**On first encounter, the study of acid-base equilibria is a little like a strange land quantum.bu/notes/GeneralChemistry/titration-recipes.pdf.
30 pages**

**129 KB – 30 Pages**

PAGE – 1 ============

Chapter 8, Acid-base equilibria Road map of acid-base equilibriaOn first encounter, the study of acid-base equilibria is a little like a strange land with seemingly confusing trails that make passage difficult. In fact, there is a road map that, once understood, allows us to navigate acid-base equilibria with confident precision and so become masters of its domain. Here is an overview of this road map. Fundamentally, aqueous acid-base equilibria are just a particular example of the ideas and techniques we have already learned in the study of gas phase chemical equilibria. However, there are two aspects that complicate the application of these ideas. First, because the autoionization of water, 2H2OlH3OaqOHaq,Kw,is always present in aqueous solution, the analysis of aqueous acid-base equilibria must always take into account at least two competing equilibria, the acid or base ionization and the water autoionization. Second, because we will be interested in how acid-base equilibria respond to changes in the system (typically by adding additional base or acid), we need also to be able to separate the chemical reactions that take place when things are combined from the subsequent equilibration of the reaction product. The key idea is to let what are combined react 100% as a limiting reagent problem as a preliminary step done before equilibration.Our overarching goal, then, is to learn to clearly distinguish and to separately master these two aspectsŠcompeting equilibria and limiting reagent reactions. Once this is achieved, we will have a framework in which any problem in aqueous acid-base equilibria can be solved in a straightforward way. The approach we take is to distinguish four regimes : 1. pure acid (or base) solution, 2. acid (or base) that has been partially consumed by addition of base (or acid), 3. acid (or base) that has been exactly consumed by addition of base (or acid), 4. acid (or base) that has been consumed by reaction with excess base (or acid). The details will be different depending on whether the acid (or base) initially present is a weak or strong acid (or base), that is, whether its ionization constant is large or small compared with 1. In all cases, the other component will be a strong base (or acid). The first regime is a straightforward equilibrium problem. The second regime involves reaction of the acid and base as a limiting reagent problem, followed by straightforward equilibration of the resulting solution. The third regime is handled differently depending on whether the acid (or base) initially present is a weak or strong acid (or base). The fourth regime, like the second, involves reaction of the acid and base as a limiting reagent problem, but then equilibration of the products is simple since it will always be the excess base (or acid) that is left after the limiting reagent reaction. You will know that you have mastered this road map of acid-base equilibria when you approach such an equilibrium as a two step process: First (always!), determine which regime applies . Then, implement the calculational procedure for that regime to determine the chemical equilibrium. That really is all there is to it. In this way you can come to find that things will seem simple; not easy, but simple.

PAGE – 2 ============

So, our plan is to learn about each of the four different regimes. We’ll do this for each one in turn. Then we’ll see how to put everything together, to construct a titration curveŠthe variation of pH throughout the four regimes. Finally, we’ll construct a compact summary of the road map. So let’s begin, by defining just what acid-base reactions are and how to characterize them. Acid-base reactionsWe are going to be working with acid-base equilibria in aqueous solution, and we will use the Brønsted-Lowry definitions that an acid is a source of H and a base is an acceptor of H .Acid ionization constant KaA convenient way to write the reaction of an acid HA in water is HAaqH2OlH3OaqAaq.Here water is acting as a base, accepting the H ; the result, H 3O, called the conjugate acid of H 2O, since H3O can donate H to reform H2O. In a similar way, A is called the conjugate base of the acid HA, since A can accept H to reform HA. The equilibrium constant is known as the acid ionization constant Ka, KaH3O1MA1M HA1M,where the reference concentration, 1 M, is written explicitly; liquid water does not appear, because its reference concentration is just the liquid water, so its activity is 1. (Strictly, we are assuming that solutes are sufficiently dilute that the concentration of the water is negligibly affected by their presence.) Usually, we will abbreviate Ka as KaH3OA HA,with the understanding that [–] stands for the numerical valueŠwithout unitsŠof the concentration in mol/L. As usual, Ka is unit-less. An example acid ionization is CH3COOHaqH2OlH3OaqCH3COO aq,KaH3OCH3COO CH3COOH.If Ka is much greater than 1, the acid is mostly dissociated and so is said to be a strong acid. If Ka is much less than 1, the acid is dissociated only to a small extent and so is said to be a weak acid. Base ionization constant KbSimilarly, we can write the reaction of a base B in water as 206Notes on General Chemistry, 2e Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 3 ============

