CHAPTER FIFTEEN APPLICATIONS OF AQUEOUS EQUILIBRIA. For Review

CHAPTER FIFTEEN APPLICATIONS OF AQUEOUS EQUILIBRIA For Review 1. A common ion is an ion that appears in an equilibrium reaction but came from a source other than that reaction. Addition of a common ion (H + or NO ) to the reaction HNO H + + NO will drive the equilibrium to the left as predicted by LeChatelier s principle. When a weak acid solution has some of the conjugate base added from an outside source, this solution is called a buffer. Similarly, a weak base solution with its conjugate acid added from an outside source would also be classified as a buffer.. A buffer solution is one that resists a change in its ph when either hydroide ions or protons (H + ) are added. Any solution that contains a weak acid and its conjugate base or a weak base and its conjugate acid is classified as a buffer. The ph of a buffer depends on the [base/[acid ratio. When H + is added to a buffer, the weak base component of the buffer reacts with the H + and forms the acid component of the buffer. Even though the concentrations of the acid and base component of the buffer change some, the ratio of [base/[acid does not change that much. This translates into a ph that doesn t change much. When OH is added to a buffer, the weak acid component is converted into the base component of the buffer. Again, the [base/[acid does not change a lot (unless a large quantity of OH is added), so the ph does not change much. The concentrations of weak acid and weak base do not have to be equal in a buffer. As long as there are both a weak acid and a weak base present, the solution will be buffered. If the concentrations are the same, the buffer will have the same capacity towards added H + and added OH. Also, buffers with equal concentrations of weak acid and conjugate base have ph = pk a. Because both the weak acid and conjugate base are major spelcies present, both equilibriums that refer to these species must hold true. That is, the K a equilibrium must hold because the weak acid is present, and the K b equilibrium for the conjugate base must hold true because the conjugate base is a major species. Both the K a and K b equilibrium have the acid and conjugate base concentrations in their epressions. The same equilibrium concentrations of the acid and conjugate base must satisfy both equilibriums. In addition, the [H + and [OH concentrations must be related through the K w constant. This leads to the same ph answer whether the K a or K b equilibrium is used. The third method to solve a buffer problem is to use the Henderson-Hasselbalch equation. The equation is: [base ph = pk a + log [acid 9

0 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA where the base is the conjugate base of the weak acid present or the acid is the conjugate acid of the weak base present. The equation takes into account the normal assumptions made for buffers. Specifically, it is assumed that the initial concentration of the acid and base component of the buffer equal the equilibrium concentrations. For any decent buffer, this will always hold true.. Whenever strong acid is added to a solution, always react the H + from the strong acid with the best base present in solution. The best base has the largest K b value. For a buffer, this will be the conjugate base (A ) of the acid component of the buffer. The H + reacts with the conjugate base, A, to produce the acid, HA. The assumption for this reaction is that because strong acids are great at what they do, they are assumed to donate the proton to the conjugate base 100% of the time. That is, the reaction is assumed to go to completion. Completion is when a reaction goes until one or both of the reactants runs out. This reaction is assumed to be a stoichiometry problem like those we solved in Chapter. Whenever a strong base is added to a buffer, the OH ions react with the best acid present. This reaction is also assumed to go to completion. In a buffer, the best acid present is the acid component of the buffer (HA). The OH rips a proton away from the acid to produce the conjugate base of the acid (A ) and H O. Again, we know strong bases are great at accepting protons, so we assume this reaction goes to completion. It is assumed to be a stoichiometry problem like the ones we solved in Chapter. When [HA = [A (or [BH + = [B) for a buffer, the ph of the solution is equal to the pk a value for the acid component of the buffer (ph = pk a because [H + = K a ). A best buffer has equal concentrations of the acid and base component so it is equally efficient at absorbing H + or OH. For a ph =.00 buffer, we would choose the acid component having a K a close to.00 10 = 1.0 10 (ph = pk a for a best buffer). For a ph = 10.00 buffer, we would want the 10.00 acid component of the buffer to have a K a close to 10 10 = 1.0 10. Of course, we can have a buffer solution made from a weak base and its conjugate acid. For a ph = 10.00 10 buffer, our conjugate acid should have K a 1.0 10 which translates into a K b value of the base close to 1.0 10 (K b = K w /K a for conjugate acid-base pairs). The capacity of a buffer is a measure of how much strong acid or strong base the buffer can neutralize. All the buffers listed have the same ph (= pk a =.7) because they all have a 1:1 concentration ratio between the weak acid and the conjugate base. The 1.0 M buffer has the greatest capacity; the 0.01 M buffer the least capacity. In general, the larger the concentrations of weak acid and conjugate base, the greater the buffer capacity, i.e., the more strong acid or strong base that can be neutralized with little ph change.. Let s review the strong acid-strong base titration using the eample (case study) covered in section 1. of the tet. The eample used was the titration of 0.0 ml of 0.00 M HNO titrated by 0.100 M NaOH. See Figure 1.1 for the titration curve. The important points are: a. Initially, before any strong base has been added. Major species: H +, NO, and H O. To determine the ph, determine the [H + in solution after the strong acid has completely dissociated as we always do for strong acid problems.

