CHAPTER FOURTEEN ACIDS AND BASES. For Review

Transcription

1 CHAPTER FOURTEEN ACIDS AND BASES For Review 1. a. Arrhenius acid: produce H + in water b. Brnsted-Lowry acid: proton (H + ) donor c. Lewis acid: electron pair acceptor The Lewis definition is most general. The Lewis definition can apply to all Arrhenius and Brnsted-Lowry acids; H + has an empty 1s orbital and forms bonds to all bases by accepting a pair of electrons from the base. In addition, the Lewis definition incorporates other reactions not typically considered acid-base reactions, e.g., BF (g) + NH (g) F BNH (s). NH is something we usually consider a base and it is a base in this reaction using the Lewis definition; NH donates a pair of electrons to form the NB bond.. a. The a reaction always refers to an acid reacting with water to produce the conjugate base of the acid and the hydronium ion (H O + ). For a general weak acid HA, the a reaction is: HA(aq) + H O(l) A (aq) + H O + (aq) where A conjugate base of the acid HA This reaction is often abbreviated as: HA(aq) H + (aq) + A (aq) b. The a equilibrium constant is the equilibrium constant for the a reaction of some substance. For the general a reaction, the a epression is: a [A [HO [HA or a [H [A (for the abbreviated a reaction) [HA c. The b reaction alwlays refers to a base reacting with water to produce the conjugate acid of the base and the hydroide ion (OH ). For a general base, B, the b reaction is: B(aq) + H O(l) BH + (aq) + OH (aq) where BH + conjugate acid of the base B d. The b equilibrium constant for the general b reaction is: b [BH [OH [B e. A conjugate acid-base pair consists of two substances related to each other by the donating and accepting of a single proton. The conjugate bases of the acids HCl, HNO, HC H O and H SO are Cl, NO, C H O, and HSO, respectively. The conjugate acids of the bases NH, C H N, and HONH are NH +, C H NH +, and HONH +, respectively. Conjugate acidbase pairs only differ by H + in their respective formulas. 8

2 8 CHAPTER 1 ACIDS AND BASES. a. Amphoteric: a substance that can behave either as an acid or as a base. b. The w reaction is also called the autoionization of water reaction. The reaction always occurs when water is present as the solvent. The reaction is: H O(l) + H O(l) H O + (aq) + OH (aq) or H O(l) H + (aq) + OH (aq) c. The w equilibrium constant is also called the ion-product constant or the dissociation constant of water. It is the equilibrium constant for the autoionization reaction of water: w [H O + [OH or w [H + [OH At typical solution temperatures of C, w d. ph is a mathematical term which is equal to the log of the H + concentration of a solution (ph log[h +. e. poh is a mathematical terem which is equal to the log of the OH concentration of a solution (poh log[oh ). f. The p of any quantity is the log of that quantity. So: p w log w. At C, p w log 1 ( ) Neutral solution at C: w 1.0 [H + [OH 1.0 Acidic solution at C: 1 10 [H + [OH and ph + poh M; ph poh log( ).00 [H + > [OH ; [H + > 1.0 Basic solution at C: 10 M; [OH < M; ph <.00; poh >.00 [OH > [H + ; [OH > M; [H + < M; poh <.00; ph >.00 As a solution becomes more acidic, [H + increases, so [OH decreases, ph decreases, and poh increases. As a solution becomes more basic, [OH increases, so [H + decreases, ph increases, and poh decreases.. The a value refers to the reaction of an acid reacting with water to produce the conjugate base and H O +. The stronger the acid, the more conjugate base and H O + produced, and the larger the a value. Strong acids are basically 100% dissociated in water. Therefore, the strong acids have a a >> 1 because the equilibrium position lies far to the right. The conjugate bases of strong acids are terrible bases; much worse than water, so we can ignore their basic properties in water.

3 CHAPTER 1 ACIDS AND BASES 8 Weak acids are only partially dissociated in water. We say that the equilibrium lies far to the left, thus giving values for a < 1. (We have mostly reactants at equilibrium and few products present). The conjugate bases of weak acids are better bases than water. When we have a solution composed of just the conjugate base of a weak acid in water, tbe resulting ph is indeed basic (ph >.0). In general, as the acid strength increases, the conjugate base strength decreases, or as acid strength decreases, the conjugate base strength increases. They are inversely related. Base strength is directly related to the b value. The larger the b value, the more OH produced from the b reaction, and the more basic the solution (the higher the ph). Weak bases have a b < 1 and their conjugate acids behave as weak acids in solution. As the strength of the base increases, the strength of the conjugate acid gets weaker; the stronger the base, the weaker the conjugate acid, or the weaker the base, the stronger the conjugate acid.. Strong acids are assumed 100% dissociated in water, and we assume that the amount of H + donated by water is negligible. Hence, the equilibrium [H + of a strong acid is generally equal to the initial acid concentration ([HA 0 ). Note that solutions of H SO can be different from this as H SO is a diprotic acid. Also, when you have very dilute solutions of a strong acid, the H + contribution from water by itself must be considered. The strong acids to memorize are HCl, HBr, HI, HNO, HClO, and H SO. a values for weak acids are listed in Table 1. and in Appendi of the tet. Because weak acids only partially dissociate in water, we must solve an equilibrium problem to determine how much H + is added to water by the weak acid. We write down the a reaction, set-up the ICE table, then solve the equilibrium problem. The two assumptions generally made are that acids are less than % dissociated in water and that the H + contribution from water is negligible. The % rule comes from the assumptions that weak acids are less than % dissociated. When this is true, the mathematics of the problem are made much easier. The equilibrium epression we get for weak acids in water generally has the form (assuming an initial acid concentration of 0.10 M): a The % rule refers to assuming The assumption is valid if is less than % of the number the assumption was made against ([HA 0 ). When the % rule is valid, solving for is very straight forward. When the % rule fails, we must solve the mathematical epression eactly using the quadratic equation (or your graphing calculator). Even if you do have a graphing calculator, reference Appendi A1. to review the quadratic equation. Appendi A1. also discusses the method of successive approimations which can also be used to solve quadratic (and cubic) equations.. Strong bases are soluble ionic compounds containing the OH anion. Strong bases increase the OH concentration in water by just dissolving. Thus, for strong bases like LiOH, NaOH, OH, RbOH, and CsOH, the initial concentration of the strong base equals the equilibrium [OH of water. The other strong bases to memorize have + charged metal cations. The soluble ones to know are Ca(OH), Sr(OH), and Ba(OH). These are slightly more difficult to solve because they donate moles OH for every mole of salt dissolved. Here, the [OH is equal to two times the initial concentration of the soluble alkaline earth hydroide salt dissolved.

