6.01 Basics of Chemical Equilibrium

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1 6.01 Basics of Equilibrium Constant Equilibrium and Thermodynamics Solubility Products Complex Formation Protic Acids and Bases Strength of Acids and Bases Solving Equilibrium Problems Recommended Problem Set: 12, 19, 25, 36, 50 Dr. Fred Omega Garces Chemistry 251 Miramar College 1

2 Equilibrium: Mass Action Expression When equilibrium is establish, A B illustrates that rate forward = rate reverse or K f [A] = K b [B] Rearranging this equation yields K f /K b = [B] / [A] which yields The mass action expression : K eq = [B] / [A] For any generic chemical process at equilibrium aa + bb pp + qq A mass action expression can be written: [ ] p [ Q] q K eq = P [ A] a B [ ] b This is also refer red to as the Law of Mass Action 2

3 Equilibrium: Meaning of K eq Review of Fractions x y < 1 : If x < y x y > 1 : If x > y x y = 1 : If x = y A) For x < y, the denominator dominates. B) For x > y, the numerator dominates. C) For x = y, numerator and denominator are equal. 3

4 Equilibrium Effects of K eq and Variation of a Chem Equation Consider the variation of a chemical reaction: A B K eq = [B] / [A] B A K rev = [A] / [B] K rev = 1 / K eq 1/2 A 1/2 B K = [B] 1/2 / [A] 1/2 K = (K eq ) 1/2 2B 2A K = [A] 2 / [B] 2 K = (K eq ) -2 Note: K rev = 1 / K eq or K eq = 1 / K rev K = (K eq ) 1/2 or K eq = K 2 K = (K eq ) -2 or K eq = (K ) -1/2 4

5 Thermodynamics: Tabulation of Thermodynamic Parameters For any Thermodynamic function: Function ΔX rxn = Σ n ΔX prod - Σ n Δ X react Enthalpy: ΔH rxn = Σ n ΔH prod - Σ n ΔH react Entropy: ΔS rxn = Σ n S prod - Σ n S react Free Energy: ΔG rxn = Σ n Δ G prod - Σ n Δ G react Example: Be(OH) 2 (s) BeO (s) + H 2 O (g) S J/mol K ΔS rxn = Σ n S prod - Σ n S react ΔS rxn = [ ] ΔS rxn = J / mol K

6 Thermodynamics: Free Energy and Spontaneity Defining a new State function ΔG: -T ΔS univ = ΔT, ΔP= 0 Consider ΔG = - T ΔS univ ΔG < 0 Spontaneous ΔS univ > 0 ΔG = equilibrium ΔS univ = 0 ΔG > 0 nonspontaneous ΔS univ < 0 (rev is spontaneous) 6

7 Thermodynamics: Relationship Between K eq and ΔG What is the relationship between ΔG and a system at Equilibrium? ΔG = ΔG + RT ln Q at equilibrium, Q = K eq and ΔG = 0 but ΔG 0 0 = ΔG + RT ln K eq therefore, ΔG = -RT ln K eq - ΔG = ln Keq RT K eq = Exp {-ΔG /RT} 7

8 Thermodynamics: ΔG Equations ΔG ΔH - Τ ΔS - RT ln K eq Σ n ΔG prod - Σ n ΔG rxn ΔG - RT ln Q 8

9 Solubility Product: Solution: Ionic Vs Covalent Compounds: Electrolyte Vs. Nonelectrolyte Substance when dissolve can break-up to ions (NaCl) or stay intact (sugar). Type: % ionization: Solubility Electrolyte: conducts electricity. Strong electrolyte 100 % ionization very soluble weak electrolyte less 100% ionization slightly soluble Nonelectrolyte: no conduction zero ionization insoluble. 9

