CHAPTER 14 ACIDS AND BASES. Questions

Transcription

1 CHAPTER 1 ACIDS AND BASES Questions 19. Acids are proton (H donors, and bases are proton acceptors. HCO as an acid: HCO (aq H O(l CO (aq H O (aq HCO as a base: HCO (aq H O(l H CO (aq OH (aq H PO as an acid: H PO H O(l HPO (aq H O (aq H PO as a base: H PO H O(l H PO (aq OH (aq 0. Acidic solutions (at C have an [H > 7 M, which gives a ph < 7.0. Because [H [OH 1 and ph poh 1.00 for an aqueous solution at C, an acidic solution must also have [OH < 7 M and poh > From these relationships, the solutions in parts a, b, and d are acidic. The solution in part c will have a ph > 7.0 (ph and is therefore not acidic (solution is basic. 1. Basic solutions (at C have an [OH > 7 M, which gives a poh < 7.0. Because [H [OH 1 and ph poh 1.00 for any aqueous solution at C, a basic solution must also have [H < 7 M and ph > From these relationshis, the solutions in parts b, c, and d are basic solutions. The solution in part a will have a ph < 7.0 (ph and is therefore not basic (solution is acidic.. 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.; 6.78 ( S.F.; 0.78 ( S.F.; a ph value is a logarithm. The numbers to the left of the decimal point identify the power of 10 to which [H is epressed in scientific notation, for eample, 10 11, , 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. 96

2 CHAPTER 1 ACIDS AND BASES 97. A Lewis acid must have an empty orbital to accept an electron pair, and a Lewis base must have an unshared pair of electrons.. a. These are strong acids like HCl, HBr, HI, HNO, H SO, and 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 are conjugate bases of strong acids; this is so 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 because 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. 6. a b w,!log( a b!log w!log a! log b!log w, p a p b p w 1.00 ( w 1 at C 7. 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

3 98 CHAPTER 1 ACIDS AND BASES 8. Only statement a is true (assuming the species is not amphoteric. You cannot add a base to water and get an acidic ph (ph < 7.0. 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 [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. 9. a. This epression holds true for solutions of strong acids having a concentration greater 6 than M M HCl, 7.8 M HNO, and.6 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 make a difference, 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 0.7 M HCN. c. This epression holds true for strong bases that donate OH ions per formula unit. As 7 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 6 H NH, and 1.1 M C H N. 0. H CO is a weak acid with a and a The [H concentration in solution will be determined from the a 1 reaction because a 1 >> a. Because a 1 << 1, the [H < 0.10 M; only a small percentage of the 0.10 M 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 as well as a very good weak acid ( a1 >> 1, a All the 0.10 M H SO solution will dissociate into 0.10 M H and 0.10 M HSO. However, because HSO is a good weak acid due to the relatively large a value, 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 M because only some of 0.10 M HSO will dissociate. Statement c is best for a 0.10 M H SO solution. 1. 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. 7 11

4 CHAPTER 1 ACIDS AND BASES 99. 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. Eercises 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. Nature of Acids and Bases. a. HClO (aq H O(l H O (aq ClO (aq. Only the forward reaction is indicated because 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 because the equilibrium constant for this reaction is designated as 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. 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. HOC 6 H (aq H (aq OC 6 H (aq a c. C 6 H NH (aq H (aq C 6 H NH (aq a [H [OC6H [HOC H [H [C [C 6 H 6 6 H NH NH. 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

5 00 CHAPTER 1 ACIDS AND BASES 6. Conjugate Conjugate Acid Base Base of Acid Acid of Base a. Al(H O 6 H O Al(H O (OH H O b. HONH H O HONH H O c. HOCl C 6 H NH OCl - C 6 H NH 7. 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 c. H SO is a strong acid. d. H SO is a weak diprotic acid because the a1 and a values are less than 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 9. 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 1. NH : a ; H O: a w 1 From the a values, the ordering is HClO > HClO > NH > H O. 0. Ecept for water, these are the conjugate bases of the acids in the previous eercise. In general, the weaker the acid, the stronger is 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 1.0. much stronger acid than H O. HCl is a 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.