H2OlBaqHBaqOHaq.Here water is acting as an acid, donating the H ; the result, OH, is called the conjugate base of H2O, since OH can accept H to reform H 2O. Analogously, HB is called the conjugate acid of the base B, since HB can donate H to reform B. The equilibrium constant is known as the base ionization constant Kb,KbHBOH B.,where we have used the abbreviated form. Here is an example base ionization, H2OlNH3aqNH4aqOHaq,KbNH4OH NH3.If Kb is much greater than 1, the base reacts nearly completely with water and so is said to be a strong base. If Kb is much less than 1, the base reacts hardly at all with water and so is said to be a weak base. Water autoionization constant KwIn this analysis of acid and base ionization we see that water in one case plays the role of a base and the other plays the role of an acid. Indeed, the role that water plays in an aqueous equilibrium can be used as another definition of acid or base. A consequence of this dual role of water is that its equilibrium with H 3O and OH is the reference standard against which aqueous acidity and basicity are defined. Here is how this works. A general base ionization reaction is H2OlBaqHBaqOHaq,Kb,and the ionization reaction of its conjugate acid is HBaqH2OlH3OaqBaq,Ka.The sum of these two reactions is 2H2OlH3OaqOHaq,Kw.This equation is called the autoionization of water and its equilibrium constant is known as the water autoionization constant K w. At 25°C it is equal to KwH3OOH1.01014Now, we have seen that the equilibrium constant of a sum of two reactions is the product of the equilibrium constants of the summed reactions. Therefore, we know that water autoionization constant can be expressed as KwKaKb.Chapter 8, Acid-base equilibria 207Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 4 ============

This means we can compute the base ionization constant from the ionization constant of its conjugate acid, KbKwKa1014Ka,where the last equality is for 25°C. It is for this reason that base ionization constants are generally not tabulated. pHBecause Kw is so small, water is ionized only very slightly. We can compute the concentration of H3O and OH by solving the autoionization equilibrium. H3OMOHMinitial00 equilibriumxx Initial and equilibrium activities for the autoionization of liquid water. The equilibrium expression is then Kw1.1014x2and so the concentration of H 3O and OH are each 1107 M at 25°C. Now, as we will see in a moment, strong acids can have H 3O concentrations of 1 M or more. This means that H 3O varies over many powers of 10 (orders of magnitude), and so it is convenient to measure [H 3O] on a logarithmic scale. Also, for weak acids and bases, which is what we will be interested in primarily, concentrations of H 3O and OH are generally much less than 1 M, which means their logarithms are negative, and so it is more convenient to work with the negative logarithms so that we have a positive quantity. For this reason, the notation pAnything is defined aspAnythinglog10Anythingand in particular pH and pOH are defined as pHlog10H3O1Mlog10H3O,pOHlog10OH1Mlog10OH,where in each case the second equality is written with understanding that [–] stands for the numerical valueŠwithout unitsŠof the concentration in mol/L. Show that pure water at 25°C has pHpOH7.To get a feeling for the amounts involved, how many OH ions are present in 1 liter of pure water at 25°C? How many H 2O molecules are present? The molecules and ions present in water are constantly buffeting one another. On average, how many collisions does a water molecule make before it encounters an hydroxide ion? Since acids produce additional H 3O, their pH is always less than 7. Similarly, since bases produce additional OH, their pOH is always less than 7. We can relate pH and pOH, by calculating pKw. 208Notes on General Chemistry, 2e Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 5 ============

