Chemistry part 28, Julia Burdge,2e (2009) pps

682 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria The percent ioniza t ion of acetic acid is 13~~~~ 3 M X 100% = 1.3% By adding so dium acetate, we also add so d ium ions to the solutio n, However, sodium ions do not interact with water or with any of the other species present [ ~ Section [6 . 10]. Remember that prior to the ionization of a wea k acid, the concentrat io n of hydrogen ion in water at 25°( is 1.0 x 1 0- 7 M. H owever, becau se this concentration is insignificant comp ared to the conc entration re su lting from the i oniza ti on, we can neglect it in our equilibrium table. The Common Ion Effect Up until now, we have discussed the properties of solutions containing a single solute. In this sec- tion, we will examine how the properties of a solution change when a second solute is introduced. Recall that a system at equilibrium will shift in response to being stressed and that stress can be applied in a variety of ways, including the addition of a reactant or a product [I •• Section 15.5]. Consider a liter of solution containing 0.10 mole of acetic acid. Using the Kafor acetic acid (1.8 X 10- 5 ) and an equilibrium table [ ~. Section 16.5], the pH of this solution at 25°C can be deter mined: CH 3 COOH(aq) +:. ===z:. H +(aq) + CH 3 COO- (aq) Initial concentration (M): 0.10 0 0 Change in concentration (M): -x +x +x Equilibrium concentration (M): 0.10 - x x x Assuming that (0.10 - x) M = 0.10 M and solving for x, we get 1.34 X 10- 3 . Therefore, . . . . . . . . . . . . . . . . , ., 3 [CH 3 COOH] = 0.09866 M, [H+] = [CH 3 COO - ] = 1.34 X 10- M and pH = 2.87. Now consider what happens when we add 0.050 mole of sodium acetate (CH 3 COONa) to the solution. Sodium acetate dissociates completely in aqueous solution to give sodium ions and acetate ions: CH 3COONa(aq) H2 0 , Na \aq) + CH 3 COO- (aq) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., Thus, by adding sodium acetate, we have increased the concentration of acetate ion. Because acetate ion is a product in the ionization of acetic acid, the addition of acetate ion causes the equi- librium to shift to the left. The net result is a reduction in the percent ionization of acetic acid. Addition / H+(aq) + CH3COO- (aq) Equilibrium is driven toward reactant. Shifting the equilibrium to the left consumes not only some of the added acetate ion, but also some ofthe hydrogen ion. This causes the pH to change (in this case the pH increases). Sample Problem 17.1 shows how an equilibrium table can be used to calculate the pH of a solution of acetic acid after the addition of sodium acetate. Sample Problem 17.1 ", " Determine the pH at 25°C of a solution prepared by adding 0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.) Strategy Construct a new eqUilibrium table to solve for the hydrogen ion concentration . Setup ' We' ~se ' the ' s ' tat~d concei;t~~iion ci ~~etic ' a c i'd : o.io M: ~nd [H +] = 0 M as the initial concentrations in the table: CH 3 COOH(aq) :.;::. =::t, H+(aq) + CH 3 COO- (aq) Initial concentration (M): 0.10 o 0.050 Change in concentration (M): - x +x +x Equilibrium concentration (M): 0.10 - x x 0.050 + x Solution Substituting the equilibrium concentrations, in terms of the unknown x, into the equilibrium expression gives 1.8 X 10-5 = (x)~o.