4 88 CHAPTER 1 ACIDS AND BASES Neutrally charged organic compounds containing at least one nitrogen atom generally behave as weak bases. The nitrogen atom has an unshared pair of electrons around it. This lone pair of electrons is used to form a bond to H +. Weak bases only partially react with water to produce OH. To determine the amount of OH produced by the weak acid (and, in turn, the ph of the solution), we set-up the ICE table using the b reaction of the weak base. The typical weak base equilibrium epression is: b (assuming [B 0 0. M) Solving for gives us the [OH in solution. We generally assume that weak bases are only % reacted with water and that the OH contribution from water is negligible. The % assumption makes the math easier. By assuming an epression like 0. M 0. M, the calculation is very straight forward. The % rule applied here is that if (/0.) 100 is less than %, the assumption is valid. When the assumption is not valid, then we solve the equilibrium epression eactly using the quadratic equation (or by the method of successive approimations).. Monoprotic acid: an acid with one acidic proton; the general formula for monoprotic acids is HA. Diprotic acid: an acid with two acidic protons (H A) Triprotic acid: an acid with three acidic protons (H A) H SO (aq) HSO (aq) + H + (aq) a 1 >> 1; this is a strong acid. HSO (aq) SO (aq) + H + (aq) a 0.01; this is a weak acid. When H SO is dissolved in water, the first proton is assumed 100% dissociated because H SO is a strong acid. After H SO dissociates, we have H + and HSO present. HSO is a weak acid and can donate some more protons to water. To determine the amount of H + donated by HSO, one must solve an equilibrium problem using the a reaction for HSO. H PO (aq) H + (aq) + H PO (aq) a H PO (aq) H + (aq) + HPO (aq) a HPO (aq) H + (aq) + PO (aq) a When H PO is added to water, the three acids that are present are H PO, H PO, and HPO. H PO, with the largest a value, is the strongest of these weak acids. The conjugate bases of the three acids are H PO, HPO, and PO. Because HPO is the weakest acid (smallest a value), its conjugate base (PO ) will have the largest b value and is the strongest base.

5 CHAPTER 1 ACIDS AND BASES 89 See Sample Eercises on the strategies used to solve for the ph of polyprotic acids. The strategy to solve most polyprotic acid solutions is covered in Sample Eercise 1.1. For typical polyprotic acids, a 1 >> a (and a if a triprotic acid). Because of this, the dominant producer of H + in solution is just the a 1 reaction. We set-up the equilibrium problem using the a 1 reaction and solve for H +. We then assume that the H + donated by the a (and a if triprotic) reaction is negligible that is, the H + donated by the a 1 reaction is assumed to be the H + donated by the entire acid system. This assumption is great when a 1 >> a (roughly a 1000 fold difference in magnitude). Sample Eercises 1.1 and 1.1 cover strategies for the other type of polyprotic acid problems. This other type is solutions of H SO. As discussed above, H SO problems are both a strong acid and a weak acid problem in one. To solve for the [H +, we sometimes must worry about the H + contribution from HSO. Sample Eercise 1.1 is an eample of an H SO solution where the HSO contribution of H + can be ignored. Sample Eercise 1.1 illustrates an H SO problem where we can t ignore the H + contribution from HSO. 8. a. H O and CH CO b. An acid-base reaction can be thought of as a competiton between two opposing bases. Since this equilibrium lies far to the left ( a < 1), CH CO is a stronger base than H O. c. The acetate ion is a better base than water and produces basic solutions in water. When we put acetate ion into solution as the only major basic species, the reaction is: CH CO + H O CH CO H + OH Now the competition is between CH CO and OH for the proton. Hydroide ion is the strongest base possible in water. The equilibrium above lies far to the left, resulting in a b 1 value less than one. Those species we specifically call weak bases ( 10 < b < 1) lie between H O and OH in base strength. Weak bases are stronger bases than water but are weaker bases than OH. The NH + ion is a weak acid because it lies between H O and H O + (H + ) in terms of acid strength. Weak acids are better acids than water, thus their aqueous solutions are acidic. They are weak acids because they are not as strong as H O + (H + ). Weak acids only partially 1 dissociate in water and have a values between 10 and 1. For a strong acid HX having a 1 10, the conjugate base, X, has b w / a / The conjugate bases of strong acids have etremely small values for b ; so small that they are worse bases than water ( b << w ). Therefore, conjugate bases of strong acids have no basic properties in water. They are present, but they only balance charge in solution and nothing else. The conjugate bases of the si strong acids are Cl, Br, I, NO, ClO, and HSO.

6 90 CHAPTER 1 ACIDS AND BASES Summarizing the acid-base properlties of conjugates: a. The conjugate base of a weak acid is a weak base ( 10 < b < 1) b. The conjugate acid of a weak base is a weak acid ( 10 < a < 1) c. The conjugate base of a strong acid is a worthless base ( b << d. The conjugate acid of a strong base is a worthless acid ( a << ) 1 10 ) Identifying/recognizing the acid-base properties of conjugates is crucial in order to understand the acid-base properties of salts. The salts we will give you will be salts containing the conjugates discussed above. Your job is to recognize the type of conjugate present, and then use that information to solve an equilibrium problem. 9. A salt is an ionic compound composed of a cation and an anion. Weak base anions: these are the conjugate bases of the weak acids having the HA general formula. Table 1. lists several HA type acids. Some weak base anions derived from the acids in Table 1. are ClO, F, NO, C H O, OCl, and CN. Garbage anions (those anions with no basic or acidic properties): these are the conjugate bases of the strong acids having the HA general formula. Some neutral anions are Cl, NO, Br, I, and ClO. Weak acid cations: these are the conjugate acids of the weak bases which contain nitrogen. Table 1. lists several nitrogen-containing bases. Some weak acid cations derived from the weak bases in Table 1. are NH +, CH NH +, C H NH +, C H NH +, and C H NH +. Garbage cations (those cations with no acidic properties or basic properties): the most common ones used are the cations in the strong bases. These are Li +, Na +, +, Rb +, Cs +, Ca +, Sr +, and Ba +. We mi and match the cations and anions to get what type of salt we want. For a weak base salt, we combine a weak base anion with a garbage cation. Some weak base salts are NaF, NO, Ca(CN), and RbC H O. To determine the ph of a weak base salt, we write out the b reaction for the weak base anion and determine b ( w / a ). We set-up the ICE table under the b reaction, and then solve the equilibrium problem to calculate [OH and, in turn, ph. For a weak acid salt, we combine a weak acid cation with a garbage anion. Some weak acid salts are NH Cl, C H NHNO, CH NH I, and C H NH ClO. To determine the ph, we write out the a reaction for the weak acid cation and determine a ( w / b ). We set-up the ICE table under the a reaction, and then solve the equilibrium problem to calculate [H + and, in turn, ph. For a neutral (ph.0) salt, we combine a garbage cation with a garbage anion. Some eamples are NaCl, NO, BaBr, and Sr(ClO ).