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 1 b. After some strong base has been added, up to the equilivance point. For our eample, this is from just after 0.00 ml NaOH added up to just before 100.0 ml NaOH added. Major species before any reaction: H +, NO, Na +, OH, and H O. Na + and NO have no acidic or basic properties. In this region, the OH from the strong base reacts with some of the H + from the strong acid to produce water (H + + OH H O). As is always the case when something strong reacts, we assume the reaction goes to completion. Major species after reaction: H +, NO, Na +, and H O: To determine the ph of the solution, we first determine how much of the H + is neutralized by the OH. Then we determine the ecess [H + and take the log of this quantity to determine ph. From 0.1 ml to 99.9 ml NaOH added, the ecess H + from the strong acid determines the ph. c. The equivalence point (100.0 ml NaOH added). Major species before reaction: H +, NO, Na +, OH, and H O. Here, we have added just enough OH to neutralize all of the H + from the strong acid (moles OH added = moles H + present). After the stoichiometry reaction (H + + OH H O), both H + and OH have run out (this is the definition of the equivalence point). Major species after reaction: Na +, NO, and H O. All we have in solution are some ions with no acidic or basic properties (NO and Na + in H O). The ph = 7.00 at the equivalence point of a strong acid by a strong base. d. Past the equivalence point (volume of NaOH added > 100.0 ml). Major species before reaction H +, NO, Na +, OH, and H O. After the stoichiometry reaction goes to completion (H + + OH H O), we have ecess OH present. Major species after reaction: OH, Na +, NO, and H O. We determine the ecess [OH and convert this into the ph. After the equivalence point, the ecess OH from the strong base determines the ph. See Figure 1. for a titration curve of a strong base by a strong acid. The stoichiometry problem is still the same, H + + OH H O, but what is in ecess after this reaction goes to completion is reverse of the strong acid-strong base titration. The ph up to just before the equivalence point is determined by the ecess OH present. At the equivalence point, ph = 7.00 because we have added just enough H + from the strong acid to react with all of the OH from the strong base (mole base present = mole acid added). Past the equivalence point, the ph is determined by the ecess H + present. As can be seen from Figures 1.1 and 1., both strong by strong titrations have ph = 7.00 at the equivalence point, but the curves are the reverse of each other before and after the equivalence point.. In section 1., the case study for the weak acid-strong base titration is the titration of 0.0 ml of 0.10 M HC H O by 0.10 M NaOH. See Figure 1. for the titration curve. As soon as some NaOH has been added to the weak acid, OH reacts with the best acid present. This is the weak acid titrated (HC H O in our problem). The reaction is: OH + HC H O H O + C H O. Because something strong is reacting, we assume the reaction goes to completion. This is the stoichiometry part of a titration problem. To solve for the ph, we see what is in solution after the stoichiometry problem and decide how to proceed. The various parts to the titration are:

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA a. Initially, before any OH has been added. The major species present is the weak acid, HC H O, and water. We would use the K a reaction for the weak acid and solve the equilibrium problem to determine the ph. b. Just past the start of the titration up to just before the equivalence point (0.1 ml to 9.9 ml NaOH added). In this region, the major species present after the OH reacts to completion are HC H O, C H O, Na +, and water. We have a buffer solution because both a weak acid and a conjugate base are present. We can solve the equilibrium buffer problem using the K a reaction for HC H O, the K b reaction for C H O, or the Henderson-Hasselbalch equation. A special point in the buffer region is the halfway point to equivalence. At this point (.0 ml of NaOH added), eactly one-half of the weak acid has been converted into its conjugate base. At this point, we have [weak acid = [conjugate base so that ph = pk a. For the HC H O titration, the ph at.0 ml NaOH added is log (1.8 10 ) =.7; the ph is acidic at the halfway point to equivalence. However, other weak acid-strong base titrations could have basic ph values at the 7 halfway point. This just indicates that the weak acid has K a < 1 10, which is fine. c. The equivalence point (0.0 ml NaOH added). Here we have added just enough OH to convert all of the weak acid into its conjugate base. In our eample, the major species present are C H O, Na +, and H O. Because the conjugate base of a weak acid is a weak base, we will have a basic ph (ph > 7.0) at the equivalence point. To calculate the ph, we write out the K b reaction for the conjugate base and then set-up and solve the equilibrium problem. For our eample, we would write out the K b reaction for C H O. d. Past the equivalence point (V > 0.0 ml). Here we added an ecess of OH. After the stoichiometry part of the problem, the major species are OH, C H O, H O, and Na +. We have two bases present, the ecess OH and the weak conjugate base. The ecess OH dominates the solution and thus determines the ph. We can ignore the OH contribution from the weak conjugate base. See the titration curve after Figure 1. that compares and contrasts strong acid-strong base titrations to weak acid-strong base titration. The two curves have the same ph only after the equivalence point where the ecess strong base added determines the ph. The strong acid titration is acidic at every point before the equivalence point, has a ph = 7.0 at the equivalence point, and is basic at every point after the equivalence point. The weak acid titration is much more complicated because we cannot ignore the basic properties of the conjugate base of the weak acid; we could ingore the conjugate base in the strong acid titration because it had no basic properties. In the weak acid titration, we start off acidic (ph < 7.0), but where it goes from there depends on the strength of the weak acid titrated. At the halfway point where ph = pk a, the ph may be acidic or basic depending on the K a value of the weak acid. At the equivalence, the ph must be basic. This is due to the presence of the weak conjugate base. Past the equivalence point, the strong acid and weak acid titrations both have their ph determined by the ecess OH added. 6. The case study of a weak base-strong acid titration in section 1. is the titration of 100.0 ml of 0.00 M NH by 0.10 M HCl. The titration curve is in Figure 1..

