1/28 2/28 Electrochemistry co04 Solutions of electrolytes co04 Subject of electrochemistry: strong electrolyte: (almost) completely ionizes (dissociates) no uncharged molecules in the solution dissociation (solutions of electrolytes, melts of salts) H2SO4, KOH, Ca(OH)2, NaCl, BaSO4,... conductivity not necessarily to max. degree: 100 % partially 2 phenomena at interfaces s/l (electrolysis, galvanic cells) H2SO4 H+ + HSO4 2 H+ + SO4 → − → − weak electrolyte: contains neutral (not dissociated) molecules Conductors: organic acids and bases, NH3,H2O, . electrons (or holes) are moving: standard states: solvent (water): ; in dilute 1 • water = metals, graphite, graphene, semiconductors c st ions: [ ] ( = γc/c ) ions are moving (or jumping): dissociation constant = equilibrium constant of the dissociation reaction solutions of electrolytes , molten salts, ionic liquids + CH3 COOH CH3 COO− + H both electrons and ions are moving: → + NH3 + H2O NH4 + OH− plasma → + (COOH)2 HOOC-COO− + H → pH log = log10, p = log − γ + =1 c + c + H H H ? + ? + pH = log H+ = log = log = log[H ] = log{H } cst mol dm 3 − − − − − − 3/28 4/28 Dissociation of water co04 Dissociation of water and the acidity constant co04 Dissociation of water: Acidity constant (ionization constant) Ka is common in tables = equilibrium con- + stant of deprotonisation, often given as pKa H2O H + OH− → acids: dissociation constant of the acid Ionic product (autoionization constant) of water Kw: + H+ OH cH+ cOH . CH3 COOH CH3 COO + H Ka = Kd K − − H+ OH 1.00 10 14 25 C − w = st 2 [ ][ −] = − ( ◦ ) → H2O ≈ (c ) ≡ × bases: dissociation constant of the conjugated acid Equivalently (at 25 C): ◦ NH4+ NH3 + H+ (Ka) ( 1) pH + pOH = pKw = 14 more accurate 13.997 → × − H O H+ OH K 1 2 + − ( w) (+ ) → × depends on temperature: pK (100 C) = 12.29 w ◦ NH H O NH + OH K K /K heavy water: pK (25 C) = 14.87 3 + 2 4 + − d = w a w ◦ → (isotopic effect: D is more strongly bound) Example. Acidity constant of ammonium is pKa = 9.25 (at 25 C). Calculate× the ◦ − 10 78 . 1 dissociation constant of ammonium hydroxide (water solution of NH3). 5 5/28 6/28 pH of strong acids and bases co04 Dissociation of a weak acid I co04 Calculate pH of aqueous HCl of concentration 0.01 mol dm 3. Example. − + HA H + A− 100 % dissociation to the 1st degree: → Determine pH of a weak acid from known concentration. cH+ = cHC pH = log H+ log cH+ = 2 ⇒ − ≈ − (analytic) concentration: c0 Calculate pH of aqueous H SO of concentration 0.001 mol dm 3. acidity constant: Ka Example. 2 4 − It dissociates completely to the 1st degree, partially to the 2nd degree, but since Assumptions: here c K 1.3 10 2, we can assume also 100 % dissociation. 