7 CHAPTER 1 ACIDS AND BASES 91 For salts that contain a weak acid cation and a weak base anion, we compare the a value of the weak acid cation to the b value for the weak base anion. When a > b, the salt produces an acidic solution (ph <.0). When b > a, the salt produces a basic solution. And when a b, the salt produces a neutral solution (ph.0). 10. a. The weaker the XH bond in an oyacid, the stronger the acid. b. As the electronegativity of neighboring atoms increases in an oyacid, the strength of the acid increases. c. As the number of oygen atoms increases in an oyacid, the strength of the acid increases. In general, the weaker the acid, the stronger the conjugate base and vice versa. a. Because acid strength increases as the XH bond strength decreases, conjugate base strength will increase as the strength of the XH bond increases. b. Because acid strength increases as the electronegativity of neighboring atoms increases, conjugate base strength will decrease as the electronegativity of neighboring atoms increases. c. Because acid strength increases as the number of oygen atoms increases, conjugate base strength decreases as the number of oygen atoms increases. Nonmetal oides form acidic solutions when dissolved in water: SO (g) + H O(l) H SO (aq) Metal oides form basic solutions when dissolved in water: Questions CaO(s) + H O(l) Ca(OH) 1. When a strong acid (HX) is added to water, the reaction HX + H O H O + + X basically goes to completion. All strong acids in water are completely converted into H O + and X. Thus, no acid stronger than H O + will remain undissociated in water. Similarly, when a strong base (B) is added to water, the reaction B + H O BH + + OH basically goes to completion. All bases stronger than OH - are completely converted into OH - and BH +. Even though there are acids and bases stronger than H O + and OH, in water these acids and bases are completely converted into H O + and OH ( S.F.);.8 ( S.F.); 0.8 ( S.F.); A ph value is a logarithm. The numbers to the left of the decimal point identify the power of ten to which [H + is epressed in scientific notation, e.g., 11 10, 10 1, 10. The number of decimal places in a ph value identifies the number of significant figures in [H +. In all three ph values, the [H + should be epressed only to two significant figures because these ph values have only two decimal places. 18. A Lewis acid must have an empty orbital to accept an electron pair, and a Lewis base must have an unshared pair of electrons.

8 9 CHAPTER 1 ACIDS AND BASES 19. a. These are strong acids like HCl, HBr, HI, HNO, H SO or HClO. b. These are salts of the conjugate acids of the bases in Table 1.. These conjugate acids are all weak acids. NH Cl, CH NH NO, and C H NH Br are three eamples. Note that the anions used to form these salts (Cl, NO, and Br ) are conjugate bases of strong acids; this is because they have no acidic or basic properties in water (with the eception of HSO, which has weak acid properties). c. These are strong bases like LiOH, NaOH, OH, RbOH, CsOH, Ca(OH), Sr(OH), and Ba(OH). d. These are salts of the conjugate bases of the neutrally charged weak acids in Table 1.. The conjugate bases of weak acids are weak bases themselves. Three eamples are NaClO, C H O, and CaF. The cations used to form these salts are Li +, Na +, +, Rb, Cs +, Ca +, Sr +, and Ba + since these cations have no acidic or basic properties in water. Notice that these are the cations of the strong bases you should memorize. e. There are two ways to make a neutral salt. The easiest way is to combine a conjugate base of a strong acid (ecept for HSO ) with one of the cations from a strong base. These ions have no acidic/basic properties in water so salts of these ions are neutral. Three eamples are NaCl, NO, and SrI. Another type of strong electrolyte that can produce neutral solutions are salts that contain an ion with weak acid properties combined with an ion of opposite charge having weak base properties. If the a for the weak acid ion is equal to the b for the weak base ion, then the salt will produce a neutral solution. The most common eample of this type of salt is ammonium acetate, NH C H O. For this salt, a for NH + b for C H O This salt at any concentration produces a neutral solution. 0. a b w, log ( a b ) log w log a log b log w, p a + p b p w 1.00 (at C) 1. a. H O(l) + H O(l) H O + (aq) + OH (aq) or H O(l) H + (aq) + OH (aq) w [H + [OH b. HF(aq) + H O(l) F (aq) + H O + (aq) or HF(aq) H+(aq) + F [H [F (aq) a [HF c. C H N(aq) + H O(l) C H NH + (aq) + OH (aq) b [CHNH [OH [C H N. Only statement a is true (assuming the species is not amphoteric). You cannot add a base to water and get an acidic ph (ph <.0). For statement b, you can have negative ph values. This just indicates an [H + > 1.0 M. For statement c, a dilute solution of a strong acid can have a higher ph than a more concentrated weak acid solution. For statement d, the Ba(OH) solution will have an

