Week 8 of Dr. Ma's notes
Week 8 of Dr. Ma's notes CHEM 1200
Popular in Chemistry II
Popular in Chemistry
This 6 page Class Notes was uploaded by Alexi Martin on Tuesday March 15, 2016. The Class Notes belongs to CHEM 1200 at Rensselaer Polytechnic Institute taught by Dr. Alexander Ma in Spring 2016. Since its upload, it has received 16 views. For similar materials see Chemistry II in Chemistry at Rensselaer Polytechnic Institute.
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Date Created: 03/15/16
Lewis AcidBase Transfer electrons from a lone pair to create a bond or from a bond to a lone pair Does not require H+ Electron donor (pair) Lewis base, nucleophilic (electron rich) Electron pair acceptor lewis acid, electrophile (electron deficient) octet violator Lewis Base Electron willing to give away or share Lone pair of electrons can donate Anions act better than a neutral atom N:<N: More electronegative the less likely it is to donate an electron O:<S: Lewis Acid Electron deficient (not being attached to the electronegative atom, not having an octet Empty orbitals to accept electron pair H+ has an empty orbital B in BF3 empty 2p orbital an incomplete octet many small highly charged metal cations empty orbitals accept electrons Al Cu Cr Atoms attracted increase in electronegativity atoms have multiple bonds= lewis acids CH2O Lewis Acid/Base Reaction Base donates electron pair to acid Formation of a covalent bond H3N:+BF3>H3NBF3(adduct, coordinate (supplied by the same atom) or a dative bond Arrhenius and Bronsted Lowry acid/base can be lewis H3COCH3+HCl>H3COCH3H+Cl NH3+BF3>NH3BF3(adduct) Ag++2NH3>Ag(NH3)2 *practice identifying lewis acids and bases* Organic acids and bases strength of acids, depends on ability to give up protons Ha>A conjugate base anion (A) therefore the stability pfo the charge determines strength of the organic acid Stability of the conjugate anion (ARIO) * look in powerpoint for examples * 1 What atom is the charge on? electronegativity for the atoms in the same row size for atoms in the same atom 2 Resonance formation of resonance structures stabilizes a negative charge 3 Induction more stable: electron withdrawing groups stabilizes negative charges, electron donating groups destabilizes charges 4 Orbitals Different hybridizes orbitals have different shapes 1 Sp more stable than sp2, sp2 more stable than sp3 Sp held closer to nucleus and is therefore more stable Review hybridization Chapter 18 Aqueous Ion Equilibrium Buffers defined as solutions that resist pH when an acid/base is added neutralize an acid or base limit and pH can change made by mixing a solution of a weak acid soluble salt with a conjugate base anion (example is blood) How Acid Buffers Work 1.addition of base HA+ H2O> A+H3O+ Le Chatelier's Principle to weak acid equilibrium Significant amounts of HA React with added base to Neutralize it > Example 1: H A+AO>HA+H3O+ pH will not shift too much A> 2.Addition of acid HA+H2O←A+H3O+ significant amounts of A make more HA HA>A+H3O+ H3O+ ← Common Ion effect HA+H2O>A+H3O+ Add salt containing anion NACl, conjugate base of the acid (the common ion, shifts equilibrium to the left pH increases pH acid , decreases H3O+ Example 2: .1 HC2H3O2 +.1NAC2H3O2 H2H3O2>CH3O2 +H3O+ HA A 0.1 0.1 Ka=(0.1+x)x/0.1x the x is small so it cancels x=1.8x10^5 x +x +x log(1.8x10^5)=4.7 Example 3: pH=Pka if [HA]=[A] pH? 0.14 M HF pka=3.14 0.071 M KF HF>F+H30+ ka=10^pka=ka=7.0x10^4 0.071x/0.14=1x10^4 0.14 0.071 log(1.4x10^3)=2.85 x +x +x HendersonHasselbalch Equation 2 calculate pH of a buffer using Ka expression calculate pH of buffer from pKa and initial concentration of conjugate base pH=pKa+log[conj base anion]/[weak acid] Example 4: .05 MHC7H5O2 .15NaC7H5O2 Ka= 6.5x10^5 HC7H5O2+H2O>C7H5O2+H3O+ log(ka)=4.187 4.187+log(.15/.05)=.66 Equilibrium of equation If x is small initial concentration are not dilute Ka small Initial 100 to 100x bigger than kA How much does pH change with a buffer calculate new pH add acid/base 1st 2nd stoichiometric calculation new initial [HA] and [A] new Example 5: .1 HCH3O2 0.1 NAC2H3O2 1 L 0.010 mol HC2H3O2+OH > C2H3O2+H2O Mol B 0.1 0.10.01 0.01 mol 0.090/L 0.01/1 0.11 0.14 Mol A 0.01 +0.01 0.01 End 0.090 0.100 0 0.11x/0.090=1.8x10^5 x=1.42 x10^5 log(1.42x10^).83 Example 6: .14 HF pKa= 