10 Solubility Product: Solubility Rules How can one predict if a precipitate forms when mixing solutions? Solubility rules - Soluble Substances containing Exceptions Insoluble substances containing Exceptions nitrates, (NO 3- ) chlorates (ClO 3- ) perchlorates (ClO 4- ) acetates (CH 3 COO - ) None carbonates (CO 3 2- ) phosphates (PO 4 3- ) Slightly soluble halogens (X - ) X - = Cl -, Br -, I - sulfates (SO 4 2- )* G-1, Heavy G-2, NH 4 + Ag, Hg, Pb hydroxides (OH - ) alkali, NH 4+, Ca, Sr, Ba Ba, Hg, Pb None Solubility rule provides information on the specie that will form a precipitate. In general, a concentration of 0.01M or greater for a substance is considered soluble. 10

11 Solubility Product: Precipitation Reaction Consider the reaction - i AgNO 3 (aq) + HCl (aq) AgCl (s) + HNO 3(aq) Net ionic equation: Ag + (aq) + Cl- (aq) AgCl (s) Write the reverse, (standard convention) AgCl (s) Ag + (aq) + Cl- (aq) Then K eq = K sp = [Ag + ] [Cl - ] K sp - solubility product constant An indicator of the solubility of substance of interest. Yields info. on the amount of ions allowed in solution. (Must be determine experimentally) If written in this form, then mass action is: K eq = 1 [Ag + ][Cl ] NaOCH 2 CH 3 (s) + H 2 O (l) D Na+ (aq) + OH- (aq) + CH 3 CH 2 OH (aq) K sp = [Na + ] [OH - ] [CH 3 CH 2 OH] 11

12 Solubility Product: Common Ion Effect (LeChatelier Principle) For a system containing a precipitate in equilibrium with its ions, solubility of solid can be reduced if an ionic compound is dissolved in the solution. 13_07 13_07 13_07 Ag + I - Ag + I - Ag + I - AgI (s) AgI (s) Add I - via NaI AgI (s) AgI (s) D Ag + (aq) + I- (aq) add common ion i.e., NaI to solution Direction of Reaction 12 Reduce solubility i.e., less AgI will dissolve in soln which means more ppt. forms in solution

13 Solubility Product An example of no common ion What is the solubility of Ag 4 Fe(CN) 6 (s), K sp = Ag 4 Fe(CN) 6 (s)! 4 Ag + (aq) + Fe(CN) 6 4- i Excess 0 0 Δ -s + 4s + s [e] Excess 4s s (aq ) = [Ag + ] 4 [Fe(CN) 4-6 ] = [4s] 4 [s] = 256 s = [s] 5 s = = M 13

14 Solubility Product An example of common ion What concentration of Fe(CN) 4-6 (ferrocyanide) is in equilibrium with 1.0µM Ag + and Ag 4 Fe(CN) 6 (s), K sp = Ag 4 Fe(CN) 6 (s)! 4 Ag + (aq) + Fe(CN) 6 4-(aq ) i Excess Δ - s + 4s + s [e] Excess s s Assume 4 s << = [Ag + ] 4 [Fe(CN) 6 4- ] = [ s] 4 [s] = [ ] 4 [s] s = [ ] 4 = M = [Fe(CN) 6 4- ] 14

15 Quotient: Review For reactions not yet at equilibrium, the Law of Mass action yield information in terms of the Reaction Quotient. Consider the following chemical process not at equilibrium. aa + bb rr + pp A reaction quotient expression can be written: Q = [R]r [P] p [A] a [B] b Where the numerical value of Q will determine the direction the reaction will proceed. Q < K eq Reaction shifts to Right Q > K eq Reaction shifts to Left 15

16 Reaction Quotient: Direction of Reaction Consider, T =532 C, K c = 0.19 what direction will the reaction proceed if the initial concentration is change to- N 2 (g) + 3H 2 (g) 2NH 3 (g) i) 0.30 M 0.20 M 0.10 M Q = [ 0.10M] 2 [ 0.30M] 0.20M [ ] 3 = 4.2 Q = 4.2 > K c = K c 4.2 Q Direction of Reaction: Proceeds to the left 16