9 CHAPTER 1 ACIDS AND BASES 9 [OH twice of the same concentration of OH, but this does not correspond to a poh value twice that of the same concentration of OH (prove it to yourselves).. a. This epression holds true for solutions of strong acids having a concentration greater than M M HCl,.8 M HNO, and. 10 M HClO are eamples where this epression holds true. b. This epression holds true for solutions of weak acids where the two normal assumptions hold. The two assumptions are that water does not contribute enough H + to solution to matter and that the acid is less than % dissociated in water (from the assumption that is small compared to some number). This epression will generally hold true for solutions of weak acids having a a value less than 1 10, as long as there is a significant amount of weak acid present. Three eample solutions are 1. M HC H O, 0.10 M HOCl, and 0. M HCN. c. This epression holds true for strong bases that donate OH ions per formula unit. As long as the concentration of the base is above 10 M, this epression will hold true. Three eamples are.0 10 M Ca(OH),.1 10 M Sr(OH), and M Ba(OH). d. This epression holds true for solutions of weak bases where the two normal assumptions hold. The assumptions are that the OH contribution from water is negligible and that and that the base is less than % ionized in water (for the % rule to hold). For the % rule to hold, you generally need bases with b < 1 10 and concentrations of weak base greater than 0.10 M. Three eamples are 0.10 M NH, 0. M C H NH, and 1.1 M C H N.. H CO is a weak acid with a and a. 10. The [H + concentration in solution will be determined from the a reaction since 1 a >> 1 a. Since a << 1, then the 1 [H + < 0.10 M; only a small percentage of H CO will dissociate into HCO and H +. So statement a best describes the 0.10 M H CO solution. H SO is a strong acid and a very good weak acid ( a >> 1, 1 a ). All of the 0.1 M H SO solution will dissociate into 0.10 M H + and 0.10 M HSO. However, since HSO is a good weak acid due to the relatively large a value, then some of the 0.10 M HSO will dissociate into some more H + and SO. Therefore, the [H + will be greater than 0.10 M, but will not reach 0.0 since only some of 0.10 M HSO will dissociate. Statement c is best for a 0.10 M H SO solution.. One reason HF is a weak acid is that the HF bond is unusually strong and is difficult to break. This contributes significantly to the reluctance of the HF molecules to dissociate in water.. a. Sulfur reacts with oygen to produce SO and SO. These sulfur oides both react with water to produce H SO and H SO, respectively. Acid rain can result when sulfur emissions are not controlled. Note that in general, nonmetal oides react with water to produce acidic solutions. b. CaO reacts with water to produce Ca(OH), a strong base. A gardener mies lime (CaO) into soil in order to raise the ph of the soil. The effect of adding lime is to add Ca(OH). Note that in general, metal oides react with water to produce basic solutions. 11

10 9 CHAPTER 1 ACIDS AND BASES Eercises Nature of Acids and Bases. a. HClO (aq) + H O(l) H O + (aq) + ClO (aq). Only the forward reaction is indicated since HClO is a strong acid and is basically 100% dissociated in water. For acids, the dissociation reaction is commonly written without water as a reactant. The common abbreviation for this reaction is: HClO (aq) H + (aq) + ClO (aq). This reaction is also called the a reaction as the equilibrium constant for this reaction is called a. b. Propanoic acid is a weak acid, so it is only partially dissociated in water. The dissociation reaction is: CH CH CO H(aq) + H O(l) H O + (aq) + CH CH CO (aq) or CH CH CO H(aq) H + (aq) + CH CH CO (aq). c. NH + is a weak acid. Similar to propanoic acid, the dissociation reaction is: NH + (aq) + H O(l) H O + (aq) + NH (aq) or NH + (aq) H + (aq) + NH (aq) 8. The dissociation reaction (the a reaction) of an acid in water commonly omits water as a reactant. We will follow this practice. All dissociation reactions produce H + and the conjugate base of the acid that is dissociated. a. HCN(aq) H + (aq) + CN [H [CN (aq) a [HCN b. C H OH(aq) H + (aq) + C H O (aq) a c. C H NH + (aq) H + (aq) + C H NH (aq) a [H [CHO [C H OH [C H NH [H [C H NH 9. An acid is a proton (H + ) donor and a base is a proton acceptor. A conjugate acid-base pair differs by only a proton (H + ). Conjugate Conjugate Acid Base Base of Acid Acid of Base a. H CO H O HCO H O + b. C H NH + H O C H N H O + c. C H NH + HCO C H N H CO 0. Conjugate Conjugate Acid Base Base of Acid Acid of Base a. + Al(H O) H O Al(H O) (OH) + H O + b. + HONH H O HONH H O + c. HOCl C H NH OCl - + C H NH

11 CHAPTER 1 ACIDS AND BASES 9 1. Strong acids have a a >> 1 and weak acids have a < 1. Table 1. in the tet lists some a values for weak acids. a values for strong acids are hard to determine so they are not listed in the tet. However, there are only a few common strong acids so if you memorize the strong acids, then all other acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO, HClO and H SO. a. HClO is a strong acid. b. HOCl is a weak acid ( a. 10 ). c. H SO is a strong acid. d. H SO is a weak diprotic acid with a1 and a values less than one. 8. The beaker on the left represents a strong acid in solution; the acid, HA, is 100% dissociated into the H + and A ions. The beaker on the right represents a weak acid in solution; only a little bit of the acid, HB, dissociates into ions, so the acid eists mostly as undissociated HB molecules in water. a. HNO : weak acid beaker b. HNO : strong acid beaker c. HCl: strong acid beaker d. HF: weak acid beaker e. HC H O : weak acid beaker. The a value is directly related to acid strength. As a increases, acid strength increases. For water, use w when comparing the acid strength of water to other species. The a values are: HClO : strong acid ( a >> 1); HClO : a NH + : a. 10 ; H O: a w From the a values, the ordering is: HClO > HClO > NH + > H O.. Ecept for water, these are the conjugate bases of the acids in the previous eercise. In general, the weaker the acid, the stronger the conjugate base. ClO is the conjugate base of a strong acid; it is a terrible base (worse than water). The ordering is: NH > ClO > H O > ClO. a. HCl is a strong acid and water is a very weak acid with a w HCl is a much stronger acid than H O. 1 b. H O, a w ; HNO, a.0 10 ; HNO is a stronger acid than H O because a for HNO > w for H O. 10 c. HOC H, a ; HCN, a. 10 ; HCN is a stronger acid than HOC H because a for HCN > a for HOC H.. a. H O; The conjugate bases of strong acids are terrible bases ( b < 10 )