3.15 0.071 HF 1 L 0.020 HCl F H3O+ HF 0.071 0.020 0.140 0.051/1 =0.051 M 0.020 0.020 +0.020 pH=3.15+ log(0.051/0.160)= 2 .65 0.051 0 0.160 Basic Buffers B:+H2O> H:B+ +OH Weak base (B:) with soluble salt of conjugate acid H:B+Cl Henderson Hasselbalch pH=pKa+log(B:/HB) pKa+Kb=14 Example 7: NH3+H2O>NH4++OH pKa= 14 4.75=9.25 pH=9.25+log(0.50/0.20)= 9.65 Example 8:NH3+H2O>NH4++OH pOH=4.75log(0.20/0.50)=4 .35 Buffering Effectiveness A good buffer should be able to neutralize moderate amounts or acid/base Limit that can be added before pH changes Buffering capacity amount of acid/base can be neutralized Buffering range pH range the buffer can be effective 1.relative amount of acid/base 2.absolute concentration of acid/base ● Check % change to determine which one is more effective* ● Buffer most effective with equal concentrations of acid/base Effect of Absolute Concentrations of Acid and Conjugate Base 3 Buffer 1 Buffer 2 0.50 mol HA+0.50 A 0.50 mol HA 0.50 mol A 5.00 5.00 HA A OH HA A OH 0.50 0.50 0 0.050 0.50 0.010 0.010 0.49 0.51 0.040 0.060 5.02 pH 5.18 pH Effectiveness of Buffers [base]:[acid]=1 most effective 0.1<[base]:[acid] Buffering Range 0.1<[base]:[acid]<10 lowest pH maximum pH=pKa pH=Ka+1 effective range pKa= plus or minus one (closest to pH) Example 9: pH 4.25 formic acid HCHO2 pKa=3.74 Ratio? base>acid 0.51=log[CHO2]/[HCHO2] [CHO2]/[HCHO2]= 3.24 Example 10: benzoic acid HC7H5O2 pKa= 4.19 pH=3.75 3.75=4.19 0.363= [C2H5O2]/[HC7H5O2] Buffering Capacity Run out when a large change in pH happens Increase with increase abs concentration of buffer [base]:[acid]=1 buffer increases Acid [base]>[acid] base [base]<[acid] Titration Unknown acid/base (titrant) added to concentration until endpoint is reached Indicator determines endpointphenolphthalein H3O+=OH equivalence point nH3O+=nOH Titration Curve pH vs titration Acidic salt pH<7 basic salt pH>7 neutral salt=7 Strong acid strong base titration is not on exam Example 11: pH 0.15 M NaOH add 50 mL 0.25 M HNO3 HNO3+NaOH>NaNO3+H2O Initial pH log(0.250)=0.60 Mol HNO3 =0.05 L (0.25)=1.25x10^2 H3O+ 1.5x10^3 mol NaOH nH+>nOH HNO3 NaNO3 NaOH pH=1.74 1.25x10^2 1.5x10^3 1.5x10^3 1.5x10^3 1.1x10^3 1.5x10^2 0.0183 M 0.025 M Titration weak acid with strong base 4 excess in acid initial pH using Ka of weak acid pH equivalence point Kb of conjugate base of weak acid pH after equivalence strong base pH in acid region determined like buffer Example 12: titration 25 mL 0.100 M HCHO2 with 0.100 NaOH HCHO2+NaOH>NaCHO2+H2O Initial pH 0.100 1.8x10^4=x^2/0.100 = log(4.24x10^3) =2 .37 x +x +x HA A OH Before equivalence 0.0250(0.1)=2.5x10^3 5mL NaOH 2.5x10^3 3.24 ___ =3.14 5x10^4 mol NaOH henderson hasselbalch log(1.8x10^4) 5.0x10^4 At equivalence HA A OH 2.5x10^3 mol NaOH 0 5.6x10^11= x^2/0.0500 0 2.5x10^3 0 x=1.7x10^6 M= log(1.7x10^6)= 8.23 +x 0.05 x After equivalence 0.0250 L x0.100=2.5x10^3 3x10^3 mol NaOH 0.0050/0.025+0.30=0.00909 M NaOH HA A NaOH log(0.00909)=14 ______ =1 1.96 2.510^3 3.0x10^3 0 2.5x10^3 5x10^4 Half Equivalence > pH=pKa Titrating a weak acid with a strong base I. Initial pH weak acid solution [HA]>[H3O+[>pH II. Before equivalence point, buffer mol Hai+ mol Ai stoich pH=pKa=log[A]/[HA] III. at equivalence point mol HA=mol base, only has conjugate base anion before equilibrium mol A =mol HA> readjusttotal volume [A]i=mol A/total L [A]>[OH]>pOH>pH IV. beyond OH is in excess OH=mol OH/total V [H3O+][OH]=1x10^14 Half neutralization pKa=pH 1:1 buffer Want more practice look on powerpoint! Titration polyprotic acid H2SO3 Ka1>Ka2 2 equivalence points, closer the Kas the less stable the acid H2SO3+H2O> HSO3+H3 or HSO3 +H2O>SO2+H3O+ Monitoring pH during titration Measure H3O+ > endpoint at equivalence point Monitor using indicate Indicators Weak acid dyes change color depending on pH HInd+H2O>Ind +H3O+ Color depends on concentration of Ind:HInd 5 Ind:HInd about equals 1 mix, Ind:HInd>10 Ind:HInd<1 HInd Titration with Indicator Curve large change in pH near equivalence point Endpoint titration changes color pKa #Ind about pH Solubility Equilibria Ionic compounds dissolve, can be insoluble in comparison to equilibrium Solubility product solid salt > aqueous ions Ksp Ksp increases more soluble ionic solid MnXm MnXm>nMm++mXn Ksp=[Mm+]^n[Xn]^m Example 13: PbCl3>Pn2++2Cl K sp=[Pb2+][Cl]^2 6