6 CHAPTER 1 ACIDS AND BASES 01 c. HOC 6 H, a ; HCN, a ; HCN is a slightly stronger acid than HOC 6 H because a for HCN > a for HOC 6 H.. a. H O; the conjugate bases of strong acids are terrible bases ( b < 1 b. NO ; the conjugate bases of weak acids are weak bases ( < b < 1. c. OC 6 H ; for a conjugate acid-base pair, a b w. From this relationship, the stronger the acid, the weaker is the conjugate base ( b decreases as a increases. Because HCN is a stronger acid than HOC 6 H ( a for HCN > a for HOC 6 H, OC 6 H will be a stronger base than CN. Autoionization of Water and the ph Scale. At C, the relationship [H [OH w always holds for aqueous solutions. When [H 7 is greater than M, the solution is acidic; when [H is less than M, the solution is basic; when [H 7 M, the solution is neutral. In terms of [OH, an acidic solution has [OH 7 < M, a basic solution has [OH > M, and a neutral solution has [OH 7 M. 1 a. [OH [H w 1-7 M; the solution is neutral. 7 b. [OH c. [OH d. [OH M; the solution is basic M; the solution is acidic M; the solution is acidic.. a. [H [OH w M; basic b. [H c. [H d. [H M; acidic M; neutral M; basic

7 0 CHAPTER 1 ACIDS AND BASES. 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.7 In pure water [H [OH, so.7 6. a. H O(l H (aq OH (aq w.9 In pure water: [H [OH,.9 b. ph!log[h!log(1.71 c. [H w /[OH ( [H [OH at 0. C 1 10 [H, [H M [OH 1 10 [H [OH 1 10 [H, [H M [OH / M; ph!log(.9 7. ph!log[h ; poh!log[oh ; at C, ph poh 1.00; for Eercise : a. ph!log[h!log( ; poh 1.00! ph 1.00! b. ph!log( ; poh 1.00! 1.08!1.08 c. ph!log(1!1.08; poh 1.00! (! d. ph!log(. 10.7; poh 1.00! Note that ph is less than zero when [H is greater than 1.0 M (an etremely acidic solution. For Eercise : a. poh!log[oh!log(1.!0.18; ph 1.00! poh 1.00! (! b. poh!log( ; ph 1.00! 1.!0. c. poh!log( ; ph 1.00! d. poh!log( ; ph 1.00! Note that ph is greater than 1.00 when [OH is greater than 1.0 M (an etremely basic solution. 8. a. [H 10 ph, [H M poh 1.00 ph ; [OH 10 poh M or [OH 1 w M; this solution is basic since ph > [H.0 10

8 CHAPTER 1 ACIDS AND BASES 0 b. [H M; poh ; [OH 10 (1. 0 M; basic c. [H 10 ( M; poh 1.0 ( ; [OH M; acidic d. [H M; poh ; [OH M; acidic e. [OH M; ph 1.0 poh ; [H M; basic f. [OH M; ph ; [H M; acidic 9. a. poh 1.00! ; [H [OH b. [H M; acidic 1 1 poh ; acidic c. ph ; [H [OH M 0.1 M; ph!log( M; basic M d. ph!log ( ; poh [OH M; neutral 0. a. poh ; [H [OH M; basic M b. [H M; ph!log( poh ; basic c. ph!log( ; poh [OH M; acidic d. ph ; [H [OH M; acidic M

9 0 CHAPTER 1 ACIDS AND BASES 1. poh 1.0! ph 1.0! ; [H ph M (1 sig. fig. [OH [H w M or [ OH poh M The sample of gastric juice is acidic because the ph is less than 7.00 at C.. ph 1.00! poh 1.00!.7 8.6; [H ph M [OH [H w M or [OH poh M The solution of baking soda is basic because the ph is greater than 7.00 at C. Solutions of Acids. 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( b. Major species H, NO and H O; ph Both are strong acids, which are assumed to completely dissociate in water L 0.00 mol/l L 0.10 mol/l mol HBr. 10 mol HI mol H. 10 mol H mol Br 10 mol I [H ( L mol M; [OH w M [H [Br. 10 mol L 0.01 M; [I mol L 0.07 M. Strong acids are assumed to completely dissociate in water; for eample; HCl(aq H O(l H O (aq Cl (aq or HCl(aq H (aq Cl (aq. a. A 0.10 M HCl solution gives 0.10 M H and 0.10 M Cl because HCl completely dissociates. The amount of H from H O will be insignificant. ph log[h log(