12 9 CHAPTER 1 ACIDS AND BASES b. NO ; The conjugate bases of weak acids are weak bases ( 10 < b < 1). c. OC H ; For a conjugate acid-base pair, a b w. From this relationship, the stronger the acid the weaker the conjugate base ( b decreases as a increases). Because HCN is a stronger acid than HOC H ( a for HCN > a for HOC H ), OC H will be a stronger base than CN -. Autoionization of Water and the ph Scale. At C, the relationship: [H + [OH 1 w always holds for aqueous solutions. When [H + is greater than M, the solution is acidic; when [H + is less than M, the solution is basic; when [H M, the solution is neutral. In terms of [OH, an acidic solution has [OH < M, a basic solution has [OH > M, and a neutral solution has [OH M. 1 a. [OH w M; The solution is neutral. [H b. [OH M; The solution is basic. c. [OH M; The solution is acidic. 1 d. [OH M; The solution is acidic. 8. a. [H + 1 w M; basic [OH 1. 1 b. [H M; acidic c. [H d. [H M; neutral M; basic 9. a. Because the value of the equilibrium constant increases as the temperature increases, the reaction is endothermic. In endothermic reactions, heat is a reactant so an increase in temperature (heat) shifts the reaction to produce more products and increases in the process. b. H O(l) H + (aq) + OH (aq) w. In pure water [H + [OH, so [H + [OH at 0. C 1 10 [H +, [H M [OH

13 CHAPTER 1 ACIDS AND BASES 9 0. a. H O(l) H + (aq) + OH (aq) w.9 In pure water [H + [OH, so.9 b. ph log [H + log (1.1 c. [H + w /[OH.9 10 ) / [H + [OH 1 10 [H +, [H M [OH 1 10 M; ph log (.9 10 ) ph log [H + ; poh log [OH - ; At C, ph + poh 1.00; For Eercise 1.: a. ph log [H + log ( ).00; poh 1.00 ph b. ph log ( ) 1.08; poh c. ph log (1) 1.08; poh 1.00 (1.08) 1.08 d. ph log (. 10 ).; poh Note that ph is less than zero when [H + is greater than 1.0 M (an etremely acidic solution). For Eercise 1.8: a. poh log [OH log (1.) 0.18; ph 1.00 poh 1.00 (0.18) b. poh log (. 10 ) 1.; ph c. poh log ( ).00; ph d. poh log (. 10 ).1; ph Note that ph is greater than 1.00 when [OH is greater than 1.0 M (an etremely basic solution). ph. a. [H + 10, [H M poh 1.00 ph ; [OH 1 or [OH w [H.0 10 b. [H + 1 poh M 10 M; This solution is basic since ph > M; poh ; [OH c. [H + ( 1.0) M; poh 1.0 (1.0) 1.0; [OH d. [H + acidic M; poh ; [OH ( 1.) M; basic 1 10 M; acidic M;

14 98 CHAPTER 1 ACIDS AND BASES e. [OH f. [OH acidic M; ph 1.0 poh ; [H a. poh ; [H + [OH b. [H M; ph ; [H M; acidic 1 poh ; acidic M 0.1 M; ph log(0.1) c. ph ; [H M [OH M; basic d. ph log ( ).00; poh [OH M; neutral. a. poh ; [H M [OH M; basic M; basic 10 M; 1 b. [H M; ph log (. 10 ) 8.9 poh ; basic c. ph log (0.0) 1.; poh [OH M; acidic d. ph ; [H M [OH M; acidic. poh 1.0 ph ; [H + 1 [OH w [H 810 ph M or [OH poh M (1 sig fig) The sample of gastric juice is acidic since the ph is less than.00 at C M

15 CHAPTER 1 ACIDS AND BASES 99. ph 1.00 poh ; [H + 1 [OH w [H. 10 ph M or [OH poh M The solution of baking soda is basic since the ph is greater than.00 at C. Solutions of Acids 10 M. All the acids in this problem are strong acids that are always assumed to completely dissociate in water. The general dissociation reaction for a strong acid is: HA(aq) H + (aq) + A (aq) where A is the conjugate base of the strong acid HA. For 0.0 M solutions of these strong acids, 0.0 M H + and 0.0 M A are present when the acids completely dissociate. The amount of H + donated from water will be insignificant in this problem since H O is a very weak acid. a. Major species present after dissociation H +, ClO and H O; ph log [H + -log (0.0) 0.0 b. Major species H +, NO and H O; ph Strong acids are assumed to completely dissociate in water: HCl(aq) H + (aq) + Cl (aq) a. A 0.10 M HCl solution gives 0.10 M H + and 0.10 M Cl since HCl completely dissociates. The amount of H + from H O will be insignificant. ph log [H + log (0.10) 1.00 b..0 M H + is produced when.0 M HCl completely dissociates. The amount of H + from H O will be insignificant. ph log (.0) -0.0 (Negative ph values just indicate very concentrated acid solutions). 11 c M H + 11 is produced when M HCl completely dissociates. If you take the 11 negative log of this gives ph This is impossible! We dissolved an acid in water and got a basic ph. What we must consider in this problem is that water by itself donates M H +. We can normally ignore the small amount of H + from H O ecept when we have a very dilute solution of an acid (as is the case here). Therefore, the ph is that of neutral water (ph.00) since the amount of HCl present is insignificant. 9. Both are strong acids L 0.00 mol/l L 0.10 mol/l mol HCl. 10 mol HNO mol H mol H mol Cl 10 mol NO [H + ( L ) mol M; [OH w [H M

16 00 CHAPTER 1 ACIDS AND BASES [Cl mol 0.000L 10 L.00mol L 0.01 M; [NO mol 0.000L 0.0 mol H + from HCl 0.0 M L 8.00mol L 0.0 mol H + from HNO [H + 0.0mol 0.0mol 1.00L 0.90 M; ph log (0.90) [H + poh ; [OH M M (carryiung one etra sig fig); M 1 V 1 M V V 1 M V M mol/ L 1mol/ L 1. L. 10 L. [H + To. ml of 1 M HCl, add enough water to make 100 ml of solution. The resulting solution will have [H M and ph M; HNO (aq) H + (aq) + NO (aq) Because HNO is a strong acid, we have a L.9 10 M HNO solution. 10 mol HNO.0g HNO 1. L mol HNO 10 g HNO. a. HNO ( a.0 10 ) and H O ( a w ) are the major species. HNO is a much stronger acid than H O so it is the major source of H +. However, HNO is a weak acid ( a < 1) so it only partially dissociates in water. We must solve an equilibrium problem to determine [H +. In the Solutions Guide, we will summarize the initial, change and equilibrium concentrations into one table called the ICE table. Solving the weak acid problem: HNO H + + NO Initial 0.0 M ~0 0 mol/l HNO dissociates to reach equilibrium Change + + Equil a [H [NO [HNO ; If we assume << 0.0, then:.0 10,.0 10 (0.0) M