17 Solubility Product: Selective Precipitation Qualitative Analysis Q < K sp, solid dissolve until Q=K sp Q = K sp, equilibrium Q > K sp, ppt until Q=K sp Grp1: Insoluble Chlorides Ag +, Pb 2+, Hg 2+ 2 Grp2: Acid-Insoluble sulfides Cu 2+, Bi 3+, Cd 2+, Pb 2+, Hg 2+, As 3+, Sb 3+, Sn 4+ Grp3: Base-Insoluble sulfides Cu 2+, Al 3+, Fe 2+, Fe 3+, Co 2+, Ni 2+, Cr 3+, Zn 2+, Mn 2+ Grp4: Insoluble Phosphates Ba 2+, Ca 2+, Mg 2+ Grp5: Alkali Metals and NH + 4 Na +, K +, NH

18 Solubility Product Selective Precipitation(2) A solution contains M Ag + and M Pb 2+. If NaI is added, will AgI (K sp = ) or PbI 2 (K sp = ) precipitate first? Specify the concentration of I - needed to begin precipitation. AgI (s)! Ag + (aq) + I - (aq ) i Excess Δ -s s s [e] Excess s s Assume s << and PbI 2 (s)! Pb 2+ (aq) + 2I - (aq ) i Excess Δ -x x 2x [e] Excess x 2x Assume x << and = [Ag + ] [I - ] = [ ] [s] = [ ] [s] = [Pb 2+ ] [I - ] 2 = [ x ] [2x] = [ ] [4x 2 ] s = = M = [I - ] x = = M = [I - ] 18 AgI will precipitate first. The [I - ] concentration needs to be M, PbI 2 will not precipitate until the [I - ] concentration reaches M

19 Complex Formation: Complex Ion Process Metals forming Complexes: Metals by virtue of electron pairs will act as Lewis acids and bind a substrate to form a Complex Consider the following: Ag(NH 3 ) 2 + (aq) Complex ion AgCl (s) Ag + (aq) + Cl- (aq) Ag + (aq) + 2NH 3(aq) Ag(NH 3 ) 2 + (aq) AgCl (s) + 2NH 3(aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) Normally insoluble AgCl can be made soluble By the addition of NH 3. The presence of NH 3 drives the top reaction to the right and increase the solubility of AgCl An assembly of metal ion and the Lewis base (NH 3 ) is called a complex ion. The formation of this complex is describe by Formation Constant Ag + (aq) + 2NH 3(aq) Ag(NH 3 ) 2 + (aq) K f = [Ag(NH 3 ) 2 + ] [Ag + = ] [NH 3 ] 19 Solubility of metal salts affected presence of Lewis Base: e.g., NH 3, CN -, OH -

20 Complex Formation: Effect on Solubility Consider the solubility of PbI2: Pbl2 (s) Pb2+(aq) + 2 I- (aq) Ksp = High concentration of I- however increases the solubility of PBI2. Pb+2(aq) + l- (aq) Pb+2(aq) + 2l- (aq) Pbl2 (aq) β2 = Pb+2(aq) + 3l- (aq) Pbl3- (aq) β3 = Pb+2(aq) + 4l- (aq) Pbl4-2 (aq) β4= Pbl+ (aq) K1 = β1 = Total solubility of PbI2 and solubilities of individual species as a function of the concentration of the concentration of free iodide. β are known as formation (Kf) constants. 20

