Chemical Equilibrium. What is the standard state for solutes? a) 1.00 b) 1 M c) 100% What is the standard state for gases? a) 1 bar b) 1.

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1 Chemical Equilibrium Equilibrium constant for the reaction: aa + bb + cc + dd + [C ] c [D ] d... equilibrium constant K = [ A] a [B ] b... [] = concentration relative to standard state molarity (M): for dilute solute concentration in bar: for a gas (i.e., [ ]= P) 1: pure liquid or solid as a second phase 1: solvent (H 2 O) equilibrium constant of the reverse reaction would be 1/K. isf n reactions are added, the new K is the product of the individual equilibrium constants for each of the n reactions What is the standard state for solutes? a) 1.00 b) 1 M c) 100% What is the standard state for gases? a) 1 bar b) 1.00 c) 100% What is the standard state for solids and liquids? a) 1.00 b) pure solid or liquid c) 100% A reaction is thermodynamically favored if: a) K<1 b) K=1 c) K>1 Thermodynamics K = e- ΔGo/RT = e- ΔHo/RT e ΔSo/R because ΔG = ΔH - TΔS ΔG = Gibbs free energy H = enthalpy change (+ = endothermic, - = exothermic) S = entropy change ( + = more disorder, - = less disorder) A reaction is spontaneous if ΔG o <0 or if K>1 (thermodynamically). The size of K does not say anything about the kinetics (i.e. rate of reaction)

2 Le Châtelier s principle when a system at equilibrium is disturbed by a change, the system will proceed back to equilibrium in the direction that will partially offset the change compare the reaction quotient Q to K [C ] c [D ] d... [ A] a [B ] b... where [] = existing concentration Q = if Q > K, excess product, reaction goes left if Q < K, excess reactant, reaction goes right heat = product for exothermic reaction heat = reactant for endothermic reaction Solubility product Ag 2 CO 3 (s) 2 Ag + (aq) + CO 3 (aq) solubility product constant: K sp = [CO 3 ][Ag + ] 2 true only if solid is present (independent of amount of solid) common ion effect = application of Le Châtelier s principle if a solid is added to a solution containing one of its constituent ion, less will dissolve The mathematical equation which represents the solubility product when the insoluble compound Mn 2 S 3 is dissolved in water is a) [Mn 3+ ][S ] = K sp. b) [Mn 3+ ]3[S ] 2 = K sp. c) [Mn 3+ ] 2 [S ] 3 = K sp. What is the solubility of Ag2CO3 in water? K sp = 8.1 x each mol of Ag 2 CO 3 dissolving generates 2 mols Ag + and one mol of CO 3 solubility = ½ [Ag+] = [CO3] K sp = [CO 3 ][Ag + ] 2 = ½ [Ag + ] [Ag + ] 2 = [Ag + ] = 2.5x10-4 M solubility = ½ (2.5x10-4 M) = 1.3x10-4 M What is the solubility of Ag 2 CO 3 in M AgNO3? molar solubility ½ [Ag + ] BUT for each mol dissolving 1 mol of CO 3 formed molar solubility = [CO 3 ] K sp =[CO 3 ]( M+2[CO 3 ]) 2 = 8.1x10-12 [CO 2 ]( M) 2 molar solubility= [CO 3 ] = 2.0 x 10-8 M much smaller than when no AgNO 3 is present

3 Which of these statements concerning the solubility is correct? a) A salt is less soluble if one of its ions is already present in solution. b) A salt is no more soluble if one of its ions is already present in solution. c) A salt is more soluble if one of its ions is already present in solution. A beaker contains ml of molar silver ion (Ag + ). To this beaker is added ml of M bromide ion (Br - ). What is the concentration of Ag + in the final solution? Ksp for AgBr = 5.0 x a) 7.1 x 10-7 M b) 6.7 x M c) 3.3 x M solution: 250.0ml x M Ag = mol Ag ml x 0.300M Br = mol Br final volume is 500 ml Q = [Ag][Br] = (0.0375mol/0.5L)(0.0750/0.5) = > Ksp so will have precipitation at eqm, [Br] = mol (left) / L = [Ag+] = Ksp / [Br] = (5.0x10-13) / x10-12M = solubility (have an excess of bromine Factors affecting solubility Temperature in general, solubility increases with temperature (exception: exothermic rxns) precipitations carried out at low temperature (if precipitate forms too slowly, an increase in temperature may facilitate its formation). Solvent addition of a miscible solvent (MeOH, EtOH) to aqueous solution to decrease solubility. Composition of the solution common ion effect ex.: Solubility of AgCl depends on the concentration of Ag + and Cl - since [Ag + ] [Cl - ]=K sp Remember each ion has only one concentration in a solution this concentration must confirm to all equilibrium constants every solution is electrically neutral