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA As HCl is added, the H + from the strong acid reacts with the best base present, NH. Because something strong is reacted, we assume the reaction goes to completion. The reaction used for the stoichiometry part of the problem is: H + + NH NH +. Note that the effect of this reaction is to convert the weak base into its conjugate acid. The various parts to a weak basestrong acid titration are: a. Initially before any strong acid is added. We have a weak base in water. Write out the K b reaction for the weak base, set-up the ICE table, and then solve. b. From 0.1 ml HCl added to just gefore the equivalence point (9.9 ml HCl added). The major species present in this region are NH, NH +, Cl, and water. We have a weak base and its conjugate acid present at the same time; we have a buffer. Solve the buffer problem using the K a reaction for NH +, the K b reaction for NH, or the Henderson- Hasselbalch equation. The special point in the buffer region of the titration is the halfway point to equivalence. Here, [NH = [NH + 10, so ph = pk a = log (.6 10 ) = 9.. For this titration, the ph is basic at the halfway point. However, if the K a for the weak acid 7 component of the buffer has a K a > 1 10 7, (K b for the base < 1 10 ) then the ph will be acidic at the halfway point. This is fine. In this review question, it is asked what is the K b value for the weak base where the halfway point to equivalence has ph = 6.0 (which is acidic). Be-cause ph = pk a at this point, pk a = 6.0 so pk b = 1.00 6.0 = 8.0; 8.0 K b = 10 8 = 1.0 10. c. The equivalence point (0.0 ml HCl added). Here, just enough H + has been added to convert all of the NH into NH +. The major species present are NH +, Cl and H O. The only important species present is NH +, a weak acid. This is always the case in a weak base-strong acid titration. Because a weak acid is always present at the equivalence point, the ph is always acidic (ph < 7.0). To solve for the ph, write down the K a reaction for the conjugate acid and then determine K a (= K w /K b ). Fill in the ICE table for the problem, and then solve to determine ph. d. Past the equivalence point (V > 0.0 ml HCl added). The ecess strong acid added determines the ph. The major species present for our eample would be H + (ecess), NH +, Cl, and H O. We have two acids present, but NH + is a weak acid. Its H + contribution will be negligible compared to the ecess H + added from the strong acid. The ph is determined by the molarity of the ecess H +. Eamine Figures 1. and 1. to compare and contrast a strong base-strong acid titration with a weak base-strong acid titration. The points in common are only after the equivalence point where ecess strong acid determines the ph. The two titrations before and at the equivalence point are very different. This is because the conjugate acid of a weak base is a weak acid, whereas, when a strong base dissolves in water, only OH is important; the cation in the strong base is garbage (has no acidic/basic properties). There is no conjugate acid to worry about. For a strong base-strong acid titration, the ecess OH from the strong base determines the ph from the initial point all the way up to the equivaldnce point. At the equivalence point, the added H + has reacted with all of the OH and ph = 7.0. For a weak base-strong acid, we initially have a weak base problem to solve in order to calculate the ph. After H + has been

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA added, some of the weak base is converted into its conjugate acid and we have buffer solutions to solve in order to calculate the ph. At the equivalence point, we have a weak acid problem to solve. Here, all of the weak base has been converted into its conjugate acid by the added H +. 7. An acid-base indicator marks the end point of a titration by changing color. Acid-base indicators are weak acids themselves. We abbreviate the acid form of an indicator as HIn and the conjugate base form as In. The reason there is a color change with indicators is that the HIn form has one color associated with it while the In form has a different color associated with it. Which form dominates in solution and dictates the color is determined by the ph of the solution. The related quilibrium is: HIn H + + In. In a very acidic solution, there are lots of H + ions present which drives the indicator equilibrium to the left. The HIn form dominates and the color of the solution is the color due to the HIn form. In a very basic solution, H + has been removed from solution. This drives the indicator equilibrium to the right and the In form dominates. In very basic solutions, the solution takes on the color of the In form. In between very acidic and very basic solutions, there is a range of ph values where the solution has significant amounts of both the HIn and In forms present. This is where the color change occurs and we want this ph to be close to the stoichiometric point of the titration. The ph that the color change occurs is determined by the K a of the indicator. Equivalence point: when enough titrant has been added to react eactly with the substance in the solution being titrated. Endpoint: indicator changes color. We want the indicator to tell us when we have reached the equivalence point. We can detect the end point visually and assume it is the equivalence point for doing stoichiometric calculations. They don t have to be as close as 0.01 ph units since, at the equivalence point, the ph is changing very rapidly with added titrant. The range over which an indicator changes color only needs to be close to the ph of the equivalence point. The two forms of an indicator are different colors. The HIn form has one color and the In form has another color. To see only one color, that form must be in an approimately ten-fold ecess or greater over the other form. When the ratio of the two forms is less than 10, both colors are present. To go from [HIn/[In = 10 to [HIn/[In = 0.1 requires a change of ph units (a 100-fold decrease in [H + ) as the indicator changes from the HIn color to the In color. From Figure 1.8, thymol blue has three colors associated with it: organge, yellow, and blue. In order for this to happen, thymol blue must be a diprotic acid. The H In form has the orange color, the HIn form has the yellow color, and the In form has the blue color associated with it. Thymol blue cannot be monoprotic; monoprotic indicators only have two colors associated with them (either the HIn color or the In color). 8. K sp refers to the equilibrium reaction where a salt (ionic compound) dissolves into its ions to form a saturated solution: For eample, consider Ca (PO ) (s). The K sp reaction and K sp epression are: Ca (PO ) (s) Ca + (aq) + PO (aq) K sp = [Ca + [PO

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA K sp is just an equilibrium constant (called the solubility product constant) that refers to a specific equilibrium reaction. That reaction is always a solid salt dissolving into its ions. Because the ionic compound is a solid, it is not included in the K sp epression. The solubility of a salt is the maimum amount of that salt that will dissolve per liter of solution. We determine solubility by solving a typical equilibrium problem using the K sp reaction. We define our solubility (usually called s) as the maimum amount of the salt that will dissolve, then fill in our ICE table and solve. The two unknowns in the equilibrium problem are the K sp value and the solubility. You are given one of the unknowns and asked to solve for the other. See Sample Eercises 1.1-1.1 for eample calculations. In Sample Eercise 1.1 and 1.1, you are given the solubility of the salt and asked to calculate K sp for the ionic compound. In Sample Eercise 1.1, you are given the K sp value for a salt and asked to calculate the solubility. In both types of problem, set-up the ICE table in order to solve. If the number of ions in the two salts is the same, the K sp s can be compared directly to determine relative solubilities, i.e., 1:1 electrolytes (1 cation:1 anion) can be compared to each other; :1 electrolyes can be compared to each other, etc. If the number of ions is the same, the salt with the largest K sp value has the largest molar solubility. If the number of ions are different be-tween two salts, then you must actually calculate the solubility of each salt to see which is more or less soluble. A common ion is when either the cation or anion of the salt in question is added from an outside source. For eample, a common ion for the salt CaF would be if F (or Ca + ) was added to solution by dissolving NaF [or Ca(NO ). When a common ion is present, not as much of the salt dissolves (as predicted by LeChatelier s Principle), so the solubility decreases. This decrease in solubility is called the common ion effect. Salts whose anions have basic properties have their solubiities increase as the solution becomes more acidic. Some eamples are CaF, Ca (PO ), CaCO, and Fe(OH). In all of these cases, the anion has basic properties and reacts with H + in acidic solutions. This removes one of the ions from the equilibrium so more salt dissolves to repenish the ion concentrations (as predicted by LeChatelier s Principle). In our eample, F, PO, and CO are all conjugate bases of weak acids so they are all weak bases. In Fe(OH), OH is a strong base. For salts with no ph solubility, the ions in the salt must not have any acidic or basic properties. Eamples of these salts have anions that are the conjugate bases of strong acids so they are worthless bases. Eamples of these types of salts are AgCl, PbBr, and Hg I. The solubility of these salts do not depend on ph because Cl, Br, and I have no basic (or acidic) properties. 9. Q and K sp have the same epression; the difference is the type of concentrations used. K sp always uses equilibrium concentrations for the ions. Q, however, uses initial concentrations of the ions given in a problem. The purpose of Q is to see if a reaction is at equilibrium. If Q = K sp, then the salt is at equilibrium with ion concentrations given in the problem. If Q K sp, then the reaction is not at equilibrium; these concentrations must change in order for the salt