21 Complex Formation: Effect on Solubility Given the equilibria below, calculate the concentration of each Zn +2 in solution, with Zn(OH) 2 (s) containing fixed concentration [OH-] = M Zn(OH) 2 (s) Zn 2+ (aq) + 2 OH- (aq) K sp = Zn +2 (aq) + OH- (aq) ZnOH + (aq) K 1 = Zn +2 (aq) + 2OH- (aq) Zn(OH) 2 (aq) β2 = Zn +2 (aq) + 3OH- (aq) Zn(OH) - 3 (aq) β3 = Zn +2 (aq) + 4OH- (aq) Zn(OH) -2 4 (aq) β4= [OH - ] = M, and from K sp calculation, see first equation below, [Zn 2+ ] = M Zn(OH) 2 (s) Zn 2+ (aq) + 2 OH - (aq) K sp = K sp = [Zn 2+ ] [OH - ] 2 [Zn 2+ ] = ( ) = M Zn 2+ (aq) + 1 OH - (aq) ZnOH + (aq) β 1 = β 1 = [ZnOH + ] [OH - ] [Zn +2 ] [ZnOH + ] = ( M) ( M) = M Zn 2+ (aq) Zn 2+ (aq) Zn 2+ (aq) + 2 OH - (aq) + 3 OH - (aq) + 4 OH - (aq) Zn(OH) Zn(OH) 2 (aq) β 2 =K 1 = β 2 = K 1 = 2 [Zn(OH) [OH - ] 2 [Zn +2 2 ] = ( M) 2 ( M) = M ] Zn(OH) - Zn(OH) 3 β (aq) 3 = β 3 = 3 [Zn(OH) [OH - ] 3 [Zn +2 3 ] = ( M) 3 ( M) = M ] -2 Zn(OH) -2 Zn(OH) 4 β (aq) 4 = β 4 = 4 [Zn(OH) -2 [OH - ] 4 [Zn +2 4 ] = ( M) 4 ( M) = M ] [Zn 2+ ] Total = Zn +2 + [ZnOH + ] + [Zn(OH) 2 ] + [Zn(OH) 3 ] + [Zn(OH) 4-2 ] = M M M M M = M 21

22 Complex Formation: Effect on Solubility Given the equilibria below, calculate the concentration of each Zn +2 in solution, with Zn(OH) 2 (s) containing fixed concentration [OH-] = M Zn(OH) 2 (s) Zn 2+ (aq) + 2 OH- (aq) K sp = Zn +2 (aq) + OH- (aq) ZnOH + (aq) β 1 = Zn +2 (aq) + 2OH- (aq) Zn(OH) 2 (aq) β2 = K 1 = Zn +2 (aq) + 3OH- (aq) Zn(OH) - 3 (aq) β3 = Zn +2 (aq) + 4OH- (aq) Zn(OH) -2 4 (aq) β4= [OH - ] = M, and from K sp calculation, see first equation below, [Zn 2+ ] = M Zn(OH) 2 (s) Zn 2+ (aq) + 2 OH - (aq) K sp = K sp = [Zn 2+ ] [OH - ] 2 [Zn 2+ ] = ( ) = M Zn 2+ (aq) + 1 OH - (aq) ZnOH + (aq) β 1 = β 1 = [ZnOH + ] [OH - ] [Zn +2 ] [ZnOH + ] = ( M) ( M) = M Zn 2+ (aq) Zn 2+ (aq) Zn 2+ (aq) + 2 OH - (aq) + 3 OH - (aq) + 4 OH - (aq) Zn(OH) Zn(OH) 2 (aq) β 2 =K 1 = β 2 = K 1 = 2 [Zn(OH) [OH - ] 2 [Zn +2 2 ] = ( M) 2 ( M) = M ] Zn(OH) - Zn(OH) 3 β (aq) 3 = β 3 = 3 [Zn(OH) [OH - ] 3 [Zn +2 3 ] = ( M) 3 ( M) = M ] -2 Zn(OH) -2 Zn(OH) 4 β (aq) 4 = β 4 = 4 [Zn(OH) -2 [OH - ] 4 [Zn +2 4 ] = ( M) 4 ( M) = M ] [Zn 2+ ] Total = Zn +2 + [ZnOH + ] + [Zn(OH) 2 ] + [Zn(OH) 3 ] + [Zn(OH) 4-2 ] = M M M M M = M 22

23 Selected Complex Formation Formation Constants Table 23

24 Acids-Bases: Brønsted-Lowry Brønsted - Lowry definition Acid - Proton H + (H 3 O + ) donor Base - Proton H + (H 3 O + ) acceptor. example: acids: HCl (aq) H+ (aq) + Cl - (aq) Bases: NH 3 (aq) NH 4 + (aq) HCl (aq) + NH 3 (aq) NH 4 + (aq) + Cl - (aq) In an acid - base reaction, H + & OH - always combine together to form water and an ionic compound (a salt): HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq) 24

