Ionization of acids and bases - PDF Free Download

ionization equation Ionization of acids and bases Acid Base AH + H 2 O H 3 O + + A B + H 2 O OH + BH + simpler eq. AH H + + A B + H + BH + ionization K A = [H 3 O + ][A - ]/[AH] K B = [OH - ][BH + ]/[B] constant (acidity* constant) (base ionization constant) *not to be confused with association (binding) constant Water self-ionization: 2H 2 O H 3 O + + OH K W = [H 3 O + ][OH - ] = [H + ][OH - ] = 10 14 at 25 C For a base, K B = K W [BH + ]/[B][H 3 O + ] = K W /K A K A is the acidity constant of the conjugate acid

ph, poh, pk A and pk B Acid pk A = log 10 K A Base pk B = log 10 K B = log 10 K W + log 10 K A = log 10 K W + log 10 K A = 14 pk A of the conjugate acid ph = log 10 [H + ] poh = log 10 [OH ] ph = 14 poh

Exercise: find ionizable groups Amphetamine Menthol Salicylate Clodronate Aminohippurate Guanadrel

Henderson-Hasselbalch Eq. Derivation step Reason K A = [H 3 O + ][A ]/[AH] definition K A /[H 3 O + ]= [A ]/[AH] (rearranged) log 10 K A log 10 [H 3 O + ] = log 10 ([A ]/[AH]) (logs taken on both sides) ph - pk A = log 10 ([A ]/[AH]) (HH equation) 1 unit ph = 10-fold change in [A - ]/[AH] ratio Solute dissociation changes ph! Can only assume that ph = ph 0 when either concentration C=[A - ]+[AH] is very low: C < 10% of [H+] i.e. 1+log C < ph or the solution is buffered (resist ph changes)

Ratio vs Fraction of Species [A - ]/[AH] %[A - ] 10 3 = 0.001 0.001/(1+0.001) ~ 0.1% 10 2 = 0.01 0.01/(1+0.01) ~ 0. 99% 10 1 = 0.1 0.1/(1+0.1) ~ 9.09% 10 0 = 1 1/(1+1) = 50% 10 1 = 10 10/(10+1) ~ 90.91% 10 2 = 100 100/(100+1) ~ 99.01 % 10 3 = 1000 1000/(1000+1) ~ 99.9%

Henderson-Hasselbalch Eq. Problem: An acidic drug with pk A of 3.5 is dissolved in stomach at ph=2. What fraction of the drug molecules are ionized? Solution: log([a-]/[ah]) = ph pk A = -1.5 [A-] = 0.0316[AH] fraction ionized is 0.0316/(1+0.0316) ~ 3% Answer: Only about 3% is ionized. Bonus Q: How and where will this drug absorb?

ph of a solution Strong acids and bases dissociate completely when dissolved in water ph = -log c for acids; 0 for 1M solution ph ~ 14+log c for bases (though even strongest bases as not as strong as acids) But most proteins, drugs, and chemicals are weak acids and bases

ph of a solution A weakly acidic substance (pk A given) is dissolved in water (ph=7) at the concentration c What is happening? Acid dissociation: AH H + + A Water dissociation: H 2 O H + + OH AH A H + OH Start (no equilibrium) c 0 10 7 10 7 Equilibration c a a x K w /x What are the rules? Acid mass-balance: [AH] + [A ] = c Water dissociation equilibrium: [H + ] [OH ] = K W Acid dissociation equilibrium: [H + ] [A ] = K A [AH] Charge balance: [H + ] = [OH ] + [A ] Finding ph requires solving for x Solving for x requires a CUBIC equation Applicable to bases (K A K B, ph poh)

ph of a solution - simpler A weakly acidic substance (pk A given) is dissolved in water at the concentration c Dissociation: AH + H 2 O H 3 O + + A This method neglects re-association of water: [OH ] and its changes are negligibly small e.g., will not work for an acidic drug in a basic solution AH A H 3 O + Start (no equilibrium) c 0 10 7 Equilibration/dissociation c x x x + 10 7 (c x)k A = x(x+10 7 ) x 2 + x(k A +10 7 ) ck A = 0 x = ( K A 10 7 + Sqrt((K A +10 7 ) 2 + 4cK A ))/2 Convenient to multiply through by 10 14 Applicable to bases (K A K B, ph poh)