~o + x) .1 - x Because we expect x to be very small (even smaller than 1.34 X 10- 3 M- see above), because the • ionization of CH 3 COOH is suppressed by the presence of CH 3 COO -, we assume ' (0.10 - x) M = 0.10 M and (0.050 - x) M = 0.050 M SECTION 17.2 Buffer Solutions 683 Therefore, the equilibrium expression simplifies to ~5 (x)(0.050) 1.8 X 10 = 0.10 and x = 3.6 X 1O ~5 M. According to the equilibrium table, [H+] = x, so pH = - log (3.6 X 1O ~5) = 4.44 . . Think About It The equilibrium concentrations of CH 3 COOH, C H 3 COO ~, and H+ are the same regardless of whether we add sodium acetate to a solution of acetic acid, add acetic acid to a solution of sodium acetate, or dissolve both s pe cies at the same tim e. We could have constructed an equilibrium table starting with th e equilibrium concentrations in the 0.10 M acetic acid solution: • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • In this case, the percent i oniza t ion of ace tic acid is 3.6 OX1~0~s M x 1 00% = 0.036% Th is is cons i derably sma lle r than the per cent ionization prior to the addition of so diu m acetate. Initial concentration (M): 0.09866 1.34 X 1O ~3 5.134 X 1O ~2 This is the sum of the equ ilib rium concentra tio- Change in concentration (M): t I of acetate ion in a 0.10 M so lu tion of ace tic aoc + ::.y + y" , t :o.y___ (1.3 4 x 1 0 ~3 M) and the added ace t ate ion Equilibrium concentration (M): 0.09866 + Y 1.34 X 1O ~3 - Y 5.134 X 1O~ 2 - Y (0.050 M). In this case, the reaction proceeds to ,the left. (The acetic acid concentration increases, and the concentrations of hydrogen and acetate ions decrease.) Solving for y gives 1.304 X 1O~ 3 M. [H+] = 1.34 X 1O~ 3 - Y = 3.6 X 1O ~ 5 M and pH = 4.44. We get the same pH either way . . • • • • • •• ••• Practice Problem A Determine the pH at 25°C of a solution pr epared by dissolving 0.075 mole of sodium acetate in 1.0 L of 0.25 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.) Practice Problem B Determine th e pH at 25°C of a solution prepared by dissolving 0.35 mole of sodium acetate in 1.0 L of 0.25 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution .) An aqueous solution of a weak: electrolyte contains both the weak: electrolyte and its ioniza- tion products, which are ion s. If a soluble salt that contains one of those ions is added, the equi- librium shifts to the left, thereby suppressing the ionization of the weak: electrolyte. In general, when a compound containing an ion in common with a dissolved substance is added to a solution at equilibrium, the equilibrium shifts to the left. This phenom enon is known as the common ion effect. Checkpoint 17.1 The Common Ion Effect 17.1.1 Which of the following would cause a decrease in the percent ionization of nitrous acid ( HN0 2 ) when added to a solution of nitrous acid at equilibrium? (Select all that apply.) a) NaN0 2 b) H 2 0 c) Ca(N0 2 h d) HN0 3 e) NaN0 3 Buffer Solutions 17 .1.2 What is the pH of a solution prepared by adding 0.05 mole of NaF to 1.0 L of 0.1 M HF at 2YC? (Assume that the addition of NaF does not change the volume of the solution.) (Ka for HF = 7.1 X 1O ~4 . ) a) 2.1 b) 2.8 c) 1.4 d) 4.6 e) 7.3 . . . . . A solution that contains a weak acid and its conjugate ba se (or a weak: base and its conjugate acid) is a buffer solution' or 'slmpiy a ' buffer: B' uff er 'soiutioiis; by ' vli-tiie ' o(i h e1r 'comPOSItIon; ' r(dst ' . changes in pH upon addition of small amounts of either an acid or a base. The ability to re sist pH change is very important to chemical and biological systems, including the human body. The pH of blood is about 7.4, whereas that of gastric juices is about 1.5. Each of these pH values is crucial for proper enzyme function and the balance of osmotic pressure, and each is maintained within a very narrow pH range by a buffer. Trea ti ng the problem as though both CH 3 COm- a nd CH 3 COO- are added at the same time a nd • the reaction proceeds to the ri ght simp li fies the solut ion. The common ion can also be H+ or O H ~ . For example, addition of a strong acid to a so luti on of a weak acid supp r esses io niza tion of the ~ acid. Similarly, addition of a st rong base to a solu t ion of weak base suppre sses ionizati on