2 = − × cOH cH+ − cH+ = 2cH2SO4 pH = log H+ log(2cH+ ) = 2.7 ⇒ − ≈ − γ 1 (approximation of infinite dilution) more accurate 2.75 (partial dissociation + Debye–Hückel) = compound 0 eq. Example. Calculate pH of aqueous NaOH of concentration 0.01 mol dm 3 at 25 C. − ◦ H+ A 2 3 14 12 3 HA c0 c0 [ ][ −] cOH = 0.01 mol dm , cH+ = 10 /0.01 = 10 mol dm , pH = 12 Balance: − Equation: = = Ka − − − − − A 0 HA c − [ ] 0 Calculate pH of 0.001 M aqueous Ca(OH) at 25 C. + − Example. 2 ◦ H 0 3 cOH = 0.002 mol dm , pOH = 2.7, pH = 14 2.7 = 11.3 2 − − st − Correctly st = Ka. We will neglect c (concentrations should be inserted in (c0 )c mol dm 3). − − 7/28 8/28 Dissociation of a weak acid II co04 Weak acid in a buffer co04 H+ A 2 [ ][ −] + Equation: = = Ka HA H + A− [HA] c0 → − Solution: How much of deprotonated form can we find in a solution of given pH? 10 2 compound 0 eq. Ka Ka c0 Ka A K cH+ = = + Kac0 Kac0 HA c0 c0 [ −] a v 2 − 2 ≈ u − = + -- Æ A 0 [HA] [H ] [CH3COOH] [CH3COO ] Summary of approximations: t − + -3 H cH+ cH+ maintained by buffer dm the acid is stronger than water (K K ) ← a w 5 c0Ka CH COOH concentration c0 is high enough (c0 Ka), then most of the acid is not ionized = cA = 3 − c + K c/mmol -3 H + a 10 mmol dm c0 is not too high so that we can use the infinite dilution approximations (γ = 1) Degree of dissociation: pKa = 4.76 Alternate form: 1 1 α = pH = (pKa + pcA) c + /Ka 1 0 2 H + 2 3 4 5 6 7 8 Degree of dissociation: We have assumed the infinite dilution ap- pH proximations 1 Ka α for pH=pK α = 2 a ≈ vc0 u t 9/28 10/28 C Calculate pH of rainwater at 25 ◦ and 1 bar. co04 pH of a weak base co04 3 1 Example. Calculate pH of aqueous solution of ethylamine of concentration c0 = Henry constant of dissolution of CO2: Kh = 0.033 mol dm bar − − 0.01 mol dm 3 at 25 C. The acidity constant of ethylammonium is K 1.6 10 11. − ◦ a = − Acidity constant of CO2: pKa1 = 6.37. This is total for reactions × C H NH H O C H NH + OH 2 5 2 + 2 2 5 3 + − CO H O H CO H+ HCO → 2 + 2 2 3 + 3− K 1 10 14 → → w − Dissociation constant of ethylamine: Kd = = × = 0.000 625 CO in the air (2019/2 Mauna Loa): y 411 ppm (18th century: y 280 ppm) K 1.6 10 11 2 = = a − 2 + × CO3 can be neglected (pKa2 = 10.32) Assumptions: [H ] [OH ], γ = 1. And in the same way as for weak acids: − − OH can be neglected − compound 0 eq. 2 3 1 C H NH + OH c K y p 0.033 mol dm bar 0.000411 1 bar C2H5NH2 c0 c0 [ 2 5 3 ][ −] CO2 = h CO2 = − − Balance: Equil.: K 5 3 × × − = = d = 1.36 10 mol dm C2H5NH3+ 0 [C2H5NH2] c0 − − − × OH 0 ? − Balance: c + c (Not cCO cCO c + because CO is given by equilib- H = HCO3− 2 = 2.0 H 2 rium, not initial concentration.) − 2 Kd Kd c0 Kd Equation: = [OH ] = + Kdc0 Kdc0 [H+][HCO3 ] − v 2 2 − K log 1.36 10 5 u − ≈ = a1 − Æ [CO2] − × or: t . 1 + 1 1 pH = (pKa + pKw pc0) [H ] = Ka1[CO2], pH = 2(pKa1 + p[CO2]) = 2(6.37 + 4.87) = 5.62 2 − = 22 . 