4 Other factors affecting solubility Other ion effect K sp is defined as a product of activities the presence of other ions will change μ and the activity coefficients change in concentration ph effect if solid = salt from weak acid: solubility increases as ph decreases ex.: CaCO 3 (s) + H + Ca HCO 3 Mg(OH) 2(s) + 2H + Mg H 2 O NiS (s) + H + Ni 2+ + HS s Complex formation increases solubility ex.: AgI (s) + 2CN - - Ag(CN) 2 + I - + AgCl (s) + 2NH 3 Ag(NH 3 ) 2 + Cl - What is the definition of ph? a) -log [H + ] b) -log [H 3 O + ] Strictly, answer c) c) -log a H3O+ For most purposes d is good d) a or b lewis acid Which of these compounds should increase in solubility in acid solution? a) KCl: no basic propertie b/c Cl is the conjugate base of a strong acid b) CaF 2 : F is the conjugate base of a weak acid basic properties c) Ba(OH) 2 : is a strong base d) NaClO 4 : perchlorate is the conjugate base of a strong acid e) two of the above lewis base adduct Example: dissolution of precipitate by complex formation AgCl (s) + 2NH 3 Ag(NH 3 ) 2+ + Cl - K sp =[Ag + ] [Cl - ] = K f =[Ag(NH 3 ) 2+ ]/{[Ag + ][NH 3 ] 2 } = which concentration of NH 3 should be used to dissolve AgCl so as to obtain a 1.0 x 10-2 M solution at the end of dissolution? Solution [Cl - ]= 10-2 M at end of the dissolution [Cl - ]=c Ag =[Ag + ] + [Ag(NH 3 ) 2+ ] [Ag(NH 3 ) 2+ ] From K sp,[ag + ]= /[Cl - ]= /10-2 = M From K f =[Ag(NH 3 ) 2+ ]/{[Ag + ][NH 3 ] 2 } = 10-2 /{ [NH 3 ] 2 }= [NH 3 ] = 0.18 M AgCl would not precipitate if 10-2 M Cl - was added to a solution of 10-2 M Ag + containing 0.20 M of NH 3

5 Precipitation of hydroxides most metallic hydroxides are not very soluble cations separation can be carried out based on the difference in solubility of the corresponding hydroxides example: Mg OH - Mg(OH) 2(s) Ksp = [Mg 2+ ][OH - ] 2 = if OH - is added to M Mg 2+, Mg(OH) 2 will start to precipitate at: [OH - ] 2 = /0.010= [OH - ] = or ph = 9.55 if addition of OH - continues, Mg(OH) 2 precipitates Separation by precipitation precipitation is considered complete when 0.01% of initial [Mg 2+ ] is left ie. when [Mg 2+ ]= 10-5 M (ph = 11.05) precipitation in the range:9.55<ph<11.05 little change in ph (buffer effect) during precipitation precipitation of Fe(OH) 3 ([Fe 3+ ][OH - ] 3 = ) occurs at 2.00 <ph< 3.00 separation: Fe(OH) 3(s) /Mg 2+ at 3.0<pH<9.5 buffers are usually employed to fix ph Prediction of separation results can be summarized in graphical form of solubility S = [M n+ ] vs ph this graph allows prediction of possibility of separation. coprecipitation may occur. only an experiment can show whether a separation actually works or not. If a M solution of NaOH is added to a solution containing M Ni 2+, M Ce 3+, and M Cu 2+, which metal hydroxide will precipitate first? Ksp for Ni(OH) 2 = 6.0 X 10-16, K sp for Ce(OH) 3 = 6.0 X 10-22, and K sp for Cu(OH) 2 = 4.8 x a) Ni(OH) 2 b) Ce(OH) 3 c) Cu(OH) 2 Can't just look at Ksp because of stoichiometry coefficients [Ni][OH] 2 = 6.0x10-16 [OH - ] = sqrt (6x10-16 /0.2M) = 5.5 x10-22m Ce = 1.4x10-7 Cu = 4.9x10-10 Precaution Check for possible complex formation between metallic ions and buffer constituents ex.: Fe(III) 0.1 M with 1 M C 2 O 4 Fe(OH) 3 will not precipitate before ph 7.7 (instead of 2.0 without C 2 O 4 ) neutralization curves of acidic solutions of cations can also be obtained where the length of the precipitation plateau is a function of the cation concentration. in general, precipitation is preferred since dissolution is often slow