6 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA to get to equilibrium with its ions. If Q > K sp, the ion concentrations are too large. The K sp reaction shifts left to reduce the ion concentrations and to eventually get to equilibrium. If Q < K sp, the K sp reaction shifts right to produce more ions in order to get to equilibrium. Selective precipitation is a way to separate out ions in an aqueous miture. One way this is done is to add a reagent that precipitates with only one of the ions in the miture. That ion is removed by forming the prcipitate. Another way to selectively precipitate out some ions in a miture is to choose a reagent that forms a precipitate with all of the ions in the miture. The key is that each salt has a different solubility with the ion in common. We add the common ion slowly until a precipitate starts to form. The first precipitate to form will be the least soluble precipitate. After precipitation of the first is complete, we filter off the solution and continue adding more of the common ion until precipitation of the net least soluble salt is complete. This process is continued until all of the ions have been pricipitated out of solution. For the 0.10 M Mg +, Ca +, and Ba +, adding F (NaF) slowly will separate the ions due to their solubility differences with these ions. The first compound to precipitate is the least 11 soluble CaF compound (it has the smallest K sp value,.0 10 ). After all of the CaF has precipitated and been removed, we continue adding F until the second least soluble flouride 9 has precipitated. This is MgF with the intermediate K sp value (6. 10 ). Finally, the only ion left in solution is Ba + which is precipitated out by adding more NaF. BaF precipitates last because it is the most soluble fluoride salt in the miture (K sp =. 10 ). For the Ag +, Pb +, Sr + miture, because the phosphate salt of these ions do not all have the same number of ions, we cannot just compare K sp values to deduce the order of precipitation. This would work for Pb (PO ), K sp = 1 10 1 and Sr (PO ), K sp = 1 10 because these contain the same number of ions (). However, Ag PO (s) only contains ions per formula unit. We must calculate the [PO necessary to start precipitation. We do this by setting Q = K sp and determining the [PO necessary for this to happen. Any [PO greater than this calculated concentration will cause Q > K sp and precipitation of the salt. A sample calculation is: Q = 1 Sr (PO ) Sr + (aq) + PO (aq) K sp = 1 10 1 1 10 = [Sr + [PO = (1.0 M) [PO, [PO 16 = 10 M When the [PO > 16 10 M, precipitation of Sr (PO ) will occur. Doing similar calculations with the other two salts leads to the answers: Ag PO (s) will precipitate when [PO > 1.8 Pb (PO ) (s) will precipitate when [PO > 1 18 10 M. 7 10 M. From the calculations, Pb (PO ) (s) needs the smallest [PO to precipitate, so it precipitates first as we add K PO. The net salt to precipitate is Ag PO (s), and Sr (PO ) (s) is last to precipitate because it requires the largest [PO.

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 7 10. A comple ion is a charged species consisting of a metal ion surrounded by ligands. A ligand is a molecule or ion having a lone pair of electrons that can be donated to an empty orbital on the metal ion to form a covalent bond. A ligand is a Lewis base (electron pair donor). Questions Cu + (aq) + NH (aq) Cu(NH ) + (aq) K 1 1 10 Cu(NH ) + (aq) + NH (aq) Cu(NH ) + (aq) K 1 10 Cu(NH ) + (aq) + NH (aq) Cu(NH ) + (aq) K 1 10 Cu(NH ) + (aq) + NH (aq) Cu(NH ) + (aq) K 1 10 Because the K values are much greater than one, all of these reactions lie far to the right (products are mostly present at equilibrium). So most of the Cu + in a solution of NH will be converted into Cu(NH ) +. Therefore, the [Cu(NH ) + will be much larger than the [Cu + at equilibrium. Cu(OH) (s) Cu + (aq) + OH (aq) As NH is added, it removes Cu + from the above K sp equilibrium for Cu(OH). It removes Cu + by reacting with it to form Cu(NH ) + mostly. As Cu + is removed, more of the Cu(OH) (s) will dissolve to replenish the Cu +. This process continues until equilibrium is reached or until all of the Cu(OH) dissolves. H + (aq) + NH (aq) NH + (aq) The added H + from the strong acid reacts with the weak base NH to form NH +. As NH is removed from solution, the comple ion equilibrium will shift left to produce more NH. This has the effect of also producing more Cu +. Eventually enough Cu + will be produced to react with OH to form Cu(OH) (s). The general effect of common ion formation is to increase the solubility of salts that contain cations which form common ions. 1. When an acid dissociates or when a salt dissolves, ions are produced. A common ion effect is observed when one of the product ions in a particular equilibrium is added from an outside source. For a weak acid dissociating to its conjugate base and H +, the common ion would be the conjugate base; this would be added by dissolving a soluble salt of the conjugate base into the acid solution. The presence of the conjugate base from an outside source shifts the equilibrium to the left so less acid dissociates. For the K sp reaction of a salt dissolving into its respective ions, a common ion would be one of the ions in the salt added from an outside source. When a common ion is present, the K sp equilibrium shifts to the left resulting in less of the salt dissolving into its ions. [base 1. ph = pk a + log ; When [acid > [base, then [acid [base [acid [base < 1 and log < 0. [acid