0 α the real pH isÆ usually lower (HNO3,H2SO4) (18th cent.: pH = 5.70) 11.40; approx.: pH=11.34, 11/28 12/28 Simultaneous equilibria: very dilute solutions co04 Very dilute solutions co04 Dissociation of water has to be taken into account. CH3COOH, pKa = 4.76 + 8 HA H + A− (Ka) → + H2O H + OH− (Kw) → “exact” 7 “Explicit” balance “Implicit” balance 2 6 compound 0 eq. compound balance + Ka Ka [H ] = 2 + Kac0 2 HA c0 c0 A [A ] + [HA] = c0 s − − − pH 5 A 0 charge H+ A OH 0 − [ ] ([ −] + [ −]) = + − H 0 + y + Equations [H ] = Kac0 4 OH 0 y − (approximate formula) H+ A p [ ][ −] 3 = Ka Equations HA [ ] In a realistic region of concentrations the ( + y) H+ OH K [ ][ −] = w simplified formula is sufficient 2 = Ka -1- 0 1 2 3 4 5 6 7 8 9 10 c0 − 4 equations, 4 unknowns: p[C H COOH] ( + y)y = Kw 2 5 H+ , HA , A , OH [ ] [ ] [ −] [ −] 2 equations, 2 unknowns: , y [cd show; blend -g lysine]13/28 [cd ../maple; xmaple lysine.mws]14/28 Simultaneous equilibria: ion speciation co04 Lysine co04 Aminoacids, ionization degree according to pH. E.g., HIS, LYS: NH -(CH ) -CH(NH )-COOH, pK 2.15, pK 9.16, pK 10.67 2 2 2 4 2 a1 = a2 = a3 = AH3 + AH2+ + H+ (Ka1) → AH2+ AH + H+ (Ka2) 1 AH → A H+ K pKa1 pKa2 pKa3 − + ( a3) → 2 H+ easily detaches from AH3 +, hardly from AH Ka1 > Ka2 > Ka3 or pKa1 < pKa2 < pKa3. − ⇒ Lys Balance: 3 Lys − LysH+ + 2+ dm 2+ [A ] + [AH] + [AH2 ] + [AH3 ] = c0 LysH − 0.5 Equation: mmol / + + i [AH2 ][H ] c K 2 = a1 [AH3 +] [AH][H+] = Ka2 [AH2+] A H+ 0 [ −][ ] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 = Ka3 [AH] pH 15/28 16/28 From real life co04 Salt of weak acid + strong base + co04 Titanium dental implants are treated (e.g.) in a phosphoric E.g., M=Na, A=CH3COO. acid solution. Calculate the needed concentration (in pure + MA M + A− (100 %) water) for pH = 3.42. Data: pK1 = 2.18, pK2 = 7.198, → A H O AH OH (hydrolysis) pK 12.319. − + 2 + − 3 = → 3 4 neglected) PO .8mo/ (H mmol/L 0.38 compound 0 eq. conditions − 2 3 4 4 ) PO +H PO .0mo/ (H mmol/L 0.40 M+ c0 c0 Kw A c c H+ A c0 − 0 0 [ ][ −] ( ) − Ka = − OH 0 HA − [ ] ≈ HA 0 H+ 0 Kw/ for [OH ] [H+] − Solution: 2 c0Kw Kw Kw c0Kw 1 = + pH 2(pKw + pKa pc0) v Ka 2Ka − 2Ka ≈ v Ka ⇒ ≈ − credit: Wikipedia u u t t Kw where the last approximation holds for c0 K & Kw, i.e., c0 Ka a p 17/28 18/28 Little soluble salts of strong electrolytes co04 Little soluble salts: more ions co04 Solubility product = equilibrium constant of dissociation. 9 Calcium oxalate: Ks(CaC2O4) = 3.9 10 . × − activities of salts (s) are salt = 1 a) how much dissolves in pure water? b) how much dissolves in blood? γ = 1 assumed for ions ( Ca2+ 2.4 mmol dm 3) [ ] = − Ba2+ SO 2 2+ 2 4 − 2+ 2 BaSO4 Ba + SO4 Ks = = [Ba ][SO4 ] 2+ 2 − − CaC2O4 Ca + C2O4 → BaSO4 − → 2 2 2+ 2+ 2 Ks = [Ca +][C2O4 ] Mg(OH)2 Mg + 2 OH− Ks = [Mg ][OH ] − → − 3+ 2 3 2 2 3 Solution: As2S3 2 As 3 S Ks As + S + − = [ ] [ −] 2 2+ 2 → 3+ 2 a) [C2O4 ] = [Ca ] = Ks but: S + H2O HS + OH ..