1 Lecture 8 EA02 November 4, 2002 Today I am going to review fundamentals of acids and bases and acid/base equilibrium constants. Although acids and bases can be defined in other solvents, our focus in this program is almost exclusively on water. Thus we will talk about acids and bases in aqueous solutions. XXXXX Here a re a number of terms dealing with acids and bases. An electrolyte is a substance that forms ions when dissolved in water to produce solutions that conduct electricity. Strong electrolytes ionize essentially completely in a solvent, producing lots of ions in solution than therefore high conductivity. Weak electrolytes do not dissociate completely and thus ionize only partially. Therefore, weak electrolytes impart less conductivity per mole to a solvent than do strong electrolytes. XXXXXXX Here is a list of strong and weak electrolytes. What is a thiocyanate? - salt of an SCNanion. This concept that substances dissociated in solution to form ions was first proposed by a Swedish chemist, Svante Arrhenius in the 1880s. His ideas were not accepted at first. In fact he received the lowest possible passing grade on his Ph.D. examination for proposing such revolutionary and obviously wrong ideas. He received the Nobel Prize for these ideas in 1903. He was also one of the first scientists to suggest that there was a relationship between the amount of carbon dioxide in the atmosphere and global temperature. In 1923, two chemists, J. N. Bronsted in Denmark and J. M. Lowry in England, proposed independently a theory of acid/base behavior that is particularly useful in analytical chemistry. XXXXXX According to the Bronsted-Lowry theory, an acid is a proton donor and a base is a proton acceptor. 2 When an acid and a base react, they produce water and a salt. In order for a species to behave as an acid, a proton acceptor (or base) must be present. In order for a species to behave as a base, a proton donor (or acid) must be present. In other words, a substance acts as an acid only in the presence of a base, and vice versa. An important feature of the Bronsted-Lowry concept is the idea that the entity produced when an acid gives up a proton is itself a potential proton acceptor and is called a conjugate base of the parent acid. For example, when acetic acid gives up a proton, the acetate ion is formed, as shown by the reaction: HOAc < == > OAc- + proton Acidic acid is the acid and acetate is its conjugate base and the two are known as a conjugate acid/base pair. I tend to use the symbol OAc- for acetate which stands for CH3CO2-, or as Harris writes C2H3O2-. Similarly, every base produces a conjugate acid as a result of accepting a proton. For example NH3 + proton < == > NH4+ When these two processes are combined, the result is an acid/base, or neutralization, reaction: HOAc + NH3 < == > OAc- + NH4+ The extent to which this reaction proceeds depends upon the relative tendencies of the two bases to accept a proton (or the two acids to donate a proton). As some of you know, Bronsted-Lowry acids and bases are a subset of Lewis acids and bases, where a base is defined as an electron pair donor and an acid an electron pair acceptor. This is given on page 108. XXXXX Some solutes have both acidic and basic properties and are called amphiprotic species. An example is dihydrogen phosphate ion, H2PO4-, which behaves as a base in the presence of a strong proton donor. H2PO4- + H3O+ < == > H3PO4 + H2O base1 acid2 acid1 base2 3 Here, H3PO4 is the conjugate acid of the original base. In the presence of a strong proton acceptor, such as hydroxide ion, however, H2PO4- behaves as an acid and forms the conjugate base HPO42-. H2PO4- + OH- < == > acid1 base2 HPO42- + H2O base1 acid2 Notice that water is an amphiprotic solvent. HCl + H2O < == > H3O+ + ClNH2- + H2O < == > NH3 + OHAmphiprotic solvents undergo self-ionization, or autoprotolysis, to form a pair of ionic species. Autoprotolysis is yet another example of acid/base behavior, as illustrated by the following equations. base1 H2 O CH3OH NH3 + acid2 acid1 + base2 + H2O < == > H3O+ + OH+ CH3OH < == > CH3OH2+ + CH3O+ NH3 < == > NH4+ + NH2- The extent to which water undergoes autoprotolysis is slight at room temperature. Thus, the hydronium and hydroxide ion concentrations in pure water are only about 10-7 M. Despite the small values of these concentrations, this dissociation reaction is of utmost importance in understanding the behavior of aqueous solutions. XXXXX This transparency shows the dissociation reaction of a few common acids in water. The first two are strong acids because reaction with the solvent is sufficiently complete as to leave no undissociated solute molecules in aqueous solution. The remainder are weak acids, which react incompletely with water to give solutions that contain significant quantities of both the parent acid and its conjugate base. Note that acids can be cationic, anionic, or electrically neutral. The acids in the transparency become progressively weaker from top to bottom. Perchloric acid and hydrochloric acid are completely dissociated in water. In contrast, only about 1% of acetic acid is dissociated. 