ph of a drug solution Problem: calculate ph of 1 um solution of ibuprofen in water (ibuprofen is a weak acid with pk A of 4.91) Solution: K A = 10 4.91 ; c = 10 6 M x 2 + x(k A +10 7 ) ck A = 0 x 2 + x(10 4.91 +10 7 ) 10 6 10 4.91 = 0 Let y = 10 7 x; multiply the eq. by 10 14 y 2 + 124y 1230.27 = 0 y = 9.234 [H 3 O + ] = 10 7 +x = 10 7 +10 7 y ph = 7 log(y+1) = 5.99 Answer: ph = 5.99 Shortcut: Concentration is low, ph >> pk A, all ibuprofen dissociates, i.e. [H 3 O + ] ~ 10 6 +10 7 ~ 1.1 10 6 ; ph ~ 6.

ph of a solution even simpler Weak acid (pk A > 2), high concentration, low fraction dissociated: 10 7 << x=[a ]=[H 3 O + ] << c [AH] = c x c (c x)k A = x(x+10 7 ) becomes ck A x 2 [H 3 O + ] 2 = ck A, take logs on both sides: ph = ½ pk A ½ log c Sanity check: [A ]<[AH] log 10 ([A ]/[AH])<0 ph < pk A (HH) If ph < pk A, linear approximation makes sense If ph pk A, approximation inaccurate, real ph is even higher

ph a of solution: simple vs simpler pk A = 2 quadratic ½ pk A ½ log c 100% Fraction dissociated 0% Ibuprofen problem: ½ pka ½ log c = 5.45 exact ph = 5.99 Approximation works when fraction dissociated < 20%: higher conc., weaker acids watch for ph < pk A 100% 100% pk A = 4 quadratic ½ pk A ½ log c Fraction dissociated 0% pk A = 6 quadratic ½ pk A ½ log c Fraction dissociated 0%

ph of solution, linear approximation: base Weak base (pk A < 12), high concentration, low fraction ionized: 10 7 << x=[bh + ]=[OH ] << c [B] = c x c (c x)k B = x(x+10 7 ) becomes ck B x 2 [OH ] 2 = ck B, take logs on both sides: poh = ½ pk B ½ log c OR ph = 7 + ½ pk A + ½ log c Sanity check: [BH + ]<[B] log 10 ([BH + ]/[B])<0 poh < pk B ph > pk A If ph > pk A, linear approximation makes sense If ph pk A, approximation inaccurate, real ph is even lower

ph of a drug solution: linear approximation Problem: calculate ph of 1 mm solution of ibuprofen in water (ibuprofen is a weak acid with pk A of 4.91) Solution: concentration is relatively high, acid is weak, try linear approximation ph = ½ pk A ½ log c = 4.91 / 2 + 3 / 2 = 3.955 this ph is less than pk A ; likely accurate Answer: ph = 3.955

Buffers Given a strong acid or base solution with a starting ph; a weak acid is dissolved in it at the given concentration c AH A H + OH Start (no equilibrium) c 0 10 ph 10 ph 14 Equilibration c a a x K w /x All components are appreciable: solving the cubic is required ph calculators exist There is a region where final ph does not depend on the starting ph Weakly ionizable substance in both ionized and nonionized forms system resists changes in ph: buffer acid with pk A = 4

Buffer capacity and range A single buffer pair [AH]/[A ] may only absorb X strong acid/base, and only does that around ph of Y Formalized by buffer capacity: [ ] [ ] + dn [ ] + ck A H β = = ln10 H + + OH ( ) ( [ ] + ) 2 d ph K A + H To increase β, increase c To increase range, use several buffer pairs and/or polyprotic substances pka = 4 pka = 5 pka = 6 pka = 7 c = 10 um c = 100 um c = 1 mm c = 10 mm β β ph ph

Physiological buffers Blood ph = 7.4 buffering capacity β = 0.025-0.039 mol/l per ph unit Blood plasma buffering system: H 2 CO 3 (pk A =6.35) HCO 3 ; Na + salt Important: CO 2 H 2 CO 3 reaction is very slow k f = 0.039 s 1, k r = 23 s 1 carbonic anhydrase (CAH) is an enzyme catalyst Erythrocytes buffering system: HPO 4 2 (pk A =7.20) H 2 PO 4 ; K + salts Any protein may create a buffer, pk A (His) = 6 Parenteral solutions: either not buffered or buffered at low capacity to allow blood buffers to bring them within tolerable ph range