25 Acid-Bases: Conjugate Pairs Neutralization reaction is a reaction between acids and base NH 3 + HCl NH Cl - Base 1 Acid 2 Conj Acid 1 Conj Base 2 Conjugates differ by a proton (H + ) Bases NH 3 : Bronsted-Lowery: proton acceptor (Conjugate Base 1 ) Cl - : Bronsted-Lowery: proton accept (Conjugate Base 2 ) Acids HCl: Bronsted-Lowery: proton donor (Conjugate Acid 2 ) NH 4+ : Bronsted-Lowery: proton donor (Conjugate Acid 1 ) 25

26 Acid-Bases: Protons in water, Hydronium Hydronium ion: HCl (aq) H + (aq) + Cl- (aq) Proton? H H + attacks water solvent to form hydronium H 3 O + (H 5 O 2+ ) e - p Proton in H 2 O H+ is always associated with solvent. In reality: H + + H 2 O H 3 O + H + protons & H 3 O + Hydronium ion H + (aq) H 3 O + considered the same. These terms are used interchangeably 26

27 Amphiprotic (Amphoteric) Species Specie which can act as either a base or an acid by donating or accepting protons are considered amphiprotic substances. Under certain conditions, chemicals can either give up a proton or accept a proton. i.e., H 2 O, HCO 3 -, HPO 4 2 -, H 2 PO 4 - Consider Hydrogen carbonate (bicarbonate) : Acid process: H 2 CO 3 + H 2 O g H 3 O + + HCO - 3 1st H+, K a1 = HCO H 2 O g H 3 O + + CO nd H+, K a2 = Base process: CO H 2 O g OH - + HCO - 3 1st OH-, K b1 = HCO H 2 O g OH - + H 2 CO 3 2nd OH-, K b2 = Note from the above reaction that K a1 competes against K b2 these two equilibrium constant are related by K w : K b2 = k w / K a1 To determine acidity or basicitiy of an amphiprotic solution, the K a s and K b s must be weighed against each other. 27

28 Acid-Base: Autoprotolysis of water Autoionization (Autoprotolysis) of water Why does water have a ph of 7? Water is Amphoteric (it reacts with itself) 2 in 1 billion self-ionize. H 2 O (l) + H 2 O (l) + E H 3 O (aq) + OH- (aq) Endothermic reaction ΔH rxn = kj, K eq = K w K w (ion-product constant) = 1 25 C 28

29 Acid-Base: Strong Acids and Bases Strong Acids Strong Electrolyte that ionizes (break up to H + ions) in solution 100 % of time. Examples : * H 2 SO 4, HClO 4, HClO 3, HNO 3, HX (X=Cl, Br, I) Only strong acids dissociates 100% in water, K a >>1 These are Strong Acids. Rxn: HClO 4 + H 2 O g H 3 O + + ClO 4-100% Dissociation HNO 3 + H 2 O g H 3 O + + NO 3-100% Dissociation H 2 S + H 2 O D H 3 O + + HS - Less 100% Dissociation *All other acids under goes less than 100 % Dissociation. 29

30 Acid-Base: Strength Strong Acids Conjugate Acid Conjugate Base Name Formula Formula Name Perchloric acid HClO 4 ClO - 4 Perchlorate ion Sulfuric acid H 2 SO 4 HSO - 4 Hydrogen sulfate ion Hydrochloric acid HCl Cl - Chloride ion Nitric acid HNO 3 NO - 3 Nitrate ion Hydronium ion H 3 O + H 2 O Water Hydrogen sulfate ion HSO - 4 SO 2-4 Sulfate ion Phosphoric acid H 3 PO 4 H 2 PO - 4 Dihydrogen phosphate ion Acetic acid CH 3 CO 2 H CH 3 CO - 2 Acetate ion Carbonic acid H 2 CO 3 HCO - 3 Hydrogen carbonate ion Hydrogen sulfide H 2 S HS - Hydrogen sulfide ion Dihydrogen phosphate ion H 2 PO - 4 HPO 2-4 Hydrogen phosphate ion Ammonium ion NH + 4 NH 3 Ammonia Hydrogen cyanide HCN CN - Cyanide ion Hydrogen carbonate ion HCO - 3 CO 2-3 Carbonate ion Water H 2 O OH - Hydroxide ion Ethanol C 2 H 5 OH C 2 H 5 O - Ethoxide ion Ammonia NH 3 NH - 2 Amide ion Hydrogen H 2 H - Hydride ion Methane CH 4 CH - 3 Methide ion Strong Base 30

