure than at constant volume to undergo same change in temperature Internal energy o E EK Ep 0 For n moles of a monatomic ideal gas potential energy is zero and kinetic energy is given by KMT o E n32 RT 0 A change in internal energy results in a temperature change E n32RT o In constant volume process E CVT 0 Speci c heat capacity constant volume CV 32R Enthalpy o H E PV 0 For n moles we replace E w the term on the left and use PV nRT o H n52 RTT o H qp nCpT o Cp5l2 R CpCVR R 8314Jmol K Ex what is q w E and H for one mole of an ideal monatomic gas w an initial volume of 5 L and pressure of 20 atm is heated until a volume of 10L is reached w constant pressure lnitial V 5 L p 20 atm Final v 10L p 20 atm E nCVT CV32 R H nCpT Cp 52 R q H W E q W PV PV nRT Ex 3 We have 200 moles of a monoatomic ideal gas initially at 500atm and 100l it can take the red or green path to reach its nal state at 300 atm and 1000 I calculate H E T w and q for each step in each path Pi 500 atm Pf300 atm Vi100 LVf 1000 L n 200 mol E nCVT PV nRT HnCpT TPVnR Step 1 T1 100L300atm200mol0821 LatmmolK E nCVT n32R 200nR 300atm1013J 304J W 0 q E 304J H n52R 200atmLnR 500 L atm1013 Jl atm 507J Step 2 P 0 H q nCpT wPV E w H nCVT T PV nR 300900nR 270 L atm nR E n32R 270L atmnR 405L atm 1013 J L atm 410kJ H n52 R 270nR 675 L atm 1013 M atm 684 kl Total AEtotal 380kJ AHtotal 633kJ Wtotal 274kJ Qtota 654kJ Red path E and H the same w and q must be calculated separately bc not state functions 406 x 103 use this to warm up 1 mole of water at 1 atm What would T be Cp 753 J mol K Constant pressure process H q nCpT T qpnCp 406X103J100mo753JmoK 54 K Calculate H for 1 mole of H20 I when it is heated from 25 degrees C to 100 degrees C and evaporated at 100 degrees C H1 heating liquid water from 25 to 100 H2 boiling water at 100 H1 nCpT 100 mo753JmoK75K H2 nHVap 100 mol 4066kJmol H1H2Htota