8 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA From the Henderson-Hasselbalch equation, if the log term is negative, then ph < pk a. When one has more acid than base in a buffer, the ph will be on the acidic side of the pk a value, i.e., the ph is at a value lower than the pk a value. When one has more base than acid in a buffer ([conjugate base > [weak acid), then the log term in the Henderson-Hasselbalch equation is positive resulting in ph > pk a. When one has more base than acid in a buffer, the ph is on the basic side of the pk a value, i.e., the ph is at a value greater than the pk a value. The other scenario you can run across in a buffer is when [acid = [base. Here, the log term is equal to zero and ph = pk a. 1. The more weak acid and conjugate base present, the more H + and/or OH that can be absorbed by the buffer without significant ph change. When the concentrations of weak acid and conjugate base are equal (so that ph = pk a ), the buffer system is equally efficient at absorbing either H + or OH. If the buffer is overloaded with weak acid or with conjugate base, then the buffer is not equally efficient at absorbing either H + or OH. 16. Titration i is a strong acid titrated by a strong base. The ph is very acidic until just before the equivalence point; at the equivalence point, ph = 7.00; and past the equivalence the ph is very basic. Titration ii is a strong base titrated by a strong acid. Here the ph is very basic until just before the equivalence point; at the equivalence point, ph = 7.00; and past the equivalence point the ph is very acidic. Titration iii is a weak base titrated by a strong acid. The ph starts out basic because a weak base is present. However, the ph will not be as basic as in Titration ii where a strong base is titrated. The ph drops as HCl is added, then at the halfway point to equivalence, ph = pk a. Because K b =. 10 for CH NH, CH NH + has 11 K a = K w /K b =. 10 and pk a = 10.6. So at the halfway point to equivalence for this weak base-strong acid titration, ph = 10.6. The ph continues to drop as HCl is added; then at the equivalence point the ph is acidic (ph < 7.00) because the only important major species present is a weak acid (the conjugate acid of the weak base). Past the equivalence point the ph becomes more acidic as ecess HCl is added. Titration iv is a weak acid titrated by a strong base. The ph starts off acidic, but not nearly as acidic as the strong acid titration (i). The ph increases as NaOH is added, then at the halfway point to equivalence, ph = pk a for HF = log(7. 10 ) =.1. The ph continues to increase past the halfway point, then at the equivalence point the ph is basic (ph > 7.0) because the only important major species present is a weak base (the conjugate base of the weak acid). Past the equivalence point, the ph becomes more basic as ecess NaOH is added. a. All require the same volume of titrant to reach the equivalence point. At the equivalence point for all of these titrations, moles acid = moles base (M A V A = M B V B ). Because all of the molarities and volumes are the same in the titrations, the volume of titrant will be the same (0.0 ml titrant added to reach equivalence point). b. Increasing initial ph: i < iv < iii < ii; The strong acid titration has the lowest ph, the weak acid titration is net, followed by the weak base titration, with the strong base titration having the highest ph. c. i < iv < iii < ii. The strong acid-titration has the lowest ph at the halfway point to equivalence and the strong base titration has the highest halfway point ph. For the weak acid titration, ph = pk a =.1, and for the weak base titration, ph = pk a = 10.6.

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 9 d. Equivalence point ph: iii < ii = i < iv; The strong by strong titrations have ph = 7.00 at the equivalence point. The weak base titration has an acidic ph at the equivalence point and a weak acid titration has a basic equivalence point ph. The only different answer when the weak acid and weak base are changed would be for part c. This is for the halfway point to equivalence where ph = pk a. 10 HOC 6 H ; K a = 1.6 10, pk a = log(1.6 10 ) = 9.80 1 C H NH + K w 1.0 10 6, K a = =.9 10, pk 9 a =. K 1.7 10 b, CHN From the pk a values, the correct ordering at the halfway point to equivalence would be: i < iii < iv < ii. Note that for the weak base-strong acid titration using C H N, the ph is acidic at the halfway point to equivalence, while the weak acid-strong base titration using HOC 6 H is basic at the halfway point to equivalence. This is fine; this will always happen 7 when the weak base titrated has a K b < 1 10 (so K a of the conjugate acid is greater 7 than 1 10 7 ) and when the weak acid titrated has a K a < 1 10 (so K b of the conjugate 7 base is greater than 1 10 ). 17. The three key points to emphasize in your sketch are the initial ph, ph at the halfway point to equivalence, and the ph at the equivalence point. For all of the weak bases titrated, ph = pk a at the halfway point to equivalence (0.0 ml HCl added) because [weak base = [conjugate acid at this point. Here, the weak base with K b = 10 has a ph = 9.0 at the halfway point 10 and the weak base with K b = 10 has a ph =.0 at the halfway point to equivalence. For the initial ph, the strong base has the highest ph (most basic), while the weakest base has the lowest ph (least basic). At the equivalence point (100.0 ml HCl added), the strong base titration has ph = 7.0. The weak bases titrated have acidic phs because the conjugate acids of the weak bases titrated are the major species present. The weakest base has the strongest conjugate acid so its ph will be lowest (most acidic) at the equivalence point. 10 strong base ph 7.0 K b = 10 - K b = 10-10 Volume HCl added (ml)

