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• Aqueous Equilibria: Acid-Base Reaction Bronsted_Lowry or proton Transfer Reactions

Chaps(14-15) Hetero/Homogeneous Equilibria. • Aqueous Equilibria: Acid-Base Reaction Bronsted_Lowry or proton Transfer Reactions Acid donates H + Base accepts H + AH(aq)+H 2 O(aq)  H 3 O + (aq) + A - (aq) K a =exp{-  G°/RT} acid base conjugate conjugate

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• Aqueous Equilibria: Acid-Base Reaction Bronsted_Lowry or proton Transfer Reactions

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  1. Chaps(14-15) Hetero/Homogeneous Equilibria • Aqueous Equilibria: Acid-Base Reaction Bronsted_Lowry or proton Transfer Reactions Acid donates H+ Base accepts H+ AH(aq)+H2O(aq)H3O+(aq) + A-(aq) Ka=exp{-G°/RT} acid base conjugate conjugate acid base • pH=-log10[H3O+]: Acidic Solns pH< 7 Basic Solns pH>7 Neutral Soln pH=7 Pure water • Weak Acid/Base Equilibria: Ka<< 1 and Kb<<1 •Buffered solutions: pH=pKa – log10[HA]0/[A-]0 Final Exam: Monday, June 11th CS 50; 3:00 - 6:00 pm Ch 9, 10, (11-1-11.3, 11.5), (12.1.-.7), 13(Ex .5 & .8), 14(Ex .4 & .8), 15.-15.4 and 18 2 pages of hand written notes,

  2. Chaps(14-15) Hetero/Homogeneous Equilibria • Aqueous Equilibria: Acid-Base Reaction Bronsted_Lowry or proton Transfer Reactions Acid donates H+ Base accepts H+ AH(aq)+H2O(aq)H3O+(aq) + A-(aq) Ka=exp{-G°/RT} acid base conjugate conjugate acid base • pH=-log10[H3O+]: Acidic Solns pH< 7 Basic Solns pH>7 Neutral Soln pH=7 Pure water • Weak Acid/Base Equilibria: Ka< 1 and Kb<1 •Buffered solutions: pH=pKa – log10[HA]0/[A-]0

  3. Generalize to Solids, Liquids and Heterogeneous mixed phase reactions. Lets define the activity “a”a=P/Pref SoG= nRTln(P/Pref) goes G= nRTln(a) For none ideal gases ai=i (Pi/Pref)=i Pi Pref= I atm where I = activity the activity coefficient is different For different gases, Ar ≠ He Reactions in Solution G= nRTln(a) ai =i ([A]i/[A]ref) = i [A] [A]ref =1 Molar In any case a=1 for Pure Solids and Liquids K= aead/aaab

  4. Phase Transitions: evaporation and sublimation(?) B(l) B(g) K(T) =aB(g)/aB(l) aB(l)=1 and aB(g)= PB K(T) = PB Has as Liquid is present G°=-RTlnK Let K1 =K(T1) and K2= K(T2) lnK2 = - (H°/RT2) + S°/R assumes H and S are not a f(T) lnK1 = - (H°/RT1) + S°/R ln(K2/K1)= - H°/R{ (1/T2) - (1/T1)} van’t Hoff Eq

  5. Example H<0 H>0

  6. For B(l) B(g) ln(K2/K1)= - H°/R{ (1/T2) - (1/T1)} K2(T)=P T=T2 and T1=Tb Therefore K1(Tb)= 1 atm and H° = H°vap lnP= ln(K2/K1)= - H°/R{ (1/T2) - (1/Tb)} lnP = - (H°vap/R){ (1/T2) - (1/Tb)} P= exp{-(H°/R)[(1/T) - (1/Tb)]}

  7. P= exp{-(H°/R)[(1/T) - (1/Tb)]} P=1atm @ Tb

  8. P= exp{-(H°/R)[(1/T) - (1/Tb)]} Log scale!

  9. P= exp{-(H°/R)[(1/T) - (1/Tb)]}

  10. Chemical Reactions Acid Base Reactions AH(aq)+H2O(aq) H3O+(aq) + A-(aq) Ka=exp{-G°/RT} acid base conjugate conjugate acid base Ka= [H3O+][A-] /[AH] H2O(aq) + H2O(aq)H3O+(aq) + OH-(aq) Kw=exp{-G°/RT} acid base conjugate conjugate acid base Kw=[H3O+][OH-]= exp{-G°/RT} for H2O Ka=Kw G°= G°f(H3O+) + Gf(OH-) – 2Gf(H2O)