of the weak base. Any solution of a weak acid co nta in s some co njugate base. In a buffer so luti on, t houg h, ;;-:: amounts of weak acid and conj ug ate base m= be comparable, meaning that the con ju ga te hax must be sup plied by a dissolved salt. 684 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria Remember that sod ium acetate is a strong electrolyte [ ~ ~ Sect ion 4 ,1]. so it dissociates completely in water to give sodium ions and acetate ions: . ,, ' . Multimed ia Acids and Bases effect of addition of a strong acid and a strong base on a buffer, The forward reaction is suppressed by the presence of the common ion, CH 3 COO -, and the reverse process is suppressed by the presence of CH 3 COOH. As long as th e amount of strong acid added to the buffer does not exceed the amount of conjugate ba se originally present, all the added acid will be consumed and converted to weak acid, Calculating the pH of a Buffer ., C'onsider' a ;;o lution th'at i's 'i:6 M Ii; aceti'c ' aClC!" and 'i:6 M sodium acetate. If a small amount of acid is added to this solution, it is consumed completely by the acetate ion, thus converting a strong acid (H+) to a weak acid (CH 3 COOH). Addition of a strong acid lowers the pH of a solution. However, a buffer's ability to convert a strong acid to a weak acid minimizes the effect of the addition on the pH. Similarly, if a small amount of a base is added, it is consumed completely by the acetic acid, thus converting a strong base (OH-) to a weak base (CH 3 COO-). Addition of a strong base increases the pH of a solution. Again, however, a buffer's ability to convert a strong base to a weak base minimizes the effect of the addition on pH. To illustrate the function of a buffer, suppose that we have 1 L of the acetic acid-sodium acetate solution described previously. We can calculate the pH of the buffer using the procedure in Section 17.1: CH3COOH(aq) +:. =::z:' H+(aq) + CH 3 COO-(aq) Initial concentration (M): 1.0 0 1,0 Change in concentration (M): -x +x +x Equilibrium concentration (M): 1.0 - x x 1.0 + x The equilibrium expression is (x)(l.O + x) K = - , ::- a 1.0 - x . . . . . . . . . . . . . . . . . . . . . . . . . ., . Because it is reasonable to assume that x will be very small, (1.0 - x) M = 1.0 M and (1.0 + x) M = 1.0 M Thus, the equilibrium expression simplifies to 1.8 X 10- 5 = (x)(l .O ) = x 1.0 At equilibrium, therefore, [H+] = 1.8 X 10- 5 M and pH = 4.74. Now consider what happens when we add 0.10 mole of HCI to the buffer. (We assume that the addition of HCI causes no change in the volume of the solution.) The reaction that takes place , ' ~ . " when we add a strong acid is the conversion of H to CH 3 COOH. The added acid is all consumed" along with an equal amount of acetate ion. We keep track of the amounts of acetic acid and acetate ion when a strong acid (or base) is added by writing the starting amounts above the equation and the final amounts (after the added substance has been consumed) below the equation: Upon addition of H+: 1.0 mol 0.1 mol 1.0 mol CH 3C OO - (aq) + H +( aq) , CH 3 COOH(aq) After H+ has been consumed: 0.9 mol o mol 1.1 mol We can use the resulting amounts of acetic acid and acetate ion to construct a new equilibrium table: CH 3COOH(aq) +:. =::z:' H+(aq) + CH 3 COO-(aq) Initial concentration (M): 1.1 0 0.9 Change in concentration (M): -x +x +x Equilibrium concentration (M): 1.1 - x x 0.9 + x We can solve for pH as we have done before, assuming that x is small enough to be neglected, SECTION 17.2 Buffer Solutions 685 8 - 5 (x)(0.9 + x) (x)(0.9) 1. XlO = =- - - 1.1 - x 1.1 x = 2.2 X 10- 5 M Thus, when equilibrium is reestablished, [H+] = 2.2 X 10- 5 M and pH = 4.66 ~ . ~h~~g~ ' ~f"~ ' ~iy """ 0.08 pH unit. In the determination of the pH of a buffer