60 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 18. HIn H + + In K a = [H [In [HIn Indicators are weak acids themselves. The special property they have is that the acid form of the indicator (HIn) has one distinct color, while the conjugate base form (In ) has a different distinct color. Which form dominates and thus determines the color of the solution is determined by the ph. An indicator is chosen in order to match the ph of the color change at about the ph of the equivalence point. 19. i. This is the result when you have a salt that breaks up into two ions. Eamples of these salts (but not all) could be AgCl, SrSO, BaCrO, and ZnCO. ii. This is the result when you have a salt that breaks up into three ions, either two cations and one anion or one cation and two anions. Some eamples are SrF, Hg I, and Ag SO. iii. This is the result when you have a salt that breaks up into four ions, either three cations and one anion (Ag PO ) or one cation and three anions (ignoring the hydroides, there are no eamples of this type of salt in Table 1.). iv. This is the result when you have a salt that breaks up into five ions, either three cations and two anions [Sr (PO ) or two cations and three anions (no eamples of this type of salt are in Table 1.). 0. Some people would automatically think that an increase in temperature would increase the solubility of a salt. This is not always the case as some salts show a decrease in solubility as temperature increases. The two major methods used to increase solubility of a salt both involve removing one of the ions in the salt by reaction. If the salt has an ion with basic properties, adding H + will increase the solubility of the salt because the added H + will react with the basic ion, thus removing it from solution. More salt dissolves in order to to make up for the lost ion. Some eamples of salts with basic ions are AgF, CaCO, and Al(OH). The other way to remove an ion is to form a comple ion. For eample, the Ag + ion in silver salts forms the comple ion Ag(NH ) + as ammonia is added. Silver salts increase their solubility as NH is added because the Ag + ion is removed through comple ion formation. Eercises Buffers 1. When strong acid or strong base is added to a bicarbonate/carbonate miture, the strong acid/base is neutralized. The reaction goes to completion, resulting in the strong acid/base being replaced with a weak acid/base, which results in a new buffer solution. The reactions are: H + (aq) + CO (aq) HCO (aq); OH + HCO (aq) CO (aq) + H O(l). Similar to the HCO /CO buffer discussed in Eercise 1, the HONH + /HONH buffer absorbs added OH and H + in the same fashion.

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 61 HONH (aq) + H + (aq) HONH + (aq) HONH + (aq) + OH (aq) HONH (aq) + H O(l). a. This is a weak acid problem. Let HC H O = HOPr and C H O = OPr -. HOPr H + + OPr K a = 1. Initial 0.100 M ~0 0 mol/l HOPr dissociates to reach equilibrium Change + + Equil. 0.100 10 K a = 1. 10 = [H [OPr [HO Pr = 0.100 0. 100 = [H + = 1.1 10 M; ph =.96 Assumptions good by the % rule. b. This is a weak base problem (Na + has no acidic/basic properties). OPr - + H O HOPr + OH - K b = Initial 0.100 M 0 ~0 mol/l OPr - reacts with H O to reach equilibrium Change + + Equil. 0.100 K w = 7.7 10 10 K a K b = 7.7 10 10 = [HOPr[OH = [OPr 0.100 0.100 = [OH = 8.8 c. pure H O, [H + = [OH - = 1.0 6 10 M; poh =.06; ph = 8.9 Assumptions good. 7 10 M; ph = 7.00 d. This solution contains a weak acid and its conjugate base. This is a buffer solution. We will solve for the ph using the weak acid equilibrium reaction. HOPr H + + OPr - K a = 1. Initial 0.100 M ~0 0.100 M mol/l HOPr dissociates to reach equilibrium Change + + Equil. 0.100 0.100 + 10 1. 10 = (0.100 )( ) 0.100 ( 0.100)( ) = = [H + 0.100 [H + = 1. 10 M; ph =.89 Assumptions good.

6 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA Alternatively, we can use the Henderson-Hasselbalch equation to calculate the ph of buffer solutions. [Base ph = pk a + log [Acid (0.100) = pk a + log = log (1. 10 ) =.89 (0.100) The Henderson-Hasselbalch equation will be valid when an assumption of the type 0.1 + 0.1 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity that it will be of no use to control the ph. Note: The Henderson-Hasselbalch equation can only be used to solve for the ph of buffer solutions.. a. Weak base problem: HONH + H O HONH + + OH K b = 1.1 10 8 Initial 0.100 M 0 ~0 mol/l HONH reacts with H O to reach equilibrium Change + + Equil. 0.100 K b = 1.1 8 10 = 0.100 0.100 = [OH =. 10 M; poh =.8; ph = 9. Assumptions good. b. Weak acid problem (Cl has no acidic/basic properties); HONH + HONH + H + Initial 0.100 M 0 ~0 mol/l HONH + dissociates to reach equilibrium Change + + Equil. 0.100 K a = K w = 9.1 7 10 = K b [HONH [HONH [H = 0.100 0. 100 = [H + =.0 c. Pure H O, ph = 7.00 10 M; ph =. Assumptions good. d. Buffer solution where pk a = log (9.1 10 ) = 6.0. Using the Henderson-Hasselbalch equation: [Base [HONH (0.100) ph = pk a + log = 6.0 + log = 6.0 + log = 6.0 [Acid [HONH (0.100) 7

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 6. 0.100 M HC H O : percent dissociation = [H [HC HO 0 100 = 1.110 M 0.100M = 1.1% 1.10 0.100 M HC H O + 0.100 M NaC H O : % dissociation = 100 = 1. 10 % 0.100 The percent dissociation of the acid decreases from 1.1% to 1. 10 % when C H O is present. This is known as the common ion effect. The presence of the conjugate base of the weak acid inhibits the acid dissociation reaction. 6. 0.100 M HONH : percent ionization [OH [HONH 0 100 =.10 M 0.100M 100 =. 10 % 0.100 M HONH + 0.100 M HONH + : % ionization = 8 1.110 0.100 100 = 1.1 10 % The percent ionization decreases by a factor of 000. The presence of the conjugate acid of the weak base inhibits the weak base reaction with water. This is known as the common ion effect. 7. a. We have a weak acid (HOPr = HC H O ) and a strong acid (HCl) present. The amount of H + donated by the weak acid will be negligible as compared to the 0.00 M H + from the strong acid. To prove it let s consider the weak acid equilibrium reaction: HOPr H + + OPr K a = 1. Initial 0.100 M 0.00 M 0 mol/l HOPr dissociates to reach equilibrium Change + + Equil. 0.100 0.00 + K a = 1. 10 = (0.00 )( ) 0.100 ( 0.00)( ), = 6. 10 M 0.100 [H + = 0.00 + = 0.00 M; ph = 1.70 Assumptions good ( = 6. 10 10 which is << 0.00). b. Added H + reacts completely with the best base present, Opr -. Since all species present are in the same volume of solution, we can use molarity units to do the stoichiometry part of the problem (instead of moles). The stoichiometry problem is: OPr - + H + HOPr Before 0.100 M 0.00 M 0 Change 0.00 0.00 +0.00 Reacts completely After 0.080 0 0.00 M