10 CHAPTER 1 ACIDS AND BASES 0 b..0 M H is produced when.0 M HClO completely dissociates. The amount of H from H O will be insignificant. ph log( (Negative ph values just indicate very concentrated acid solutions. c. 11 M H is produced when 11 M HI completely dissociates. If you take the negative log of 11, 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 7 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 in the case here. Therefore, the ph is that of neutral water (ph 7.00 because the amount of HI present is insignificant. 6. HNO (aq H (aq NO (aq; HNO is a strong acid, which means it is assumed to completely dissociate in water. The initial concentration of HNO will equal the [H donated by the strong acid. a. ph log[h log( b. ph log( c. Because the concentration of HNO is so dilute, the ph will be that of neutral water (ph In this problem, water is the major H producer present. Whenever the strong acid has a concentration less than 7 M, the [H contribution from water must be considered. 7. [H 10 ph M. Because HI is a strong acid, a. 10 M HI solution will produce. 10 M H, giving a ph [H 10 ph M. Because HBr is a strong acid, a.6 10 M HBr solution is necessary to produce a ph. solution. 9. HCl is a strong acid. [H M (carrying one etra sig. fig. M 1 V 1 M V, V 1 M V mol/l 1.6 L M 1 mol/l L To. ml of 1 M HCl, add enough water to make 1600 ml of solution. The resulting solution will have [H. 10 M and ph ml conc. HCl soln 1.19 g ml 8 g HCl 1 mol HCl 0.6 mol HCl 100 g conc. HCl soln 6. g 0.0 ml conc. HNO soln 1. g ml 70. g HNO 1 mol HNO 0. mol HNO 100 g soln 6.0 g HNO HCl(aq H (aq Cl (aq and HNO (aq H (aq NO (aq (Both are strong acids.

11 06 CHAPTER 1 ACIDS AND BASES So we will have mol of H in the final solution. [H 0.9 mol 0.9 M; ph log[h log( L [OH [H w M a. HNO ( a.0 10 and H O ( a w 1 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. 0.0! a [H [NO [HNO ; if we assume << 0.0, then:.0 10,.0 10 ( We must check the assumption: M 100.0% All the assumptions are good. The H 7 contribution from water (1 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( b. CH CO H ( a and H O ( a w 1.0 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. 0.0!

12 CHAPTER 1 ACIDS AND BASES 07 a [H [CHCO, 1.8 [CH CO H (assuming << M; checking assumption: [H.1 10 M; ph log( %. Assumptions good a. HOC 6 H ( a and H O ( a w are the major species. The major equilibrium is the dissociation of HOC 6 H. Solving the weak acid problem: HOC 6 H H OC 6 H Initial 0.0 M ~0 0 mol/l HOC 6 H dissociates to reach equilibrium Change! Equil. 0.0! a [H [OC6H [HOC H (assuming << 0.0 [H M; checking assumption: is. 10 % of 0.0, so assumption is valid by the % rule. ph!log( b. HCN ( a and H O are the major species. HCN is the major source of H. HCN H CN Initial 0.0 M ~0 0 mol/l HCN dissociates to reach equilibrium Change! Equil. 0.0! a [H [CN [HCN (assuming << 0.0 [H M; checking assumption: is % of 0.0. Assumptions good. ph!log( 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. 0.00

13 08 CHAPTER 1 ACIDS AND BASES a [H [F [HF (assuming << 0.00 [H.8 10 M; check assumptions: H % The assumption << 0.00 is not good ( is more than % of We must solve /( eactly by using either the quadratic formula or the method of successive approimations (see Appendi 1 of the tet. Using successive approi-mations, we let M be a new approimation for [HF. That is, in the denominator try (the value of we calculated making the normal assumption so that ; then solve for a new value of in the numerator ,. 10 We use this new value of to further refine our estimate of [HF, that is, (carrying an etra sig. fig ,. 10 We repeat until we get a self-consistent answer. This would be the same answer we would get solving eactly using the quadratic equation. In this case it is,. 10. Thus: [H [F. 10 M; [OH w /[H M [HF M; ph.6 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 percent error is less than ~% (unless you have a graphing calculator. 6. HClO H ClO a Initial 0. M ~0 0 mol/l HClO dissociates to reach equilibrium Change Equil. 0. a [H [ClO [HClO 0.,