such as the one just described, we always neglect the small amount of weak acid that ionizes (x) because ionization is suppressed by the presence of a common ion. Similarly, we ignore the hydrolysis of the acetate ion because of the presence of acetic acid. This enables us to derive an expression for determining the pH of a buffer. We begin with the equilibrium expression Rearranging to solve for [H+] gives Taking the negative logarithm of both sides, we obtain or Thus, where + [HA] - log [H ] = -log Ka - log [A ] [A - ] - log [H +] = -log Ka + log [HA] [A - ] pH = pKa + log [HA] pKa = - log Ka Equation 17.1 Equation 17.2 Equation 17.1 is known as the Henderson-Hasselbalch equation. Its more general for III is [conjugate base] pH = pKa + log [weak acid] Equation 17.3 In the case of our acetic acid (1.0 M) and sodium acetate (1.0 M) buffer, the concentrations of weak acid and conjugate base are equal. When this is true, the log term in the Henderson-Hasselbalch equation is zero and the pH is numerically equal to the pK a . In the case of an acetic acid-acetate ion buffer, pKa = -log 1.8 X 10 - 5 = 4.74. After the addition of 0.10 mole of HCl, we determined that the concentrations of acetic acid and acetate ion were 1.1 M and 0.9 M, re spectively. Using these concentrations in the Henderson- Hasselbalch equation gives . [CH 3 COO-] pH = 4.74 + log [CH 3 COOH] = 4.74 + 1 0.9 M og 1.1 M = 4.74 + (- 0.087) = 4.65 The small difference between this pH and the 4.66 calculated using an equilibrium table is due to differences in rounding. Figure 17.1 illustrates how a buffer solution resists drastic changes in pH . • Had we added 0.1 0 mole of He I to 1 L of pU lE water, the pH would have gone from 7. 00 to 1 .00! Media Player /MP EG Animation : Figure 17.1, Buffer Solutions, pp. 686-68 7. Figure 17.1 686 0.100 MCH 3 COOH 0.100 MCH 3 COO- [CH 3 COO- ] = 4.74 + log [CH 3 COOH] = 4.74 The buffer solution is 0.100 M in acetic acid and 0.100 M in sodium acetate. 100 mL of this buffer contains (0.100 mollL)(O.lO L) = 0.010 mol each acetic acid and acetate ion. Water pH = 7.00 aO When we add 0.001 mol of strong acid, it is completely consumed by the acetate ion in the buffer. Before reaction: 0.001 mol 0.010 mol 0.010 mol • H+(aq) + CH 3 COO-(aq) • CH 3 COOH(aq) After reaction: 0 mol 0.009 mol 2S.0 ·c When we add 0.001 mol of strong base, it is completely consumed by the acetic acid in the buffer. 0.011 mol Before reaction: 0.001 mol 0.010 mol 0.010 mol OH-(aq) + CH 3 COOH(aq) • H20 (0 + CH 3 COO- (aq) After reaction: 0 mol 0.009 mol 0.011 mol N 7 • - We can calculate th€ new pH using the Henderson-Hasselbalch equation: 0.009 pH = 4.74 + log 0.011 = 4.65 We can calculate the new pH using the Henderson-Hasselbalch equation: 0.011 pH= 4.74 + log 0.009 =4.83 - ~ There is nothing in pure water to consume strong acid. Therefore, its pH drops drastically. H = -1 0.001 mol = 200 P og O.lDL . . Ther€ is nothing in pure water to Gonsume strong base. Therefore, its pH rises drastically. 0.001 mol pOH = -log 0.10 L = 2.00, pH = 12.00 What's the point? A buffer contains both a weak acid and its conjugate base. * Small amounts of strong acid or strong base are consumed by the buffer components, thereby preventing drastic pH changes. Pure water does not contain species that can consume acid or base. Even a very small addition of either causes a large change in pH. * A buffer could also be prepared using a weak base and its conjugate acid. 687 688 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria The volume of the buffer is 1 L in this ex ampl e, so the number of moles of a substance is equal to the molar con centration. In c ases where the buffer volume is something other than 1 L, however, we can still use molar amounts in the Henderson-Hasselbalch