6 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA After reaction, a weak acid, HOPr, and its conjugate base, OPr, are present. This is a buffer solution. Using the Henderson-Hasselbalch equation where pk a = log (1. 10 ) =.89: [Base ph = pk a + log [Acid =.89 + log (0.080) (0.00) c. This is a strong acid problem. [H + = 0.00 M; ph = 1.70 d. Added H + reacts completely with the best base present, OPr. OPr + H + HOPr =.9 Assumptions good. Before 0.100 M 0.00 M 0.100 M Change 0.00 0.00 +0.00 Reacts completely After 0.080 0 0.10 A buffer solution results (weak acid + conjugate base). Using the Henderson-Hasselbalch equation: ph = pk a + [Base (0.080) =.89 + log =.71 [Acid (0.10) 8. a. Added H + reacts completely with HONH (the best base present) to form HONH +. HONH + H + HONH + Before 0.100 M 0.00 M 0 Change 0.00 0.00 +0.00 Reacts completely After 0.080 0 0.00 After this reaction, a buffer solution eists, i.e., a weak acid (HONH + ) and its conjugate base (HONH ) are present at the same time. Using the Henderson-Hasselbalch equation to solve for the ph where pk a = log (K w /K b ) = 6.0: [Base ph = pk a + log [Acid = 6.0 + log (0.080) (0.00) = 6.0 + 0.60 = 6.6 b. We have a weak acid and a strong acid present at the same time. The H + contribution from the weak acid, HONH +, will be negligible. So, we have to consider only the H + from HCl. [H + = 0.00 M; ph = 1.70 c. This is a strong acid in water. [H + = 0.00 M; ph = 1.70 d. Major species: H O, Cl, HONH, HONH +, H + H + will react completely with HONH, the best base present.

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 6 HONH + H + HONH + Before 0.100 M 0.00 M 0.100 M Change 0.00 0.00 +0.00 Reacts completely After 0.080 0 0.10 A buffer solution results after reaction. Using the Henderson-Hasselbalch equation: [HONH (0.080) ph = 6.0 + log = 6.0 + log [HONH (0.10) 9. a. OH will react completely with the best acid present, HOPr. HOPr + OH OPr - + H O = 6.0 0.18 =.86 Before 0.100 M 0.00 M 0 Change 0.00 0.00 +0.00 Reacts completely After 0.080 0 0.00 A buffer solution results after the reaction. Using the Henderson-Hasselbalch equation: [Base ph = pk a + log [Acid =.89 + log (0.00) (0.080) =.9 b. We have a weak base and a strong base present at the same time. The amount of OH added by the weak base will be negligible compared to the 0.00 M OH - from the strong base.. To prove it, let s consider the weak base equilibrium: OPr - + H O HOPr + OH K b = 7.7 Initial 0.100 M 0 0.00 M mol/l OPr - reacts with H O to reach equilibrium Change - + + Equil. 0.100 0.00 + 10 10 K b = 7.7 10 10 = ( )(0.00 ) 0.100 ()(0.00) 9, =.9 10 M 0.100 [OH = 0.00 + = 0.00 M; poh = 1.70; ph = 1.0 Assumptions good. c. This is a strong base in water. [OH = 0.00 M; poh = 1.70; ph = 1.0 d. OH will react completely with HOPr, the best acid present. HOPr + OH OPr + H O Before 0.100 M 0.00 M 0.100 M Change 0.00 0.00 +0.00 Reacts completely After 0.080 0 0.10

66 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA Using the Henderson-Hasselbalch equation to solve for the ph of the resulting buffer solution: [Base ph = pk a + log [Acid =.89 + (0.10) =.07 (0.080) 0. a. We have a weak base and a strong base present at the same time. The OH contribution from the weak base, HONH, will be negligible. Consider only the added strong base as the primary source of OH. [OH = 0.00 M; poh = 1.70; ph = 1.0 b. Added strong base will react to completion with the best acid present, HONH +. OH + HONH + HONH + H O Before 0.00 M 0.100 M 0 Change 0.00 0.00 +0.00 Reacts completely After 0 0.080 0.00 The resulting solution is a buffer (a weak acid and its conjugate base). Using the Henderson-Hasselbalch equation: ph = 6.0 + log (0.00) (0.080) = 6.0-0.60 =. c. This is a strong base in water. [OH - = 0.00 M; poh = 1.70; ph = 1.0 d. Major species: H O, Cl, Na +, HONH, HONH +, OH Again, the added strong base reacts completely with the best acid present, HONH +. HONH + + OH HONH + H O Before 0.100 M 0.00 M 0.100 M Change 0.00 0.00 +0.00 Reacts completely After 0.080 0 0.10 A buffer solution results. Using the Henderson-Hasselbalch equation: [HONH ph = 6.0 + log = 6.0 + log [HONH 1. Consider all of the results to Eercises 1., 1.7, and 1.9: (0.10) = 6.0 + 0.18 = 6. (0.080) Solution Initial ph after added acid after added base a.96 1.70.9 b 8.9.9 1.0 c 7.00 1.70 1.0 d.89.71.07

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 67 The solution in Eercise 1.d is a buffer; it contains both a weak acid (HC H O ) and a weak base (C H O ). Solution d shows the greatest resistance to changes in ph when either strong acid or strong base is added; this is the primary property of buffers.. Consider all of the results to Eercises 1., 1.8, and 1.0. Solution Initial ph after added acid after added base a 9. 6.6 1.0 b. 1.70. c 7.00 1.70 1.0 d 6.0.86 6. The solution in Eercise 1.d is a buffer; it shows the greatest resistance to a change in ph when strong acid or base is added. The solution in Eercise 1.d contains a weak acid (HONH + ) and a weak base (HONH ), which constitutes a buffer solution.. Major species: HNO, NO and Na +. Na + has no acidic or basic properties. One appropriate equilibrium reaction you can use is the K a reaction of HNO which contains both HNO and NO. However, you could also use the K b reaction for NO and come up with the same answer. Solving the equilibrium problem (called a buffer problem): HNO NO + H + Initial 1.00 M 1.00 M ~0 mol/l HNO dissociates to reach equilibrium Change + + Equil. 1.00 1.00 + K a =.0 10 = [NO [H [HNO = (1.00 )( ) 1.00 ( 1.00)( ) 1.00 (assuming < < 1.00) =.0 10 M = [H + ; Assumptions good ( is.0 10 % of 1.00). ph = log (.0 10 ) =.0 Note: We would get the same answer using the Henderson-Hasselbalch equation. Use whichever method you prefer.. Major species: HF, F and K + (no acidic/basic properties). We will use the K a reaction of HF which contains both HF and F to solve the equilibrium problem. HF F + H + Initial 0.60 M 1.00 M ~0 mol/l HF dissociates to reach equilibrium Change + + Equil. 0.60 1.00 +