14 CHAPTER 1 ACIDS AND BASES 09 The assumption that is small is not good ( is % of 0.. Using the method of successive approimations and carrying etra significant figures: , ,.6 10 (consistent answer [H [ClO.6 10 M; percent dissociation % 6. HC H O ( a and H O ( a w are the major species present. HC H O will be the dominant producer of H because 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( Assumption follows the % rule ( is 1.1% of [H [C H O M; [OH w /[H M [HC H O 0.100! M Percent dissociation [H [HC HO % 66. This is a weak acid in water. We must solve a weak acid problem. Let HBz C 6 H CO H. 0.6 g HBz 1 mol HBz 1.1 g.6 10 mol; [HBz M HBz H Bz - Initial.6 10 M ~0 0 mol/l HBz dissociates to reach equilibrium Change Equil..6 10

15 10 CHAPTER 1 ACIDS AND BASES a [H [Bz [HBz ( [H ; check assumptions: % Assumption is not good ( is 1% of When assumption(s fail, we must solve eactly using the quadratic formula or the method of successive approimations (see Appendi 1 of tet. Using successive approimations: (.6 10 (.6 10 (. 10 ( , ,.1 10 M (consistent answer Thus: [H [Bz [C 6 H CO.1 10 M [HBz [C 6 H CO H M ph log( ; poh 1.00 ph 10.71; [OH M 67. Major species: HC H ClO ( a 1. HC H ClO H C H ClO Initial 0.10 M ~0 0 mol/l HC H ClO dissociates to reach equilibrium Change! Equil. 0.10! 10 and H O; major source of H : HC H ClO 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( This is a weak acid in water, so we solve the weak acid problem. HCO H H HCO Initial 0.0 M ~0 0 mol/l HCO H dissociates to reach equilibrium Change Equil. 0.0 a

16 CHAPTER 1 ACIDS AND BASES 11 a [H [HCO [HCO H [H.1 10 M; check assumptions: % The assumption that << 0.0 is not good (fails the % rule. Solving using the method of successive approimations (see Appendi 1 in tet: [H.0 10 M; ph ,.0 10, which we get consistently. 69. HF and HOC 6 H are both weak acids with a values of and , respectively. Since the a value for HF is much greater than the a value for HOC 6 H, HF will be the dominant producer of H (we can ignore the amount of H produced from HOC 6 H because it will be insignificant. HF H F Initial 1.0 M ~0 0 mol/l HF dissociates to reach equilibrium Change! Equil. 1.0! 10 a [H [F [HF [H.7 10 M; ph!log( ; assumptions good. Solving for [OC 6 H using HOC 6 H H OC 6 H - equilibrium: a [H [OC6H [HOC H 6 (.7 10 [OC6H 1.0, [OC 6 H M 9 Note that this answer indicates that only.9 10 M HOC 6 H dissociates, which confirms that HF is truly the only significant producer of H in this solution. 70. a. The initial concentrations are halved since equal volumes of the two solutions are mied. HC H O H C H O Initial M M 0 Equil a ( (

17 1 CHAPTER 1 ACIDS AND BASES.6 10 ; assumption is horrible. Using the quadratic formula: ( M; [H M; ph.80 b. [C H O M 71. 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.0 10 M; assumptions good. Percent dissociation [H [HC HO 0 H H % b. The setup for solutions b and c are similar to solution a ecept that the final equation is different because the new concentration of HC H O is different. a [H [C H O M; assumptions good. Percent dissociation % c. a [H [C H O.0 10 M; check assumptions. Assumption that is negligible is borderline (6.0% error. We should solve eactly. Using the method of successive approimations (see Appendi 1 of the tet: , ( Net trial also gives.9 10.