31 Acid -Base, Strength of Weak Acids The % ionization is dependent of the K eq constant. For acids, this is called K-acid or K a. 31

32 Acid-Base: Strong Base Strong Base: Strong Electrolyte that ionizes (break up to OH - ions) in solution 100 % of time. Examples : Group-1 hydroxide: LiOH, KOH, NaOH, RbOH Heavy Grp-2 hydroxides: Ca(OH) 2, Sr(OH) 2, Ba(OH) 2, Ionic metal oxides: Li 2 O, Na 2 O, K 2 O, CaO These are bases which dissociate 100% in water (or in the case of oxides, react with water) to produce hydroxides. Rxn: KOH OH - + K + 100% Dissociation Ca(OH) 2 2OH - + Ca % Dissociation Li 2 O + H 2 O 2OH - + Li + 100% Dissociation CaO + H 2 O Ca(OH) 2 strong base product NH 3 + H 2 O OH - +NH 4 + Less 100% Dissociation *All other bases undergoes less than 100 % Dissociation. 32

33 Acid-Base: Strength of Weak Bases Weak bases produces OH - at less than 100%? Base Examples : NH 3 + H 2 O D OH - + NH 4 + Less 100% Dissociation (Weak Base) HS - + H 2 O D OH - + H 2 S Less 100% Dissociation (Weak Base) In a base equilibrium process, hydroxides are formed. When the hydroxide formation is not stoichiometric, the strategy of equilibrium analysis must be used. 33

34 Acid-Base: Relationship between K a and K b We note that weak acid have conjugate strong base. This suggest there is a relationship between the k a of the weak acid and the K b of the strong base. That is the strength of an acid and the strength of its conjugate base is expressed by: K a K b = K w Consider (1) HF + H 2 O D H 3 O + + F - K a = (2) F - + H 2 O D OH - + HF K b =? k b = [HF] [OH- ] [F - ] k a = [H 3 O+ ] [F - ] [HF] k a = [H 3 O+ ] [F ] [OH ] k b [F ] k b [F - ] [OH - ] =! " HF # $ Plug " k a k b = H 3 O + % # $ & ' " % # $ OH &' = K w k a k b = K w 34

35 Acid-Base: Leveling Effect When dealing with Bronsted-Lowry Acid-Base systems, all reactions takes place in water and acid systems yield H 3 O +, where as base systems yield OH -. The question that comes to mind is why are strong acids and strong bases equally strong in water? Consider putting HCl or HNO 3 in water, as soon as it goes into water, H 3 O + forms. In fact all strong acid dissociate completely and react with water to form H 3 O +, in fact HCl and HNO 3 no longer exist. Any strong acid considered will simply donate it s proton to H 2 O and form H 3 O +. The same is true for strong base, any strong base will accept a proton from H 2 O to form OH -. In water, the strongest acid is H 3 O + and in base the strongest base is OH -. This phenomena is called the Leveling effect. Water equalizes (level) the strength of all strong acids by reacting with the acid to form products of water s autoionization. To rank strong acids in terms of relative strength require using a different solvent system. i.e., dissolve HCl, HNO 3 in acetic acid. 35