equati on becau se the volume would cancel in the top and bottom of the log term. Think About It Always do a "reality check" on a calculated pH. Although a buffer does minimize the effect of added base, the pH does increase. If you find that you've calculated a lower pH after the addition of a base, check for errors like mixing up the weak acid and conjugate base concentrations or losing track cif a minus sign. Weak Acid Ka pKa HF 7.1 x 10- 4 3.15 HN0 2 4.5 X 10- 4 3.35 HCOOH 1.7 x 10- 4 3.77 C 6 H s COOH 6.5 x lO - s 4.19 CH 3 C OOH 1.8 X lO - s 4.74 HCN 4.9 x 10- 10 9.31 C 6 H s OH 1.3 x 10- 10 9. 89 Sample Problem 17.2 shows how the Henderson-Hasselbalch equation is used to determine the pH of a buffer after the addition of a strong base. Sample Problem 17.2 Starting with 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the pH after the addition of 0.100 mole of NaOH. (Assume that the addition does not change the volume of the solution.) Strategy Added base will react with the acetic acid component of the buffer, converting OH- to CH 3 COO- : • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Write the starting amount of each species above the equation and the final amount of each species below the equation. Use the final amounts as concentrations in Equation 17.1. Setup Upon addition of OH - : 1.00 mol 0.10 mol 1.00 mol CH 3 COOH(aq) + OW(aq) H 2 0(/) + CH 3 COO- (aq) After OH- has been consumed: 0.90 mol o mol 1.10 mol Solution H = 4.74 + I a 1.10 M P 0 0 0.90M = 474 + I 1.10 M = 483 . og 0.90 M . Thus, the pH of the buffer after addition of 0.10 mole of NaOH is 4.83. Practice Problem A Calculate the pH of 1 L of a buffer that is 1.0 M in acetic acid and 1.0 Min sodium acetate after the addition of 0.25 mole of NaOH. Practice Problem B Calculate the pH of 1 L of a buffer that is 1.0 M in acetic acid and 1.5 Min sodium acetate after the addition of 0.20 mole of HCI. Preparing a Buffer Solution with a Specific pH A solution is only a buffer if it has the capacity to resist pH change when either an acid or a base is added. If the concentrations of a weak acid and conjugate base differ by more than a factor of 10, the solution does not have this capacity. Therefore, we consider a solution a buffer, and can use Equation 17.1 to calculate its pH, only if the following condition is met: [conjugate base] 10 > >0.1 [weak acid] Consequently, the log term in Equation 17.1 can only have values from -1 to 1, and the pH of a buffer cannot be more than one pH unit different from the pKa of the weak acid it contains. This is known as the range of the buffer, where pH = pKa + 1. This enables us to select the appropriate conjugate pair to prepare a buffer with a specific, desired pH. First, we choose a weak acid whose pKa is close to the desired pH. Next, we substitute the pH and pKa values into Equation 17.1 to obtain the necessary ratio of [conjugate base ]/[ weak acid]. This ratio can then be converted to molar quantities for the preparation of the buffer. Sample Problem 17.3 demonstrates this procedure. Sample Problem 17.3 . Select an appropriate weak acid from the table in the margin, and describe how you would prepare a buffer with a pH of 9.50. SECTION 17.2 Buffer Solutions 689 Strategy Select an acid with a pKa within one pH unit of 9.50. Use the pKa of the acid and Equation 17.1 to calculate the necessary ratio of [conjugate base]/[ weak acid]. Select concentrations of the buffer components that yield the calculated ratio. Setup Two of the acids listed in Table 16.6 have pKa values in the desired range: hydrocyanic acid (HCN, pK a = 9.31) and phenol (C 6 H s OH, pK a = 9.89). Solution Plugging the values for phenol into Equation 17.1 gives Therefore, the ratio of [C 6 H s O- ] to [C 6 H s OH] must be 0.41 