18 CHAPTER 1 ACIDS AND BASES Percent dissociation % d. As we dilute a solution, all concentrations are decreased. 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 toward 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. 7. a. HNO is a strong acid; it is assumed 100% dissociated in solution. b. HNO H NO a.0 10 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. Percent dissociation [H [ HNO % c. HOC 6 H H OC 6 H a Initial 0.0 M ~0 0 mol/l HOC 6 H dissociates to reach equilibrium Change! Equil. 0.0! a [H [OC6H [HOC H

19 1 CHAPTER 1 ACIDS AND BASES [H [OC 6 H M; assumptions good. Percent dissociation % d. For the same initial concentration, the percent dissociation increases as the strength of the acid increases (as a increases. 7. Let HA symbolize the weak acid. Set up the problem like a typical weak acid equilibrium problem. HA H A Initial 0.1 M ~0 0 mol/l HA 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. a [H [A [HA 0.1 ( ( HX H X Initial I ~0 0 where I [HX 0 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 ( Set up 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 !

20 CHAPTER 1 ACIDS AND BASES 1 a [H [OCN [HOCN ; ph.77: [H ph M a ( ( Set up the problem using the a equilibrium reaction for HOBr. HOBr H OBr Initial 0.06 M ~0 0 mol/l HOBr dissociates to reach equilibrium Change Equil a [H [OBr [HOBr ; from ph.9: [H 10 ph M 0.06 a ( Major species: HCOOH and H O; major source of H : HCOOH HCOOH H HCOO Initial C ~0 0 where C [HCOOH 0 mol/l HCOOH dissociates to reach equilibrium Change Equil. C a [H [HCOO [HCOOH , where [H C [H ; because ph.70: [H M C [H (.0 10 C ( , C ( , C. 10 M A 0.0 M formic acid solution will have ph Major species: HC H O (acetic acid and H O; major source of H : HC H O HC H O H C H O Initial C ~0 0 where C [HC H O 0 mol/l HC H O dissociates to reach equilibrium Change Equil. C

21 16 CHAPTER 1 ACIDS AND BASES a [H [C HO, where [H [HC H O C [H C [H ; from ph.0: [H M (1 10 C (1 10, C ( , C M [HA 0 A 6 10 M acetic acid solution will have ph mol.0 L 0.0 mol/l; solve using the a equilibrium reaction. 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 80. Let HSac saccharin and I [HSac 0. ( HSac H Sac a Initial I ~0 0 Equil. I a I ( I ( ; [H , I 1.6 M [HSac M 1 mol 1 L 1000 ml g HC 7 H NSO 0 ml g 1.6 mol L Solutions of Bases All b reactions refer to the base reacting with water to produce the conjugate acid of the base and OH.

22 CHAPTER 1 ACIDS AND BASES 17 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 8. a. C 6 H NH (aq H O(l C 6 H NH (aq OH (aq b b. (CH NH(aq H O(l (CH NH (aq OH (aq b [NH [OH [NH [CHNH [OH [C H N [C6HNH [OH [C H NH 6 [(CH NH [OH [(CH NH 8. NO : Because HNO is a strong acid, NO is a terrible base ( b << w. All conjugate bases of strong acids have no base strength. H O: b w 1 ; NH : b ; C H N: b Base strength NH > C H N > H O > NO - (As b increases, base strength increases. 8. Ecluding water, these are the conjugate acids of the bases in the preceding eercise. In general, the stronger the base, the weaker is 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. Acid strength HNO > C H NH > NH > H O 8. a. C 6 H NH b. C 6 H NH c. OH d. CH NH The base with the largest b value is the strongest base ( b, C6HNH OH is the strongest base possible in water. b, CH NH 10 10, 86. a. HClO (a strong acid b. C 6 H NH c. C 6 H NH The acid with the largest a value is the strongest acid. To calculate a values for C 6 H NH and CH NH, use a w / b, where b refers to the bases C 6 H NH or CH NH. 87. NaOH(aq Na (aq OH (aq; NaOH is a strong base that 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( ph 1.00! poh 1.00! 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.