36 Summary: Equilibrium constant The mass action expression is equal to the equilibrium constant. The equilibrium constant value are generally written without units. The equilibrium expression for a reaction is the reciprocal of that reaction when it is written in the reverse direction. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the factored reaction is the original expression raised to the nth power. Thus K2 = (k1) n The equilibrium constant express in terms of concentration, Kc, is related to the equilibrium constant expressed in terms of partial pressure, Kp, by the amount of gas moles changes during the course of the reaction. 36

37 Summary The following summary lists the important tools needed to solve problems dealing with acid-base equilibria. Equation / Concept Function 1 [H + ] [OH - ] = K w Permits the calculation of [H + ] or [OH - ] when the other is known. 2 p X = - log X This equation is the basis of the p-scale. 3 ph + poh = This equation shows the relationship between the ph and the poh 4 HA! H + + A - K a = [H+ ] [A ] [HA] 5 B + H 2 O! HB + OH - K b = [HB] [OH ] [B] 6 Percent ionization (α) amount ionized α = 100% initial amount This is the Mass Action Equation for the ionization of a weak acid in water. This equation yields the k a given the equilibrium concentration of all specie. The equation also yields the [H 3 O+] given the initial concentration of the weak base [HA] and the k a. This is the Mass Action Equation for the ionization of a weak base in water. This equation yields the k b given the equilibrium concentration of all specie. The equation also yields the [OH-] given the initial concentration of the weak base [B] and the k b. The percent ionization can be calculated from the initial concentration of the acid (or base) and the change in the concentration of the ions. Given the percent ionization (α) and the ph, the k a (or k b ) can be determined. 7 K a K b = K w This equation relates K a and K b for conjugate pairs in aqueous solution, 8 ID of the solute as : i) only a weak acid ii) only a weak base iii) a mixture of a weak acid and its conjugate base Identification of the function of the solute leads to the correct Mass Action expression and thereby leading to the correct equilibrium law. This is a critical first step to solve any acid-base equilibria 9 Identification of acidic cations and basic anions 10 Assumption which simplifies Mass Action 11 Reactions when H+ or OHare added to a buffer solution. Identification of function of cation and anion of a salt lead to ph of the salt solution. Given the k a or k b of the conjugates of these ions leads to the calculation of the ph or poh In order to simplify the math calculation of a Mass Action expression, assumption can be made base on the k a or k b value. Understanding the buffer reaction permits the determination of the effect of a strong acid or strong base on the ph of the solution. Adding H+ lowers the [A-] and raises [HA], adding OH- lowers [HA] and raises [A-]. 37

38 Lecture Notes Answers Slide19 [OH - ] = M, and from K sp calculation, see first equation below, [Zn 2+ ] = M Zn(OH) 2 (s) Zn 2+ (aq) + 2 OH - (aq) K sp = K sp = [Zn 2+ ] [OH - ] 2 [Zn 2+ ] = ( ) = M Zn 2+ (aq) + 1 OH - (aq) ZnOH + (aq) β 1 = β 1 = [ZnOH + ] [OH - ] [Zn +2 ] [ZnOH + ] = ( M) ( M) = M Zn 2+ (aq) Zn 2+ (aq) Zn 2+ (aq) + 2 OH - (aq) + 3 OH - (aq) + 4 OH - (aq) Zn(OH) Zn(OH) 2 (aq) β 2 =K 1 = β 2 = K 1 = 2 [Zn(OH) [OH - ] 2 [Zn +2 2 ] = ( M) 2 ( M) = M ] Zn(OH) - Zn(OH) 3 β (aq) 3 = β 3 = 3 [Zn(OH) [OH - ] 3 [Zn +2 3 ] = ( M) 3 ( M) = M ] -2 Zn(OH) -2 Zn(OH) 4 β (aq) 4 = β 4 = 4 [Zn(OH) -2 [OH - ] 4 [Zn +2 4 ] = ( M) 4 ( M) = M ] [Zn 2+ ] Total = Zn +2 + [ZnOH + ] + [Zn(OH) 2 ] + [Zn(OH) 3 ] + [Zn(OH) 4-2 ] = M M M M M = M 38

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