to 1. One way to achieve this would be to dissolve 0.41 mole of C 6 H s ONa and 1.00 mole of C 6 H s OH in 1 L of water. Practice Problem A Select an appropriate acid from Table 16.6, and describe how you would prepare a buffer with pH = 4.5. Practice Problem B What range of pH values could be achieved with a buffer consisting of nitrous acid (HN0 2 ) and sodium nitrite (N0 2 )7 Bringing Chemistry to Life Maintaining the pH of Blood There are about 5 L of blood in the average adult. Circulating blood keeps cells alive by providing them with oxygen and nutrients and by removing carbon dioxide and other waste materials. The efficiency of this enolmously complex system relies on buffer s. The two primary components of blood are blood plasma and red blood cells, or erythrocytes. Blood plasma contains many compounds, including proteins, metal ions, and inorganic phosphates. The erythrocytes contain hemoglobin molecules, as well as the enzyme carbonic anhydrase, which catalyzes both the formation and the decomposition of carbonic acid (H 2 C0 3 ) : The substances inside the erythrocytes are protected from extracellular fluid (blood plasma) by a semipermeable cell membrane that allows only certain molecules to diffuse through it. The pH of blood plas ma is maintained at about 7.40 by several buffer systems, the most important of which is the HCO ;- IH 2 C0 3 system. In the erythrocyte, where the pH is 7.25, the prin- cipal buffer systems are HCO ;- IH 2 C0 3 and hemoglobin. The hemoglobin molecule is a complex protein molecule (molar mass of 65,000 g) that contains a number of ionizable protons. As a very rough approximation, we can treat it as a monoprotic acid in the form HHb: where HHb represents the hemoglobin molecule and Hb - is the conjugate base of HHb. Oxyhe- moglobin (HHb0 2 ), fOlmed by the combination of oxygen with hemoglobin, is a stronger acid than HHb: Carbon dioxide produced by metabolic processes diffuses into the erythrocyte, where it is rapidly converted to H 2 C0 3 by carbonic anhydrase: The ionization of the carbonic acid, Think About It There is an infinite number of combinations of [conjugate base] and [weak acid] that will give the necessary ratio. Note that this pH could also be achieved using HCN and a cyanide salt. For most purpose s, it is best to use the least toxic compounds available. 690 CHAPTER 17 Acid-Base Equi libria and Solubility Equilibria ha s two important consequences. First, the bicarbonate ion diffuses out of the erythrocyte and is carried by the blood plasma to the lungs. This is the major mechanism for removing carbon dioxide. Second, the H+ ions shift the equilibrium in favor of the un-ionized oxyhemoglobin molecules: Because HHb0 2 releases oxygen more readily than does its conjugate base (Hb0 2 ), the formation of the acid promotes the following reaction from left to right: HHbO zeaq) +:. ==:!:' HHb(aq) + 0 2(a q) The O 2 molecules diffuse out of the erythrocyte and are taken up by other cells to carry out metabolism. When the venous blood returns to the lungs, the preceding processes are reversed. The bicar- bonate ions now diffuse into the erythrocyte, where they react with hemoglobin to form carbonic acid: HHb (aq) + HC0 3 (a q) +:. ==:!: ' Hb- (aq) + H 2 C0 3 (aq) Most of the acid is then converted to CO 2 by carbonic anhydrase: The carbon dioxide diffuses to the lungs and is eventually exhaled. The formation of the Hb - ions (due to the reaction between HHb and HC0 3") also favors the uptake of oxygen at the lungs, Hb -(aq) + O?