pKwlog10Kwlog10H3OOHlog10H3Olog10OHlog10H3Olog10OHpHpOHThis means that at 25°C pH14pOH.so that bases have pH greater than 7, at 25°C. Calculate the pH and pOH of the following HCl solutions, assuming the HCl dissociates 100%: 15 M, 10 M, 5 M, 1 M, 0.1 M and 0.01 M. Can pH be less than 0? Calculate the pH and pOH of the following NAOH solutions, assuming the NaOH dissociates 100%: 15 M, 10 M, 5 M, 1 M, 0.1 M and 0.01 M. Can pH be greater than 14? It is important always to be mindful of the temperature being used. The reason is that equilibrium constants in general, and the value of Kw in particular, are different at different temperatures. This means that the pH of pure water will be different at different temperatures. For this reason, saying a solution has pH = 7 does not, by itself, mean the solution is “neutral” (has equal concentrations of H3O and OH. More generally, the acidic or basic character of a solution is due to the relative concentrations of H 3O and OH , rather than a particular numerical value of pH. Like all equilibrium constants, the autoionization constant of water changes with temperature. At 10°C its value is 0.292 1014 and at 30°C its value is 1.47 1014. Calculate the pH and pOH of water at 10°C and 30°C. Based on your results, does the “acidity” of water depend on temperature? Use the dependence the values of the equilibrium constants in the previous problem to estimate the standard enthalpy change and the standard entropy change of the water autoionization reaction.Origin of the notation pHSeveral years ago a student asked where the notation “p” came from. Another student answered that it was from the French word pouvoir (in this sense, capability), referring to the power of ten. I asked a French-speaking colleague about this, and her response was that perhaps it was from the French word puissance (power) instead. Eventually we checked the Oxford English Dictionary, 2e . Here is what it says: “Introduced (in Ger.) as pH, by S. P. L. Sörensen 1909, in Biochem. Zeitschr. XXI. 134, the p repr. G. potenz power and H the hydrogen ion.” That is, it is from the German (!) word potenz (power). So, the next question is, can we get a reference tracing an earlier French use? That way we could decide which use takes precedence. Now that we have seen what acid-base reactions are and how to characterize them, let’s follow our road map to study these reactions in their four different regimes. Chapter 8, Acid-base equilibria 209Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 6 ============

Regime 1: pure strong acid (or base) HCl is a strong acid, with Ka107. What is the pH of 10 M HCl? We can begin as usual by setting up the equilibrium. The ionization reaction is HClaqH2OlH3OaqClaq.The equilibration analysis is HClMH3OMClMinitial10.1. 1070.equilibrium10. 1.×1. 107xxInitial and equilibrium activities for the aqueous ionization of 1 M HCl. Now, because Ka is so large, at equilibrium the HCl will be nearly 100% ionized. This means that x is very large. In a case such as this we have learned that it makes sense to use revised initial conditions, starting from 100% ionized acid. HClMH3OMClMinitial10.1. 1070.revisedinitial0.10.10. equilibriumy10. 1.y10. 1.y Initial, revised initial and equilibrium activities for the aqueous ionization of 1 M HCl. Here we have included the contribution 10 7 M H 3O, due to the autoionization of water, but this is negligible compared to the 10 M H 3O from the 10 M HCl. The resulting equilibrium expression is Ka10710y2 y.Now, since Ka1, most of the HCl is ionized, the change y to achieve equilibrium is small. This means that 10 y is negligibly different from 10, and so we can approximate the equilibrium constant expression as Ka107102 y.so that y1105.Confirm this result. Show that the percent of 10. M HCl that is undissociated is 1104, that is, that 10 M HCl becomes 99.9999% dissociated. The net result of all of this is that the molarity of H 3O is the same as the nominal (that is, ignoring any dissociation) molarity of the HCl. Thus, the pH of 10 M HCl is pHlog10101210Notes on General Chemistry, 2e Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 8 ============