4 Ammonium ion is an even weaker acid. Only about 0.01% of this ion is dissociated into hydronium ions and ammonia molecules. Strong acids and strong bases are discussed in Harris in section 10-1. Common Strong bases – NaOH, KOH, LiOH Another generality illustrated in the transparency is that the weakest acid forms the strongest conjugate base; that is, ammonia has a much stronger affinity for protons than any base above it. Perchlorate and chloride ions have no affinity for protons. Acid/base equilibrium deals primarily with weak acid and weak bases that do not dissociate completely in aqueous solutions. XXXXX When a weak acid or a weak base is dissolved in water, partial dissociation occurs. Thus, for nitrous acid, we can write [H3O+][NO2-] HNO2 + H2O < == > H3O+ + NO2Ka = [HNO2] Or as our book would write it: HNO2 < == > H+ + NO2- Ka = [H+][NO2-] [HNO2] where Ka, is the acid dissociation constant for nitrous acid. In an analogous way, the base dissociation constant for ammonia is [NH4+][OH-] NH3 + H2O < == > NH4+ + OHKb = [NH3] As we have noted before, the concentration of water does not appear in the denominator in either dissociation constant expression. Thus in common with the ion-product constant for water, [H2O] is incorporated in the equilibrium constants Ka and Kb. XXXXX Consider the base dissociation-constant expression for ammonia and the acid dissociation-constant expression for its conjugate acid, ammonium ion: NH3 + H2O < == > NH4+ + OHNH4+ < == > NH3 + H+ [NH4+][OH-] Kb = [NH3] Ka = [NH3][H+] [NH4+] 5 Multiplication of one equilibrium-constant expression by the other gives Ka Kb so [NH4+][OH-] = [NH3] x [NH3][H+] [NH4+] = [H+][OH-] = Kw K w = K a Kb This relationship is general for all conjugate acid/base pairs. Many compilations of equilibrium-constant data list only acid dissociation constants because it is so easy to calculate basic dissociation constants with this relationship. We can use this equation to confirm the observation that, as the acid of a conjugate acid/base pair becomes weaker, its conjugate base becomes stronger and vice versa. Thus the conjugate base of an acid with a dissociation constant of 10-2 will have a basic dissociation constant of 10-12, whereas an acid with a dissociation constant of 10-9 has a conjugate base with a dissociation constant of 10-5. Which is the stronger acid? Which is the stronger base? Let's look at an example. What is Kb for the equilibrium CN- + H2O < == > HCN + OHFirst, how do we tell that this is a base reaction? Appendix G lists a Ka value of 6.2 x 10-10 for HCN. Thus, Kb = Kw = Ka [HCN][OH-] = [CN-] 1.00 x 10-14 = 1.62 x 10-5 6.2 x 10-10 If your like Dr. Stroh and think in log terms then pKa + pKb = pKw = 14.00 pKb = 14.00 – 9.21 = 4.79 Kb = 10-4.79 = 1.62 x 10-5 Suppose we want to calculate the hydrogen ion concentration in 0.120 M solution of the weak acid, nitrous acid. The principal equilibrium is HNO2 < == > H+ + NO2From Appendix G Ka = 7 .1 x 10-4 = [H+][NO2-] [HNO2] 6 The chemical equation says that for each nitrite ion formed (the conjugate base) there will be one hydrogen ion formed. As long as we are dealing with concentrations of the acid that are large compared to the dissociation of water at 10-7 molar, we can write Charge balance and Mass balance [NO2-] = [H+] F = 0.120 M = [HNO2] + [NO2-] = [HNO2] + [H+] Note that the text uses Formal concentrations in Chapter 10. Remember, this is the total number of moles (in all forms) of the compound that is dissolved in a liter. Rearranging [HNO2] = 0.120 M - [NO2-] = 0.120 M - [H+] When these relationships are introduced into the expression for Ka, we obtain Ka = [H+]2 _ = 0.120 - [H+] 7.1 x 10-4 If we now assume [H+] << 0.120, we find [H+]2 = 7.1 x 10-4 0.120 + [H ] = (0.120 x 7.1 x 10-4)1/2 = 9.2 x 10-3 M We now examine the assumption that 0.120 - 0.0092 0.120 and see that the error is about 8%. We have two significant figure data, so this is a pretty large error. We can improve the value by using the quadratic equation or by doing another approximation. 0.120 - 0.0092 = 0.1108 and if we put that back into the square root we get 8.9 x 10-3 M for the hydrogen ion concentration which is the same answer I got using the quadratic formula. pH = 2.05 Let's use this example to bring in another concept from Chapter 10, that of fraction of dissociation which is given the symbol , and is defined as the fraction of the acid in the Aform. XXXXXXX 7 = . [A-] . [HA] + [A-] = . x . x + (F – x) = x F = 8.9 x 10-3 = 0.074 0.120 Thus nitrous acid is 7.4 % dissociated at a formal concentration of 0.120 M. Now let's look at a weak base example. Our generalized equation from chapter 6 is B + H2O < == > BH+ + OHKb = [BH+][OH-] [B] Formal concentration here is F = [B] + [BH+] Let's calculate the hydroxide ion concentration and pH of a 0.0750 M NH3 solution. The predominant equilibrium is NH3 + H2O < == > NH4+ + OHKb = [NH4+][OH-] [NH3] In Appendix G we find that the conjugate acid, the ammonium ion, has a Ka of 5.70 x 10-10. from chapter 6 KaKb = Kw we get Kb = 1.00 x 10-14 5.70 x 10-10 = 1.75 x 10-5. The chemical equation shows that [NH4+] = [OH-] and from the formal concentration we get F = [NH4+] + [NH3] = 0.0750 M [NH3] = 0.0750 M - [NH4+] = 0.0750 M - [OH-] Substituting these quantities into the expression for Kb yields . [OH-]2 0.0750 – [OH-] = 1.75 x 10-5 which is analogous to the equation we got for the weak acid problem. 8 Assume that[OH-] << 7.50 x 10-2, then this equation simplifies to [OH-]2 = 0.0750 x 1.75 x 10-5 [OH-] = 1.15 x 10-3 M pOH = 2.939 pH = 14.000 - 2.939 = 11.061 This value is less than 2% of F, 0.0750 so should be fine for most purposes. The quadratic formula gives 1.14 x 10-3 M.