(aq) :;::::. ===z:' Hb0 2 (aq) because Hb- has a greater affinity for oxygen than does HHb. ' When the arterial blood flows back to the body tissues, the entire cycle is repeated. Checkpoint 17.2 Buffer Solutions 17.2.1 Which of the following combinations ca n be used to prepare a buffer? a) HCIICI- b) HFIF- c) CH 3 COOHlOH - d) HN0 2 IN0 2 e) HN0 3 IN0 3 17.2.2 What is the pH of a buffer that is 0.76 M in HF and 0.98 Min NaF? a) 3.26 b) 3.04 c) 3.15 d) 10.85 e) 10.74 Acid-Base Titrations 17.2.3 17.2.4 Consider 1 L of a buffer that is 0.85 M in formic acid (HCOOH) and 1.4 M in sodium formate (HCOONa). Calculate the pH after the addition of 0.15 mol HCI. (Assume the addition causes no volume change.) a) 4. 11 b) 3.99 c) 3.87 d) 10.13 e) 10.01 Consider 1 L of a buffer that is 1.5 M in hydrocyanic acid (HCN) and 1.2 M in sodium cyanide (NaCN). Calculate the pH after the addition of 0.25 mol NaOH. (Assume the addition causes no volume change.) a) 9.21 b) 9.37 c) 9.04 d) 4.63 e) 4.96 In Section 4.6 we introduced acid-base titrations as a form of chemical analysis. Having discussed buffer solutions, we can now look in more detail at the quantitative aspects of acid-base titrations. We will consider three types of reactions: (1) titrations involving a strong acid and a strong base, SECTION 17.3 Acid-Base Titrations 6 9' . u " ~ I ! (2) titrations involving a weak acid 'and a strong base, and (3) titrations involving a strong acid and a weak base. Titrations involving a weak acid and a weak base are complicated by the hydrolysis of both the cation and the anion of the salt formed. These titrations will not be discussed here. Figure 17.2 shows the experimental setup for monitoring the pH over the course of an acid-base , titration. Strong Acid- Strong Base Titrations The reaction between the strong acid. HCl and the strong ba se NaOH can be represented by NaOH(aq) + HCI(aq) NaCI(aq) + H 2 0(l) or by the net ionic equation, , Consider the addition of a 0.100 M NaOH solution (from a buret) to a vessel containing 25.0 mL of 0.100 M HCI. For convenience, we will use only three significant figures for volume and con- Fig u re 17.2 A pH meter is us ed [0 monitor an acid-base titration. I Multimedia Acids and Bases - titration of He l wit h NaOH . . . . . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . centration and two significant figures for pH. Figure 17.3 shows the titration curve the plot of pH . . x . . . . . . . as a function of titrant volume added. : Before the addition of NaOH begins, the pH of the acid is given by -log(0,100), or 1.00. : • Re call that only the di gits to the right of t he de cim al point are significant in a pH va lu e. When NaOH is added, the pH of the solution increases slowly at first. Near the equivalence point, T he ti trant is th e so l ution that is added fro m the buret. . ~ ">" . the pH begins to rise steeply, and at the equivalence point, when equimolar amounts of acid and : • base have reacted, the curve rises almost vertically. In a strong acid-strong base titration, both the . · hydrogen ion and hydroxide ion concentrations are very small at the equivalence point (roughly ····· Re c all that f or an acid and base that com bine 1 X 10- 7 M); consequently; the addition of a single drop of the ba se causes a large increase in [OH- ] and a steep rise in the pH of the solution. Beyond the equivalence point, the pH again increases slowly with the continued addition of NaOH. 14 13 12 11 10 9 8 - :r: 7 0, 6 5 - 4 3 2 1 0 20 - F- Equivalence point J __ I I _ I - I - I ~ 30 Volume of NaOH added (mL) -~ 40 50 in a 1: 1 ra tio, the equi valence point is wh ere equal mol ar am o unts of ac id and base ha ve beer comb i ned [ ~~ Sec tion 4.5 ]. Figure 17.3 Titrationcurve(pH as a function of volume titrant added) of a strong acid-strong base titration. A 0.100 M NaOH solution, the titrant, is added from a buret to 25.0 mL of a 0.100 M Hel solution in an Erlenm ey er flask. [...]