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Chemistry 2000 Lecture 19: Organic acids Marc R. Roussel Ka I The strength of an acid HA is measured by Ka , the equilibrium constant for the dissociation of the acid into a proton and its conjugate base: HA H+ + A− I Equivalently, we can think of Ka as the equilibrium constant for the reaction of an acid with water: HA + H2 O H3 O+ + A− I Larger Ka =⇒ stronger acid pKa I Ka values range over many orders of magnitude, so they are not very convenient for some purposes (e.g. comparisons between acids). I Define pKa = − log10 Ka I Smaller pKa =⇒ stronger acid I Strong acids may have negative pKa values (dissociate completely in water). Organic acids Stronger acid Weaker acid Functional group Carboxylic acids Phenols Water Alcohols pKa 3–5 1–10 14 (pKw ) 15–18 Acidity of carboxylic acids O C I I OH Carboxylic acids are among the most acidic organic compounds. Acid strength depends on I I the polarity of the bond to the dissociable hydrogen (in this case the O-H bond), and the stability of the conjugate base. Acidity of carboxylic acids Polarity of the O-H bond O C I OH The carbonyl group is electron withdrawing: Because of the large electronegativity of oxygen, electron density is pulled away from the carbon atom. This in turn pulls electron density away from the hydrogen atom, increasing the partial positive charge on this atom. Acidity of carboxylic acids Stability of carboxylate ions I The carboxylate anion is resonance stabilized: O O − C C O − O The negative charge is therefore spread over a larger region of the molecule, which tends to stabilize the anion. I Substituents on the adjacent carbon to the carboxylic acid functional group can have a substantial effect on the pKa . I Consider the following series: I Molecule CH3 COOH FCH2 COOH F2 CHCOOH F3 CCOOH pKa 4.67 2.66 1.24 0.23 The fluorine atom is very electronegative and creates a partial positive charge on carbon 2: H δ− F O δ+ C H C − O I The nearness of the partial positive charge on carbon 2 and of the negative charge of the carboxylate group stabilizes the latter. I This is called an inductive effect. Alcohols I I Unlike carboxylic acids, most alcohols are only about as acidic as water itself. Why? I I I No electron-withdrawing group (in an ordinary alcohol) to increase the polarization of the O-H bond No resonance stabilization of the negative charge of the conjugate base Inductive effects can increase the acidity of an alcohol. Example: Molecule pKa CH3 CH2 OH 15.9 Cl3 CCH2 OH 12.24 Phenols I Phenols are much more acidic than ordinary alcohols. I Phenol itself has a pKa of 9.95, vs 17 for cyclohexanol. I Why? Resonance stabilization of the charge on the phenolate anion: .. − :O: H :O: H H :O: :O: H H H − H H H H H .. .. − H H H .. H H − H H H H I Some phenols are even more acidic because the substituents can participate in charge delocalization. Example: p-nitrophenol has a pKa of 7.21. .. − :O: :O: H :O: H H H H H H H .. − H H H .. O .. N + .. H − + + .. − O: .. .. O .. N .. − O: .. .. O .. :O: :O: H H H H H H − .. H .. O .. N H + .. − O: .. − .. :O .. N + .. − O: .. N .. − O: .. Acid dissociation equilibria I If we know the Ka and concentration of an acid, we can calculate the pH. I Reminder: pH = − log10 aH+ I We usually don’t need to take the autoionization of water into account unless the concentration of protons liberated from the acid is similar to the concentration of protons generated by autoionization. I We often can treat the acid as if it’s mostly undissociated. I Note that we’ll assume 25◦ C in all of the following calculations. Example: pH of a 0.32 M phenol solution I The pKa of phenol is 9.95. I pH = 5.22 Example: pH of a 4.2 × 10−5 M ethanoic (acetic) acid solution I The pKa of ethanoic acid is 4.76. I pH = 4.71 Balance of acid and conjugate base at given pH I Sometimes, we put an acid into a solution of fixed pH (a buffer) and want to know how much is in the acid and how much in the conjugate base form. I This is an easy problem because the pH fixes aH+ , which immediately gives us the ratio of the conjugate base to the acid: Ka a − = A aH+ aHA I This can easily be converted to percentages of the two forms if we add the equation [HA] + [A− ] = 100% (with a slight abuse of notation). Example: ethanoic acid at pH 4 Ka 1.74 × 10−5 [CH3 COO− ] = = 0.174 = aH+ 10−4 [CH3 COOH] and [CH3 COOH] + [CH3 COO− ] = 100% ∴ [CH3 COOH] + 0.174[CH3 COOH] = 100% ∴ [CH3 COOH] = 85% ∴ [CH3 COO− ] = 15% Distribution curves I If we repeat the above calculation at a number of different pH values and plot the results, we obtain distribution curves for the acid and its conjugate base. I Note: If pH = pKa , we have 10−pKa aA− = =1 aHA 10−pH In other words, 50% of the acid is undissociated, and 50% in the form of the conjugate base at the pKa . Distribution curve of ethanoic acid 100 acid base 90 80 70 % 60 50 40 30 20 10 0 0 2 4 6 8 pH 10 12 14 Distribution curves for polyprotic acids I The calculation is analogous for polyprotic acids except that there are two (or more) equilibria and three (or more) forms of the acid to consider. I When the pKa ’s of a polyprotic acid differ by several units, the distribution curves look like a simple superposition of distribution curves for the monoprotic case. Example: Ethanedioic (oxalic) acid at pH 3 I We want to calculate the fraction (or percentage) of the acid in each of the three possible ionization states: (COOH)2 , HOOCCOO− and (COO)2− 2 . I The pKa s of the two protons are 1.27 and 4.27. Answer: [HOOCCOO− ] = 93%, [(COO)2− 2 ] = 5.0% and [(COOH)2 ] = 1.7% Distribution curves of ethanedioic acid 100 (COOH)2 HOOCCOO(COO)22 90 80 70 % 60 50 40 30 20 10 0 0 2 4 6 8 pH 10 12 14 Take a log, have an equation named after you. . . I We are now familiar with the equation Ka = (aA− )(aH+ ) aHA I If we take the negative log of this equation, we get aA− − log10 Ka = − log10 aH+ − log10 aHA aA− ∴ pKa = pH − log10 aHA aA− or pH = pKa + log10 aHA I This last equation is called the Henderson-Hasselbach equation.