17 CHAPTER 1 ACIDS AND BASES 01 We must check the assumption: % All the assumptions are good. The H + contribution from water ( 10 M) is negligible, and is small compared to 0.0 (percent error.0%). If the percent error is less than % for an assumption, we will consider it a valid assumption (called the % rule). Finishing the problem: M [H + ; ph log(0.010).00 b. CH CO H ( a ) and H O ( a w ) are the major species. CH CO H is the major source of H +. Solving the weak acid problem: CH CO H H + + CH CO Initial 0.0 M ~0 0 mol/l CH CO H dissociates to reach equilibrium Change + + Equil a [H [CH CO [CH CO H (assuming << 0.0) M; Checking assumption: %. Assumptions good. [H M; ph -log (.1 10 ) a. HOC H ( a ) and H O ( a w ) are the major species. The major equilibrium is the dissociation of HOC H. Solving the weak acid problem: HOC H H + + OC H Initial 0.0 M ~0 0 mol/l HOC H dissociates to reach equilibrium Change + + Equil a [H [OCH [HOC H (assuming << 0.0) [H M; Checking assumption: is. 10 % of 0.0, so assumption is valid by the % rule. ph log(. 10 ).0 10 b. HCN ( a. 10 ) and H O are the major species. HCN is the major source of H +. HCN H + + CN Initial 0.0 M ~0 0

18 0 CHAPTER 1 ACIDS AND BASES mol/l HCN dissociates to reach equilibrium Change + + Equil a. 10 [H [CN [HCN [H (assuming << 0.0) 10 M; Checking assumption: is % of 0.0 Assumptions good. ph log (1. 10 ).9. [CH COOH 0 [HC H O 0 1molHCHO.00g HCHO 0.0g L M HC H O H + + C H O a Initial M ~0 0 mol/l HC H O dissociates to reach equilibrium Change + + Equil a [H [CHO [HC H O [H M; ph. Assumptions good ( is.1% of 0.018). [H + [C H O [CH COO.8 10 M; [CH COOH M. HC H O ( a ) and H O ( a w ) are the major species present. HC H O will be the dominant producer of H + since HC H O is a stronger acid than H O. Solving the weak acid problem: HC H O H + + C H O - Initial M ~0 0 mol/l HC H O dissociates to reach equilibrium Change + + Equil a [H [CHO [HC H O [H M; ph log ( ).9 Assumption follows the % rule ( is 1.1% of 0.100). 10

19 CHAPTER 1 ACIDS AND BASES 0 [H + [C H O M; [OH w /[H M [HC H O [H Percent dissociation [HC M HO %. This is a weak acid in water. Solving the weak acid problem: HF H + + F a. Initial 0.00 M ~0 0 mol/l HF dissociates to reach equilibrium Change + + Equil a. 10 [H [F [HF [H M; Check assumptions: (assuming << 0.00) % The assumption << 0.00 is not good ( is more than % of 0.00). We must solve /(0.00 ). 10 eactly by using either the quadratic formula or by the method of successive approimations (see Appendi 1. of tet). Using successive approimations, we let 0.01 M be a new approimation for [HF. That is, in the denominator, try (the value of we calculated making the normal assumption), so , then solve for a new value of in the numerator ,. 10 We use this new value of to further refine our estimate of [HF, i.e., (carry etra significant figure) ,. 10 We repeat until we get an answer that repeats itself. This would be the same answer we would get solving eactly using the quadratic equation. In this case it is:. 10 So: [H + [F. 10 M; [OH w /[H M [HF M; ph. Note: When the % assumption fails, use whichever method you are most comfortable with to solve eactly. The method of successive approimations is probably fastest when the

20 0 CHAPTER 1 ACIDS AND BASES percent error is less than ~% (unless you have a calculator that can solve quadratic equations). 8. Major species: HIO, H O; Major source of H + : HIO (a weak acid, a 0.1) HIO H + + IO Initial 0.0 M ~0 0 mol/l HIO dissociates to reach equilibrium Change + + Equil. 0.0 a , 0.18; Check assumption. 0.0 Assumption is horrible ( is 90% of 0.0). When the assumption is this poor, it is generally quickest to solve eactly using the quadratic formula (see Appendi 1. in tet). The method of successive approimations will require many trials to finally converge on the answer. For this problem, trials were required. Using the quadratic formula and carrying etra significant figures: 0.1, 0.1(0.0 ), [(0.1) (1)( 0.0) (1) 1/ , 0.1 or 0.9 Only 0.1 makes sense. 0.1 M [H + ; ph log (0.1) Major species: HC H ClO ( a 1. HC H ClO H + + C H ClO 10 ) and H O; Major source of H + : HC H ClO Initial 0.10 M ~0 0 mol/l HC H ClO dissociates to reach equilibrium Change + + Equil a , M Checking the assumptions finds that is 1% of 0.10 which fails the % rule. We must solve /(0.10 ) eactly using either the method of successive approimations or the quadratic equation. Using either method gives [H M. ph log [H + log ( ) 1.9.