... the equivalence point [part (a)], the solution contains both acetic acid and acetate ion, making the solution a buffer We can solve part (a) using Equation 17.1, the Henderson-Hasselbalch equation At the equivalence point [part (b)], all the acetic acid has been neutralized and we have only acetate ion in solution We must determine the concentration of acetate ion and solve part (b) as an equilibrium... corresponds to the steepest part of each titration curve and select an indicator (or indicators) that changes color within that range Setup (a) The titration curve in Figure 17.3 is for the titration of a strong acid with a strong base The steep part of the curve spans a pH range of about 4 to 10 (b) Figure 17.4 shows the curve for the titration of a weak acid with a strong base The steep part of the curve spans... range that coincides with the steepest part of the titration curve Consider the information in Figure 17.6, which shows the titration curves for hydrochloric acid and acetic acid each being titrated with sodium hydroxide Either of the indicators shown can be used for the titration of a strong acid with a strong base because both end points coincide with the steepest part of the HCl-NaOH titration curve... neutralized and we have only acetate ion in solution We must determine the concentration of acetate ion and solve part (b) as an equilibrium problem, using the Kb for acetate ion After the equivalence point [part (c)], all the acetic acid has been neutralized and there is nothing to consume the additional added base We must determine the concentration of excess hydroxide ion in the solution and solve for pH... conjugate base, In- 'the' end point of a titration is the point at which the color of the indicator changes Not all indicators change color at the same pH, however, so the choice of indicator for a particular titration depends on the strength of the acid (and the base) used in the titration To use the end point to determine the equivalence point of a titration, we must select an appropriate indicator... can neglect x in the denominator of the equation Solving for x, 2 x = 56 X 10- 10 0.050 x2 = (5.6 X 10- 10)(0.050) x = ~2.78 X 10 II = = 2.8 X 10- 11 5.3 X 10- 6 M gives [OH- ] = 5.3 X 10- 6 M, pOH = 5 .28, and pH = 8.72 4 After the equivalence point, the curve for titration of a weak acid with a strong base is identical to the curve for titration of a strong acid with a strong base Because all the acetic... water extracts pigments that exhibit a variety of colors at different pH values (Figure 17.7) Table 17.3 lists a number of indicators commonly used in acid-base titrations The choice of indicator for a particular titration depends on the strength of the acid and base to be titrated SECTION 17.3 14 I 12 - 10 Titration curve of a strong acid with a strong base (blue) and titration curve of a weak acid... either titration Methyl red can be used for the strong acid- strong base titration but cannot be used for the weak acid- strong base titration because its color change does not coincide with the steepest part of the curve ! r, / 11 - - ~, I Phenolphthalein (8.3 - 10) I 9, 8- :a - 7 546 3 Methyl red (4.2 - 6.3) /' - r 2- 1 • o 0 699 Figure 17.6 , 13 - Acid-Base Titrations 10 20 io 3'0 '.,~ . hydroxide is Prior to the equivalence point [part (a)], the solution contains both acetic acid and acetate ion, making the solution a buffer. We can solve part (a) using Equation 17.1, the Henderson-Hasselbalch. equivalence point [part (b)], all the acetic acid has been neutralized and we have only acetate ion in solution. We must determine the concentration of acetate ion and solve part (b) as an. base. The steep part of the curve spans a pH range of about 4 to 10. (b) Figure 17.4 shows the curve for the titration of a weak acid with a strong base. The steep part of the curve