Diprotic Acids Diprotic acids have two ionizable protons that undergo successive ionization. + H2A + H2O º H3O + + HA [H3O [HA Ka [H2A + 2 HA + H2O º H3O + + A 2 [H3O [A Ka 2 [HA In general, Ka >> Ka 2. (Harder to remove H + from something already negatively charged.) In this case, the two ionization reactions can be treated as occurring stepwise and we can proceed as follows: The calculation of [H3O +, [HA, and [H2A start exactly the same way as for monoprotic acids. From the first dissociation (Ka ), o [H3O + [HA c (solved using assumptions, successive approximations, or quadratic formula as needed) and o [H2A Facid [H3O + (i.e., Facid c as above). Then, the second ionization step can be considered: HA (aq) + H2O(l) º H3O + (aq) + A 2 (aq) (I) [HA [H3O + 0 C c + c + c [E [HA c [H3O + + c c Ka 2 + 2 [H3O [A [HA [H + 3O c c [HA c + [H3O c [HA c Ka 2 So, [H3O + TOT [H3O + [HA TOT [HA [H2A Facid [H3O + [A 2 Ka 2 Same [H3O +, [HA, and [H2A as if only the first ionization were considered. The second ionization () uses up so little HA and (2) produces so little additional H3O + that it doesn t contribute significantly to those quantities. 3. Salts of weak, monoprotic acids in water conjugate pair HA(aq) + H2O(l) º H3O + (aq) + A (aq) Ka (acid) HA & A A (aq) + H2O(l) º HA(aq) + OH (aq) Kb (conj. base) For any conjugate acid-base pair, KaKb KW As an acid gets stronger, its conjugate base gets weaker (and vice versa) Salts containing ions that hydrolyze (i.e., ionize in water) e.g., CH3COONa contains CH3COO, the conjugate base of CH3COOH (a weak acid), and Na +, the cation of NaOH (a strong base). CH3COO is basic. Na + is not acidic. CH3COONa solution is basic. e.g., NH4Cl contains NH4 +, the conjugate acid of NH3 (a weak base), and Cl, the conjugate base of HCl (a strong acid). NH4 + is acidic. Cl is not basic. NH4Cl solution is acidic. Big Question # 6 350:53 002/004/80
Quantitative treatment: Salt of a Weak, Monoprotic Acid in Water () A (aq) + H2O(l) º HA(aq) + OH (aq) Kb(A ) [OH [A (2) 2H2O(l) º H3O + (aq) + OH (aq) KW Salt of Weak Acid Rules. If KbFsalt >.0 H 0 3, then a. water dissociation unimportant b. solve equilibrium problem 2. If KbFsalt <.0 H 0 3, then a. water dissociation important b. solve using: [OH TOT KbF salt + K W e.g., what is the ph of a 0.0 M solution of sodium acetate? (Ka of acetic acid.8 H 0 5.) The treatment is analogous for salts of weak bases: BH + (aq) + H2O(l) º H3O + (aq) + B(aq) Ka Salt of Weak Base Rules 2H2O(l) º H3O + (aq) + OH (aq) KW. If KaFsalt >.0 H 0 3, then a. water dissociation unimportant b. solve equilibrium problem 2. If KaFsalt <.0 H 0 3, then a. water dissociation important b. solve using: [H3O + TOT KaF salt+ K W Salts containing highly charged metal cations e.g., AlCl3, Fe(NO3)3, CrCl3 Al(H2O)6 3+ + H2O º H3O + + Al(H2O)5(OH) 2+ Ka.0 H 0 5 (similar Ka to acetic acid!) 2+ OH 2 + H 2O º H 3O + H 2 O OH 2 + Al H 2 O O OH 2 H Big Question # 7 350:53 002/004/80
Reminder about effect of molecular structure on oxoacid strength More electron-withdrawing character in the rest of the molecule will polarize the O H bond more greatly, causing the acid strength to increase. On rest of molecule more oxygen atoms more electronegative central atom added electron-withdrawing groups highly charged aquated metal ion O H bond more polarized; increase in acid strength Salts, Conjugate acid-base pairs, and Drugs Many drugs are formulated as salts since water solubility of a salt is often greater than that of a neutral covalent compound. e.g., pseudoephedrine HCl means that the compound is present as the chloride salt of its conjugate acid. Many compounds with amine groups are prepared as hydrochlorides this way. Many weakly acidic compounds are prepared as the sodium salts of their conjugate bases (e.g., naproxen sodium). WARNING! DO NOT do the following acid-base chemistry at home. It is illegal and harmful. + HCl + Cl Salt of a Weak Acid and a Weak Base e.g., Would a solution of methylammonium benzoate, CH3NH3C6H5COO, be acidic or basic? contains CH3NH3 + and C6H5COO ions CH3NH3 + conjugate acid of CH3NH2 (Kb 4.4 H 0 4 ) 4 Ka of CH3NH3 + KW K.00 2.3 H 0 4 4.40 C6H5COO conjugate base of C6H5COOH (Ka 6.3 H 0 5 ) 4 Kb of C6H5COO KW K.00.6 H 0 0 5 6.30 PROTIP The same principle would apply to determining whether an amphoteric species (such as HCO 3 ) is acidic or basic in solution. Calculate its K a and K b, and the larger value would rule. b Big Question # 8 350:53 002/004/80 a
4. Weak, monoprotic acid + salt of the weak acid buffer solution: resists change in ph when small amounts of acid or base are added It s easy (relatively) to transform the Ka expression to relate ph to the ratio of A to HA: + [H3O [A Ka log Ka log [H3O + + log [A Ka [H3O + [A log Ka log HO 3 [A log Ka log [H3O + log [A pka ph log [A ph pka + log [A Henderson-Hasselbach Equation Add OH to a buffer: some H3O + reacts to consume it. HA(aq) + H2O(l) º H3O + (aq) + A (aq) [A ; ; [A increases; ph increases slightly Add H3O + to a buffer: some A reacts to consume it. HA(aq) + H2O(l) º H3O + (aq) + A (aq) [A ; ; [A decreases; ph decreases slightly Relative amounts of HA and A are important!!! Large amounts of HA and A will keep relative changes small when small amounts of acid or base are added. buffer capacity: measure of the ability of a buffer to resist ph change the amount of acid/base a buffer can absorb without significant change in ph usually defined as ± ph unit, buffer range Optimal buffering occurs when [A. o When [A, solution will buffer equally well against addition of acid or base. o If not equal, the buffer will be better against either acid or base than the other. ph pka + log [A, so if [A, then ph pka. e.g., if you want a ph 4 buffer with [A, you need to find an acid with pka 4, i.e., Ka 0 4. A Buffer Prepared with a Base and its Conjugate Acid B(aq) + H2O(l) º BH + (aq) + OH (aq) poh pkb + log [BH [B e.g., if you want a ph 0 buffer with [BH + [B, you need to find a base with pkb 4, i.e., Kb 0 4. CAUTION!!! It is NOT necessarily true that a basic buffer (a buffer solution with ph >7) must be made with a weak base and its conjugate acid. For example, if you used equal concentrations of HA and A where HA had Ka.0 0 8, then the ph of that buffer would be 8.00. Big Question # 9 350:53 002/004/80