23 18 CHAPTER 1 ACIDS AND BASES b. The [OH concentration donated by the NaOH is M. Water by itself donates 7 M. In this eercise, water is the major OH contributor, and [OH M. poh log( ; ph c. [OH.0 M; poh log(.0 0.0; ph 1.00! (! a. Ca(OH Ca OH ; Ca(OH is a strong base and dissociates completely. [OH ( ph 1.00 poh M; poh!log[oh b. g OH L 1 mol OH 0. mol OH/L 6.11g OH OH is a strong base, so [OH 0. M; poh log (0. 0.; ph 1.6 c g NaOH L 1 mol.70 M; NaOH is a strong base, so [OH.70 M g poh log( and ph ( Although we are justified in calculating the answer to four decimal places, in reality, the ph can only be measured to ±0.01 ph units. 89. a. Major species:, OH, H O (OH is a strong base. [OH 0.01 M, poh log( ; ph 1.00 poh 1.18 b. Major species: Ba, OH, H O; Ba(OH (aq Ba (aq OH (aq; because each mole of the strong base Ba(OH dissolves in water to produce two mol OH, [OH (0.01 M 0.00 M. poh!log( ; 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 ( M; poh!log( ; ph poh ; [OH [OH M L.6 10 mol OH L 6.11g OH mol OH 0.16 g OH

24 CHAPTER 1 ACIDS AND BASES ph 10.0; poh ; [OH M Sr(OH (aq Sr (aq OH (aq; Sr(OH donates mol OH per mol Sr(OH. [Sr(OH. 10 mol OH L 1 mol Sr(OH 1.6 mol OH 10 M Sr(OH A M Sr(OH solution will produce a ph 10.0 solution. 9. NH is a weak base with b The major species present will be NH and H O 1 ( b w. Because NH has a much larger b value than 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 using the b reaction for NH. 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. 0.10! b [NH [OH [NH (assuming << 0.10 [OH M; check assumptions: is 1.1% of 0.10, so the assumption 0.10! 0.10 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( M.80; ph 1.00 poh Major species: H NNH ( b 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..0! OH b b [HNNH [OH [H NNH.0.0 (assuming <<.0 [OH. [H NNH. 10 M; poh.6; ph 11.8; assumptions good ( is 0.1% of M; [H NNH.0 M; [H M

25 0 CHAPTER 1 ACIDS AND BASES 9. These are solutions of weak bases in water. In each case we must solve an 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. 0.0! 10 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 b. HONH H O HONH OH b Initial 0.0 M 0 ~0 Equil. 0.0! b , [OH.7 10 M; assumptions good. [H M; ph These are solutions of weak bases in water. a. C 6 H NH H O C 6 H NH OH b Initial 0.0 M 0 ~0 mol/l of C 6 H NH reacts with H O to reach equilibrium Change! Equil. 0.0! , [OH M; assumptions good. [H w /[OH M; ph 9.08 b. CH NH H O CH NH OH b.8 Initial 0.0 M 0 ~0 Equil

26 CHAPTER 1 ACIDS AND BASES 1 b , M; assumptions good. [OH 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.6 Initial 0.0 M 0 ~0 mol/l C H NH reacts with H O to reach equilibrium Change! Equil. 0.0! [CHNH [OH b [C H NH ; checking assumption: (assuming << % The assumption fails the % rule. We must solve eactly using either the quadratic equation or the method of successive approimations (see Appendi 1 of the tet. Using successive approimations and carrying etra significant figures: , M (consistent answer [OH M; [H w [OH M; ph (C H NH H O (C H NH OH b 1. Initial 0.00 M 0 ~0 mol/l (C H NH reacts with H O to reach equilibrium Change! Equil. 0.00! 10 b [(CH NH [OH [(C H NH ; assumption is bad ( is 16% of 0.0. Using successive approimations: 1. 10, , [OH (consistent answer 10 M; [H w /[OH M; ph 11.8