Physiological non-buffer Urine is not buffered; its ph may be adjusted with NH 4 Cl, NaHCO 3, chlorothiazide diuretics, CAH inhibitors, K citrate: To ensure complete ionization and easy excretion of amphetamines, barbiturates, salicylates (in case of drug overdose) To prevent ionization and promote drug reabsorption for therapeutic reasons

Solubility at different ph: buffer In a saturated solution of an ionizable substance: Acid Base HH log[a ]/[AH] = ph pk A log[b]/[bh + ] = ph pk A S 0 : solubility of neutral form All ionized species are soluble [AH] = S 0 [B] = S 0 [A ] = S S 0 [BH + ] = S S 0 log(s S 0 )/S 0 = ph pk A (S S 0 )/S 0 = 10 ph pk A log(s S 0 )/S 0 = pk A ph (S S 0 )/S 0 = 10 pk A ph S = S 0 (1+10 ph pk A ) S = S0 (1+10 pk A ph ) ph is affected by a high concentration of acid/base Approximation is accurate for buffered solutions when S S 0 << β (buffer capacity)

Solubility at different ph Problem: The saturation solubilities of a drug at different ph and T=300K are shown in the table. What type of compound is it and what is pk A? ph Saturation solubility 7.4 205 µm 9 10 µm 10 5.5 µm 12 5 µm Solution: Solubility with ph it is a base. Solubility at ph = 12 is ~ S 0 pk A = ph + log (S S 0 )/S 0 When using S=205 µm at ph=7.4, pk A =7.4+log 200/5~9 When using S=10 µm at ph=9, pk A =9+log 5/5=9

Solubility at different ph: water A weakly acidic drug (pk A given) is dissolved in water at the saturating concentration Solubility of neutral form is S 0, total = S 0 + [A ] Dissociation: AH + H 2 O H 3 O + + A AH A H 3 O + Start (no equilibrium) S 0 + x 0 10 ph 0 Equilibration/dissociation S 0 x x + 10 ph S 0 K A = x(x+10 ph ) x 2 + 10 ph x S 0 K A = 0 Convenient to multiply through by 10 2pH ; y = 10 ph x y 2 + y 10 2pH S 0 K A = 0 Applicable to bases (K A K B, ph poh)

Solubility at different ph: water and buffer In non-buffered solution at high ph (i.e. low [H 3 O + ]), solute dissociation dominates ph and thus limits solubility Buffer (within its range and capacity) cancels this effect S 0 = 1nM water buffer 100% S 0 = 1uM water buffer 100% S 0 = 1mM water buffer 100% pk A = 2 pk A = 4 pk A = 6 Fraction dissociated pk A = 2 pk A = 4 Fraction dissociated pk A = 2 Fraction dissociated pk A = 6 pk A = 4 0% 0% pk A = 6 0%

Ionizable drugs are often Commonly used salts: Weak acid / strong base Weak base / strong acid Weak acid / weak base formulated as salts Salt formulations affect solubility Salts deliver ionized components into the solution; the ph variations are favorable for dissolution Crystal state interactions vary between salts and free form Drug Acid pk A Base pk A Divalproex Sodium Valproate 4.8 Na + Penicillin G Potassium Penicillin G 2.7 K + Chlorpromazine HCl Cl Chlorpromazine H + 9.3 Codeine Phosphate PO 4 2.2,7.2,12.3 Codeine H + 8.2 Dramamine 8-Chlorotheophylline 4.6 Diphenhydramine H + 9.0

ph of a salt solution Crystal: both components are completely ionized Solution: strong component remains completely ionized weak component partially de-ionizes producing [OH ] (weak acid) or [H + ] (weak base) Linear approximations for ph: Weak A / strong B: ph > 7; ph = 7 + ½ pk A + ½ log c Weak B / strong A: ph < 7; ph = ½ pk A ½ log c 1:1 weak acid / weak base: ph = 7 + ½ pk A ½ pk B More accurate ph determination may require solving higher degree equations