6 Protic acids and bases electrolyte: dissociates partly or totally in water to give ions electrical conductivity increases General Bronsted-Lowry ACID any substance that furnishes H 3 O + or consumes OH - proton donor BASE any substance that furnishes OH - or consumes H 3 O + Proton accepor *H 3 O + = hydronium or hydroxonium ion (often written as H + ) Neutralization reaction of a base with an acid CH 3 COOH + H 2 O H 3 O + + CH 3 COO - conjugate base and acid: related by gain or loss of one proton NH 3 + H 2 O NH4 + + OH - HCl (g) + NH 3 (g) NH4 + Cl - (s) most salts are strong electrolytes + NH 4 Cl (s) NH 4 (aq) + Cl - (aq) - In the following reaction, identify a conjugate acid-base pair: NO 2 + H 2 O a) NO, OH - b) NO, H 2 O c) NO, HNO 2 Water amphiprotic solvent: can act as proton acceptor or donor other examples: MeOH, EtOH, CH 3 COOH undergoes self-ionization or autoprotolysis H 2 O + H 2 O H 3 O + + OH - K w = [H 3 O + ] [OH - ] = 1.01 x at 25.0 o C -log(k w )=pk w =-log([h 3 O + ][OH - ]) =ph+poh=14.00 at 25.0 o C [H 3 O + ]=[OH - ]=10-7 M in pure water at 25 o C ph of various substances limestone HNO2 + OH -. CO 2 H 2 CO 3

7 Strong acids and bases complete dissociation (in dilute aq. Solution) strong electrolyte acid: HCl, HBr, HI, not HF, H 2 SO 4 (1st proton), HNO 3, HClO 4 HCl + H 2 O H 3 O + + Cl - base: LiOH, NaOH, KOH, RbOH, CsOH, R 4 N + OH -, alkaline earth hydroxide (Mg 2+, Ca 2+, Sr 2+, Ba 2+ ) NaOH Na + + OH - conjugate of strong acid: no basic property conjugate of strong base: no acidic property Weak acids and bases reacts partly to produce H 3 O + or OH - weak electrolyte acid: CH 3 COOH, RCOOH, HF, NR 3 H +, NH 4+, RNH 3+, R 2 NH 2+, HCO 3-, HX - (H 2 X) base: RCOO -, CH 3 NH 2, RNH 2, RNH 2, R 2 NH, R 3 N, HCO 3-, HX -, X - Weak acids and bases HA + H 2 O H 3 O + +A - K a =[H 3 O + ][A - ]/[HA] acidity constant or acid "dissociation constant measure of acid strength relative to solvent solvent acts as base B + H 2 O BH + + OH - K b =[BH + ][OH - ]/[B] basicity constant or base "dissociation constant measure of base strength relative to solvent solvent acts as acid A M solution of nitrous acid (HNO 2 ) had a ph of What is the K a value for nitrous acid. a) 7.24 x 10-4 [H b) 7.24 x O + ] = = M = [NO ] c) 8.51 x 10-2 Ka = ( ) 2 / ( ) = 7.24x10-4 Relation between K a and K b for a conjugate acid/base system K a K b = K w FOR A CONJUGATE PAIR pk a +pk b =pk w =14.00 at 25.0 o C

8 Leveling effect the solvent determines how strong an acid or a base is in it reference = 0 Stronger acid Stronger base HCl, HBr, HClO 4 : no difference in strength in water all completely dissociated Polyprotic acids and bases can donate or accept more than one proton K a1 and K b1 for species with most # of protons and least # of protons respectively Relation between K a and K b for diprotic acid: K a1 K b2 = K w and K a2 K b1 = K w For polyprotic species: as many equations as protons: K ai x K b(n-i + 1) = K w i = i th proton from 1 to n n= total number of protons ex. for triprotic acid K a1 K b3 = K w K a2 K b2 = K w K a3 K b1 = K w

9 From the equilibria below, what is [H 3 AsO 4 ] equal to in terms of [AsO 4 3- ]? **add them all up a) [H 3 AsO 4 ] = ([H 2 AsO 4- ][H ])/K a1 b) [H 3 AsO 4 ] = ([AsO 4 3- ][H + ] 3 )/(K a3 K a2 K a1 ) c) [H 3 AsO 4 ] = K a3 K a2 K a1 /([AsO 4 3- ][H + ] 3 ) The three acid-base equilibria must b combined to get H 3 ASO 4 2H AsO 4