Weak acids, on the other hand, only donate a small portion of their hydrogen ions to the solution. This is because the Ka of a weak acid is much less than 1, and so a weak acid is mostly undissociated . This means that, while the hydronium ion concentration of a weak acid is greater than 10 7 M, it is much less than that of an equivalent amount of strong acid. Pure weak acid example Here is how to calculate the hydronium ion concentration of a weak acid. As example, let’s work with 1.712 M acetic acid, which we’ll write simply as HA. The acid ionization is HAaqH2OlH3OaqAaqand Ka1.76105. We begin by assuming that initially no dissociation occurs, so that HAca1.712 M and Acb0. The initial H3O107, due to the autoionization of water. Because cb is 0, the reaction quotient is Q0, which is less than Ka. This means that initially we have too much reactants, not enough products. Therefore, at equilibrium we must decrease the undissociated acid concentration to cax, and so increase the conjugate base concentration from 0 to x, and increase the hydronium ion concentration from its value in pure water to 107x. The amount of dissociation, x, is determined by solving the equilibrium expression Ka1.76105107xx cax107xx 1.712x.We can simplify this equation by making two approximations. First, because Ka is much larger than the autoionization constant of water, Kw, we know that the hydronium ion concentration x contributed by the acid dissociation is greater than 10 7. This in turn means that x is not negligible compared to 10 7. We will assume, in fact, that x is so much greater than 10 7 that we can ignore 107 relative to x, 107xx.Second, because Ka is much less than 1, HA dissociates only to a small extent. This means x is small compared to the initial concentration of the acid, ca. We will assume, in fact, that x is so much smaller than ca that we can ignore x relative to ca, caxca.Notice that in one case we are neglecting something compared to x, and in the other case we are neglecting x compared to something! With these two approximations, the equilibrium expression becomes Ka1.76105107xx caxx2 ca,which we can solve by taking the square root to get x caKa.Use the values of Ka and ca to show that x5.49103.This result shows that both of our approximations are justified: 5.49 103 is indeed large compared to 107, but 5.49 103 is also small compared with 1.712. Always be sure to check to see that this is so. 212Notes on General Chemistry, 2e Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 9 ============

Show that the corresponding pH is 2.260. How many undissociated acetic acid molecules are present in 1 liter of 1.712 M acetic acid at 25°C? How many acetate ions are present? On average, how many collisions does an acetate ion make before it encounters a hydronium ion? Calculate the pH of 0.100 M acetic acid at 25°C. Besure to include the correct number of siginifcant figures in your answer.Compare the ratio of acetate ions to undissociated acetic acid at 25°C that are present in a 1.712 M solution and in a 0.100 M solution. Should the ratios be different? If the ratios differ, could you have predicted which of the ratios would be larger without doing any computations? Pure weak base example The same kind of analysis works for a weak base. Now the ionization reaction is H2OlBaqHBaqOHaqand the equilibrium expression is Kbx107x cbxSince (we assume) Kb is greater than Kw, the hydroxide concentration, 10 7x, is greater than 107, and so x is not negligible compare to 10 7. As before, we assume that x is in fact so much larger than 107 then we can ignore the 10 7. But also, since (we assume) the base is weak, so that Kb is less than 1, it reacts with water only to a small extent, and so x is small compared to cb. The result is that the equilibrium expression becomes Kbx107x cbxx2 cbwhich we can solve by taking the square root to get x cbKb.For example, the Kb of NH3 is 1.8105. We get this value from the Ka of the conjugate acid, NH4, which is 5.61010. Show that Kb1.8105.Show that a 0.112 M solution of NH 3 has hydroxide ion molarity x1.41103.As before, this value of x shows that the approximations are justified. Show that the pH is 11.15. Chapter 8, Acid-base equilibria 213Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 10 ============