21 CHAPTER 1 ACIDS AND BASES 0 0. [HC 9 H O 0.g HC9H O tablets tablet 0.L 1molHC9H O 180.1g 0.01 M HC 9 H O H + + C 9 H O Initial 0.01 M ~0 0 mol/l HC 9 H O dissociates to reach equilibrium Change + + Equil a. 10 [H [C9HO [HC H O , M Assumption that fails the % rule: % Using successive approimations or the quadratic equation gives an eact answer of.1 10 M. [H M; ph log (.1 10 ).8 1. a. HCl is a strong acid. It will produce 0.10 M H +. HOCl is a weak acid. Let's consider the equilibrium: HOCl H + + OCl a. Initial 0.10 M 0.10 M 0 mol/l HOCl dissociates to reach equilibrium Change + + Equil a [H [OCl [HOCl (0.10 )( ) 0.10, M Assumptions are great ( is % of 0.10). We are really assuming that HCl is the only important source of H +, which it is. The [H + contribution from HOCl,, is negligible. Therefore, [H M; ph 1.00 b. HNO is a strong acid, giving an initial concentration of H + equal to 0.00 M. Consider the equilibrium: HC H O H + + C H O a Initial 0.0 M 0.00 M 0 mol/l HC H O dissociates to reach equilibrium Change + + Equil

22 0 CHAPTER 1 ACIDS AND BASES a [H [CHO [HC H O (0.00 ) ; Assumptions are good (well within the % rule). [H M and ph 1.0. HF and HOC H are both weak acids with a values of and 1. 10, respectively. Since the a value for HF is much greater than the a value for HOC H, HF will be the dominant producer of H + (we can ignore the amount of H + produced from HOC H since it will be insignificant). HF H + + F Initial 1.0 M ~0 0 mol/l HF dissociates to reach equilibrium Change + + Equil. 1.0 a. 10 [H [F [HF [H M; ph log (. 10 ) 1. Assumptions good. Solving for [OC H using HOC H H + + OC H - equilibrium: a [H [OCH [HOC H (. 10 )[OCH 1.0, [OC H M 9 Note that this answer indicates that only.9 10 M HOC H dissociates, which indicates that HF is truly the only significant producer of H + in this solution.. In all parts of this problem, acetic acid (HC H O ) is the best weak acid present. We must solve a weak acid problem. a. HC H O H + + C H O Initial 0.0 M ~0 0 mol/l HC H O dissociates to reach equilibrium Change + + Equil. 0.0 a [H [CHO [HC H O [H + [C H O M Assumptions good.

23 CHAPTER 1 ACIDS AND BASES 0 Percent dissociation [H [HC HO % b. The setups for solutions b and c are similar to solution a ecept the final equation is slightly different, reflecting the new concentration of HC H O. a [H + [C H O M Assumptions good. % dissociation % c. a [H + [C H O.0 10 M; Check assumptions. Assumption that is negligible is borderline (.0% error). We should solve eactly. Using the method of successive approimations (see Appendi 1. of tet): Net trial also gives % dissociation 100.8% ,.9 d. As we dilute a solution, all concentrations decrease. Dilution will shift the equilibrium to the side with the greater number of particles. For eample, suppose we double the volume of an equilibrium miture of a weak acid by adding water, then: Q [H eq [X [HX eq eq 1 a Q < a, so the equilibrium shifts to the right or towards a greater percent dissociation. e. [H + depends on the initial concentration of weak acid and on how much weak acid dissociates. For solutions a-c the initial concentration of acid decreases more rapidly than the percent dissociation increases. Thus, [H + decreases. 10. a. HNO is a strong acid; it is assumed 100% dissociated in solution. b. HNO H + + NO a.0 10

24 08 CHAPTER 1 ACIDS AND BASES Initial 0.0 M ~0 0 mol/l HNO dissociates to reach equilibrium Change + + Equil. 0.0 a.0 10 [H [NO [HNO [H + [NO M; Assumptions good. % dissociation [H [HNO % c. HOC H H + + OC H a Initial 0.0 M ~0 0 mol/l HOC H dissociates to reach equilibrium Change + + Equil. 0.0 a [H [OCH [HOC H [H + [OC H. 10 M; Assumptions good. % dissociation % d. For the same initial concentration, the percent dissociation increases as the strength of the acid increases (as a increases).. Let HX symbolize the weak acid. Setup the problem like a typical weak acid equilibrium problem. HX H + + X Initial 0.1 M ~0 0 mol/l HX dissociates to reach equilibrium Change + + Equil. 0.1 If the acid is.0% dissociated, then [H + is.0% of 0.1: 0.00 (0.1 M). 10 M. Now that we know the value of, we can solve for a.

25 CHAPTER 1 ACIDS AND BASES 09 a [H [X [HX 0.1 (. 10 ) HX H + + X Initial I ~0 0 where I [HX o mol/l HX dissociates to reach equilibrium Change + + Equil. I From the problem, 0.(I) and I 0.0 M. I 0.(I) 0.0 M, I 0.0 M and 0. (0.0 M) 0.10 M a [H [X [HX I (0.10) Setup the problem using the a equilibrium reaction for HOCN. HOCN H + + OCN Initial M ~0 0 mol/l HOCN dissociates to reach equilibrium Change + + Equil a [H [OCN [HOCN ; ph.: [H + ph M (1. 10 ) a HClO is a strong acid with [H M. This equals the [H + in the trichloroacetic acid solution. Now setup the problem using the a equilibrium reaction for CCl CO H. CCl CO H H + + CCl CO Initial 0.00 M ~0 0 Equil a [H [CCl CO [CCl CO H 0.00 ; [H M

26 10 CHAPTER 1 ACIDS AND BASES a ( (.0 10 ) ) Major species: HCOOH and H O; Major source of H + : HCOOH HCOOH H + + HCOO Initial C ~0 0 where C [HCOOH o mol/l HCOOH dissociates to reach equilibrium Change + + Equil. C a [H [HCOO [HCOOH C where [H [HA o [H ; ph.0, so: [H + C [H (.0 10 ) C ( M, C ( ), C. ) A 0.0 M formic acid solution will have ph mol 0.0 mol/l; Solve using the a equilibrium reaction..0 L HA H + + A Initial 0.0 M ~0 0 Equil. 0.0 [H [A a [HA 0.0 ; In this problem, [HA 0. M so: [HA 0. M 0.0 M, 0.0 M; a (0.0) M Solutions of Bases 1. a. NH (aq) + H O(l) NH + (aq) + OH (aq) b b. C H N(aq) + H O(l) C H NH + (aq) + OH (aq) b [NH [OH [NH [CHNH [OH [C H N