68 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA K a = 7. 10 = [F [H [HF = (1.00 )( ) 0.60 ( 1.00)( ) 0.60 (assuming << 0.60) = [H + = 0.60 (7. 10 ) =. ph = log (. 10 ) =.7 10 M; Assumptions good ( is 7. 10 % of 0.60).. Major species after NaOH added: HNO, NO, Na + and OH. The OH from the strong base will react with the best acid present (HNO ). Any reaction involving a strong base is assumed to go to completion. Since all species present are in the same volume of solution, we can use molarity units to do the stoichiometry part of the problem (instead of moles). The stoichiometry problem is: OH + HNO NO + H O Before 0.10 mol/1.00 L 1.00 M 1.00 M Change 0.10 M 0.10 M +0.10 M Reacts completely After 0 0.90 1.10 After all the OH reacts, we are left with a solution containing a weak acid (HNO ) and its conjugate base (NO ). This is what we call a buffer problem. We will solve this buffer problem using the K a equilibrium reaction. HNO NO + H + Initial 0.90 M 1.10 M ~0 mol/l HNO dissociates to reach equilibrium Change + + Equil. 0.90 1.10 + K a =.0 10 = (1.10 )( ) 0.90 ( 1.10)( ), = [H + =. 0.90 10 M; ph =.8; Assumptions good. Note: The added NaOH to this buffer solution changes the ph only from.0 to.8. If the NaOH were added to 1.0 L of pure water, the ph would change from 7.00 to 1.00. Major species after HCl added: HNO, NO, H +, Na +, Cl ; The added H + from the strong acid will react completely with the best base present (NO,). Before H + + NO 0.0mol 1.00L HNO 1.00 M 1.00 M Change 0.0 M 0.0 M +0.0 M Reacts completely After 0 0.80 1.0 After all the H + has reacted, we have a buffer solution (a solution containing a weak acid and its conjugate base). Solving the buffer problem:

CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 69 HNO NO + H + Initial 1.0 M 0.80 M 0 Equil. 1.0 0.80 + + K a =.0 10 = (0.80 )( ) 1.0 0.80)( ), = [H + = 6.0 1.0 10 M; ph =.; Assumptions good. Note: The added HCl to this buffer solution changes the ph only from.0 to.. If the HCl were added to 1.0 L of pure water, the ph would change from 7.00 to 0.70. 6. When NaOH is added, the OH - reacts completely with the best acid present, HF. Before OH + HF F + H O 0.10mol 1.00L 0.60 M 1.00 M Change 0.10 M 0.10 M +0.10 M Reacts completely After 0 0.0 1.10 We now have a new buffer problem. Solving the equilibrium part of the problem: HF F + H + Initial 0.0 M 1.10 M 0 Equil. 0.0 1.10 + K a = 7. 10 = (1.10 )( ) 0.0 ph =.8; Assumptions good. 1.10( ), = [H + =. 10 M 0.0 When HCl is added, H + reacts to completion with the best base (F ) present. H + + F HF Before 0.0 M 1.00 M 0.60 M After 0 0.80 M 0.80 M After reaction, we have a buffer problem: HF F + H + Initial 0.80 M 0.80 M 0 Equil. 0.80 0.80 + K a = 7. 10 = (0.80 )( ) 0.80, = [H + = 7. 10 M; ph =.1; Assumptions good.

70 CHAPTER 1 APPLICATIONS OF AQUEOUS EQUILIBRIA 7. [HC 7 H O = 1 molhc7ho 1. g HC7HO 1.1g 0.000L = 0.880 M 1mol NaC7HO 1molC7HO 7.7 g NaC7HO [C 7 H O 1.10g mol NaC7HO = = 1.1 M 0.000L We have a buffer solution since we have both a weak acid and its conjugate base present at the same time. One can use the K a reaction or the K b reaction to solve. We will use the K a reaction for the acid component of the buffer. HC 7 H O H + + C 7 H O Initial 0.880 M ~0 1.1 M mol/l of HC 7 H O dissociates to reach equilibrium Change + + Equil. 0.880 1.1 + K a = 6. 10 = (1.1 ) 0.880 (1.1), = [H + =. 10 M 0.880 ph = log(. 10 ) =.7; Assumptions good. Alternatively, we can use the Henderson-Hasselbalch equation to calculate the ph of buffer solutions. [Base ph = pk a + log [Acid [C7HO = pk a + log [HC H O ph = log(6. 10 1.1 ) + log =.19 + 0.17 =.6 0.880 7 Within round-off error, this is the same answer we calculated solving the equilibrium problem using the K a reaction. The Henderson-Hasselbalch equation will be valid when an assumption of the type 1.1 + 1.1 that we just made in this problem is valid. From a practical standpoint, this will almost always be true for useful buffer solutions. If the assumption is not valid, the solution will have such a low buffering capacity that it will be of no use to control the ph. Note: The Henderson-Hasselbalch equation can only be used to solve for the ph of buffer solutions. 8. 0.0 g NH Cl 1mol NH Cl = 0.9 mol NH Cl added to 1.00 L; [NH + = 0.9 M.9g NH Cl [NH 10 ph = pk a + log = log (.6 10 0.7 ) + log = 9. 0.096 = 9.1 [NH 0.9