Regime 3: Neutralized weak acid (or base): hydrolysis If we know the volume and molarity of a weak acid (or base), we can compute how many moles of the acid (or base) we have. If we add an amount of a strong base (or acid) that has the same number of moles, then we will exactly neutralize the weak acid (or base). The resulting solution will contain the conjugate base of the acid and the conjugate acid of the base. For example, if we neutralize acetic acid with sodium hydroxide, we will end up with a solution of acetate ions (the conjugate base of the acetic acid), water, (the conjugate acid of the hydroxide), and sodium ionsŠa solution of the salt sodium acetate. The surprising thing is that the pH is such solutions is not 7. Here is why. Because the acid HA is weak, its conjugate base, A , is much more likely to be found attached to H , as HA, than as unattached A. That is, the equilibrium H2OlAaqHAaqOHaqrests heavily on the side of products. A consequence of this attachment process is an increase in the concentration of OH and so a decrease in the concentration of H 3O (since their concentrations are inversely related through the water autoionization equilibrium). The result is that solutions of salts of weak acids are basic. In an analogous way we can see that solutions of salts of weak bases are acidic . The conjugate acid, BH, of the weak base, is more likely to exist as the unattached base, B. That is, the equilibrium BHaqH2OlH3OaqBaqstrongly favors products. This donation of the proton to the water makes the solution acidic. The reaction of the conjugate base (acid) of a weak acid (base) with water is called hydrolysis, because the effect is to break apart water. We can calculate the pH due to the hydrolysis by solving the hydrolysis equilibrium expression. The equilibrium constant is related to the equilibrium constant of the weak acid (base) as KbKwKaKaKwKb.For example, let’s calculate the pH of 1.000 M ammonium chloride, formed by neutralization of ammonia (a weak base) by hydrochloric acid (a strong acid). The Kb of ammonia is 1.8 105, and so the Ka of the conjugate acid NH4 is 5.61010.Note that because Kw is so small, usually the hydrolysis equilibrium constant is quite small. Here is the table of initial and equilibrium concentrations. NH4MH3OMNH3Minitial1.1. 1070.equilibrium1. 1.×1. 107xxInitial and equilibrium activities for the hydrolysis of 1.000 M NH4.Now, we can approximate x as large compared to 10 7, since the resulting solution will be acidic, and we can approximate x as small compared to 1.000, since the equilibrium constant is so small. This leads to Kax107x caxx2 ca214Notes on General Chemistry, 2e Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

PAGE – 11 ============

and sox caKaThis evaluates to x2.4105.Calculate the pH of 1.000 M ammonium chloride at 25°C. In a similar way you can calculate that solutions of the salt of a weak acid will be basic. Calculate the pH of a 1.000 M solution of sodium acetate at 25°C. Note that sodium acetate when dissolved in water dissociates into sodium ions and acetate ions, and that acetate is the conjugate base of the weak acid acetic acid. Explain quantitatively why we do not need to worry about hydrolysis of the salts of strong acid neutralized by a strong base. When hydrolysis is important: Weak can be strong! We have learned that aqueous solutions of salts of weak acids are basic , and that aqueous solutions of salts of weak bases are acidic. The reason these solutions have pH different from that of pure water is because the salts hydrolyze water. The hydrolysis equations are H2OlAaqHAaqOHaq,pH 7basicBHaqH2OlH3OaqBaq.pH 7acidicThe question is, why do such salts hydrolyze water? . The answer, as we have seen, is because HA is preferred over A , since HA is a weak acid, and because B is preferred over BH , since B is a weak base; therefore the equilibrium point of the hydrolysis equations is far to the product (HA and B) side. Now, here is another way to look at these results. Let’s consider three different acids, HCl Ka107), H2O (viewed as an acid), and a weak acid HA with Ka105. The hydrolysis reactions and equilibrium constants of the corresponding conjugate bases are H2OlClaqHClaqOHaq,KbKw1071021,H2OlH2OlH3OaqOHaq,KbKw1014,H2OlAaqHAaqOHaq,KbKw105109.The first thing to notice is that each of the conjugate bases is quite weak in the absolute sense, that is, in each case Kb1. The second thing to notice, however, in that in a relative sense, the conjugate base of the strong acid is a weaker base than water, KbHClKw, and so a lesser source of OH , while the conjugate base of the weak acid is a stronger base than water, KbAKwand so a greater source of OH .This analysis shows that we can ignore hydrolysis by the conjugate base of a strong acid, in the sense that this hydrolysis has negligible effect on OH concentration compared to the autoionization of water. On the other hand, we cannot ignore hydrolysis by the conjugate base of a weak acid, in the sense that this hydrolysis results in OH concentration that dwarfs that due to the autoionization of water. Chapter 8, Acid-base equilibria 215Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

**129 KB – 30 Pages**