  11. Auto Ionization of water Kw=exp{-G°/RT} At T=25° Kw=10-14 =[H3O+][OH-] Water has no net charge Therefore [H3O+]=[OH-] 10-14 =[H3O+][OH-] 10-14 =[H3O+]2 [H3O+] =10-7mols/L=[OH-] pH=-log10 [H3O+]=-log1010-7 pH=-(-7)=7 Neutral water

  12. + The Hydrodrium Ion IR Spectroscopy Basically like Ammonia sp3 But with a positive charge

  13. Since these are electrolyte solutions charge/voltage measurements gives ion concentrations and therefore the pH

  14. Ka=exp{-G°/RT} pKa=-log10Ka

  15. Base B(aq) + H2O(l) OH- (aq) + BH+(aq) Kb=exp{-G°/RT}} Base acid conjugate conjugate base acid Kb=[OH-][BH+]/[B] NH3(aq) + H2O(l) OH- (aq) + NH4+(aq) Base acid conjugate conjugate base acid Kb=[OH-][NH4+]/[NH3] Ka= [NH3][H3O+]/[NH4+] KbKa=[OH-][H3O+]=Kw=10-14 Always true for aqueous equilibrium

  16. Aqueous Equilibria For the overall reaction: lL + bB  eE + fF Then the Equilibrium Constant which is only a function K= (aE)e(aF)f/(aL)l(aB)b For any reaction the Rate Rate= (-1/l)d[L]/dt=(-1/b)d[B]/dt=(1/e)d[E]/dt=(1/f)d[F]/dt

  17. lL + bB  eE + fF [B]0 D[F]=[F]eq - [F]0 F D[B]=[B]eq - [B]0 [B]eq B time

  18. Aqueous Equilibria For the overall reaction: lL + bB  eE + fF For any reaction the Rate Rate= (-1/l )d[L]/dt=(-1/b)d[B]/dt=(1/e)d[E]/dt=(1/f)d[F]/dt Change of concentration from the initial reactants to the reaction at Equilibrium (-1/l )D[L]=(-1/b) D[B]=(1/e) D[E]=(1/f)D[F]=x D[B]=[B]eq - [B]0=-bx D[F]=[F]eq - [F]0=fx [B]eq = [B]0 - bx and [F]eq = [F] + fx

  19. Aqueous Equilibria For the overall reaction: lL + bB  eE + fF K= (aE)e(aF)f/(aL)l(aB)b =( [E]eq)e([F]eq)f/( [L]eq)l([B]eq)b D[B]=[B]eq - [B]0=-bx D[F]=[F]eq - [F]0=fx [B]eq = [B]0 - bx and [F]eq = [F]0 + fx K= (aE)e(aF)f/(aL)l(aB)b K=( [E]0 + ex )e([F]eq = [F]0 + fx )f/([L]0 - fx )l( [B]0 - bx )b

  20. Example of and Acid/Base application Aqueous Equilibria AH(aq) + H2O(l) H3O+(aq) + A-(aq) Ka=exp{-G°/RT} acid base conjugate conjugate Acid base K= (aE)e(aF)f/(aL)l(aB)b K=( [E]0 - ex )( [F]0 + fx )/( [L]0 - fx )( [B]0 - bx ) Assume [H3O+]0=0 K=( [A-]0 + x )(x)/( [AH]0 - x ) if [A-]0 = 0 conjugate base K=( x )(x)/([ [AH]0 - x ) = x2/( [AH]0 - x ) x2 - Kx + K[AH]0 =0 solve for x

  21. Example of and Acid/Base application Aqueous Equilibria AH(aq) + H2O(l) H3O+(aq) + A-(aq) Ka=exp{-G°/RT} acid base conjugate conjugate Acid base x2 - Kx + K[AH]0 =0 solve for x If Ka>>1 then x=[AH]0 pH=-log10 x =-log10[H3O+] strong Acids If Ka<<1; x is small therefore x2=Ka[AH]0 x=(Ka[AH]0)1/2 Then find the pH=-log10 (Ka[AH]0)1/2

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