27 CHAPTER 1 ACIDS AND BASES 11. a. C H NH (aq) + H O(l) C H NH + (aq) + OH (aq) b [CHNH [OH [C H NH b. (CH ) NH(aq) + H O(l) (CH ) NH + (aq) + OH [(CH ) NH [OH (aq) b [(CH ) NH. NO : b << w since HNO is a strong acid. All conjugate bases of strong acids have no base 1 strength. H O: b w ; NH : b ; C H N: b NH > C H N > H O > NO - (As b increases, base strength increases.). Ecluding water, these are the conjugate acids of the bases in the previous eercise. In general, the stronger the base, the weaker the conjugate acid. Note: Even though NH + and C H NH + are conjugate acids of weak bases, they are still weak acids with a values between w and 1. Prove this to yourself by calculating the a values for NH + and C H NH + ( a w / b ). HNO > C H NH + > NH + > H O. a. C H NH b. C H NH c. OH d. CH NH The base with the largest b value is the strongest base (.8 b, C HNH b, CH NH. 10. OH is the strongest base possible in water ,. a. HClO b. C H NH + c. C H NH + The acid with the largest a value is the strongest acid. To calculate a values for C H NH + and CH NH +, use a w / b where b refers to the bases C H NH or CH NH.. NaOH(aq) Na + (aq) + OH (aq); NaOH is a strong base which completely dissociates into Na + and OH. The initial concentration of NaOH will equal the concentration of OH donated by NaOH. a. [OH 0.10 M; poh log[oh -log(0.10) 1.00 ph 1.00 poh Note that H O is also present, but the amount of OH produced by H O will be insignificant compared to the 0.10 M OH produced from the NaOH. b. The [OH concentration donated by the NaOH is M. Water by itself donates M. In this problem, water is the major OH - contributor and [OH M. poh -log ( ).00; ph c. [OH.0 M; poh -log (.0) -0.0; ph 1.00 (0.0) a. Ca(OH) Ca + + OH ; Ca(OH) is a strong base and dissociates completely. 10

28 1 CHAPTER 1 ACIDS AND BASES [OH (0.0000) M; poh log [OH.10; ph 1.00 poh b. g OH 1molOH 0. mol OH/L L.11g OH OH is a strong base, so [OH 0. M; poh -log (0.) 0.; ph 1. c g NaOH L 1mol.0 M; NaOH is a strong base, so [OH.0 M. 0.00g poh log (.0) 0.0 and ph (0.0) 1.0 Although we are justified in calculating the answer to four decimal places, in reality ph values are generally measured to two decimal places, sometimes three. 9. a. Major species: +, OH, H O (OH is a strong base.) [OH 0.01 M, poh log (0.01) 1.8; ph 1.00 poh 1.18 b. Major species: Ba +, OH, H O; Ba(OH) (aq) Ba + (aq) + OH (aq); Since each mol of the strong base Ba(OH) dissolves in water to produce two mol OH, then [OH (0.01 M) 0.00 M. poh log (0.00) 1.; ph a. Major species: Na +, Li +, OH, H O (NaOH and LiOH are both strong bases.) [OH M; poh 1.000; ph b. Major species: Ca +, Rb +, OH, H O; Both Ca(OH) and RbOH are strong bases and Ca(OH) donates mol OH per mol Ca(OH). [OH (0.0010) M; poh log (0.0) 1.; ph poh ; [OH [OH M L. 10 mol OH L.11g OH mol OH 0.1 g OH 8. ph 10.0; poh ; [OH M Sr(OH) (aq) Sr + (aq) + OH (aq); Sr(OH) donates two mol OH per mol Sr(OH). [Sr(OH). 10 moloh L 1 molsr (OH) 1. mol OH 10 M Sr(OH)

29 CHAPTER 1 ACIDS AND BASES 1 A M Sr(OH) solution will produce a ph 10.0 solution. 8. NH is a weak base with b The major species present will be NH and H O ( b 1 w ). Since NH has a much larger b value compared to H O, NH is the stronger base present and will be the major producer of OH. To determine the amount of OH produced from NH, we must perform an equilibrium calculation. NH (aq) + H O(l) NH + (aq) + OH (aq) Initial 0.10 M 0 ~0 mol/l NH reacts with H O to reach equilibrium Change + + Equil b [NH [OH [NH (assuming << 0.10) [OH M; Check assumptions: is 1.1% of 0.10 so the assumption is valid by the % rule. Also, the contribution of OH from water will be insignificant (which will usually be the case). Finishing the problem, poh log [OH log (1. 10 M).80; ph 1.00 poh Major species: H NNH ( b.0 10 ) and H O ( b w ); The weak base H NNH will dominate OH production. We must perform a weak base equilibrium calculation. H NNH + H O H NNH + Initial.0 M 0 ~0 mol/l H NNH reacts with H O to reach equilibrium Change + + Equil OH b.0 10 b.0 10 [HNNH [OH [H NNH.0. 0 (assuming <<.0) [OH. [H NNH M; poh.; ph 11.8 Assumptions good ( is 0.1% of.0). 10 M; [H NNH.0 M; [H M 8. These are solutions of weak bases in water. We must solve the equilibrium weak base problem. a. (C H ) N + H O (C H ) NH + + OH b.0 Initial 0.0 M 0 ~0 mol/l of (C H ) N reacts with H O to reach equilibrium Change + + Equil

30 1 CHAPTER 1 ACIDS AND BASES b.0 10 [(CH) NH [OH [(C H ) N , [OH M Assumptions good ( is.% of 0.0). [OH M [H + 1 w [OH M; ph 11.9 b. HONH + H O HONH + + OH b Initial 0.0 M 0 ~0 Equil. 0.0 b , [OH. 10 M; Assumptions good. [H M; ph These are solutions of weak bases in water. a. C H NH + H O C H NH + + OH b Initial 0.0 M 0 ~0 mol/l of C H NH reacts with H O to reach equilibrium Change + + Equil , [OH M; Assumptions good. [H + w /[OH M; ph 8.9 b. CH NH + H O CH NH + + OH b.8 Initial 0.0 M 0 ~0 Equil. 0.0 b.8 [OH , M; Assumptions good. 10 M; [H + w /[OH M; ph This is a solution of a weak base in water. We must solve the weak base equilibrium problem. C H NH + H O C H NH + + OH b. Initial 0.0 M 0 ~0 mol/l C H NH reacts with H O to reach equilibrium Change