27 CHAPTER 1 ACIDS AND BASES 99. To solve for percent ionization, we first solve the weak base equilibrium problem. a. NH H O NH OH b Initial 0.10 M 0 ~0 Equil b , [OH M; assumptions good M Percent ionization % [NH 0.10 M 0 b. NH H O NH OH Initial M 0 ~0 Equil , [OH. 10 M; assumptions good.. 10 Percent ionization 100.% Note: For the same base, the percent ionization increases as the initial concentration of base decreases. c. CH NH H O CH NH Initial 0.10 M 0 ~0 Equil OH b , ; assumption fails the % rule ( is % of Using successive approimations and carrying etra significant figures: , (consistent answer Percent ionization % 100. C H N H O C H N OH b 1.7 Initial 0.10 M 0 ~0 Equil. 0.10! 10 9

28 CHAPTER 1 ACIDS AND BASES b , [C H N M; assumptions good M Percent C H N reacted 0.10 M % 101. Using the b reaction to solve where PT p-toluidine (CH C 6 H NH : PT H O PTH OH Initial M 0 ~0 mol/l of PT reacts with H O to reach equilibrium Change Equil b [PTH [OH [PT Because ph 8.60: poh and [OH M b ( ( HONH H O HONH OH b Initial I 0 ~0 I [HONH 0 Equil. I b I From problem, ph 10.00, so poh.00 and [OH ( I (, I 0.91 M 10 M mol HONH.0 g HONH Mass HONH 0.00 L 7. g HONH L mol HONH Polyprotic Acids 10. H SO (aq HSO (aq H (aq a 1 [HSO [H [H SO HSO (aq SO (aq H (aq a [SO [H [HSO

29 CHAPTER 1 ACIDS AND BASES 10. H C 6 H O 7 (aq H C 6 H O 7 (aq H (aq a 1 [HC6HO7 [H [H C H O 6 7 H C 6 H O 7 (aq HC 6 H O 7 (aq H (aq a [HC H O 6 [H C H O 6 7 [H 7 HC 6 H O 7 (aq C 6 H O 7 (aq H (aq [C6HO7 [H [HC H O a For H PO, a 7. 10, a 6. 10, and a Because 1 a 1 is much larger than a and a, the dominant H producer is H PO, and the H contributed from H PO and HPO can be ignored Solving the weak acid problem in the typical manner. H PO H PO H Initial M 0 ~0 Equil a [HPO [H [H PO ; assumption is horrible because is 100% of We will use the quadratic equation to solve eactly , 10 (7. 10, ( ± [(7. 10 (1( 10 [H 10 M (1 ph log( The reactions are: 1/ H AsO H H AsO a 1 10 H AsO H HAsO a HAsO H AsO a We will deal with the reactions in order of importance, beginning with the largest a, a 1.

30 CHAPTER 1 ACIDS AND BASES H AsO H H AsO Initial 0.0 M ~0 0 Equil a 1 10 [H [HAsO [H AsO , 10 M; assumption fails the % rule. 0.0 Solving by the method of successive approimations: 10 /( , 10 (consistent answer [H [H AsO 10 M; [H AsO M [H [HAsO Because a is much smaller than the a [HAsO 1 value, very little of H AsO (and HAsO dissociates compared to H AsO. Therefore, [H and [H AsO will not change significantly by the a reaction. Using the previously calculated concentrations of H and H AsO to calculate the concentration of HAsO : ( 10 [HAsO 10, [HAsO M The assumption that the repeat the process using a reaction does not change [H and [H AsO is good. We to get [AsO. a a [H [AsO [HAsO ( [AsO [AsO M; assumption good. So in 0.0 M analytical concentration of H AsO : 107. Because [H AsO 0.17 M; [H [H AsO 10 M [HAsO M; [AsO 10 1 M; [OH w /[H 10 1 M a for H S is so small, we can ignore the H contribution from the H S H HS a 7 1 Initial 0.10 M ~0 0 Equil. 0.10! a reaction.

31 6 CHAPTER 1 ACIDS AND BASES a , [H ; assumptions good. ph!log(.00 Use the a reaction to determine [S. HS H S Initial M M 0 Equil.! a 1.0 [S The relevant reactions are: ( ( ( M; assumptions good. H CO H HCO a ; HCO H CO a Initially, we deal only with the first reaction (since a 1 >> control values of concentrations in the second reaction. H CO H HCO Initial M ~0 0 Equil a, and then let those results a [H [HCO [H CO M [H [HCO ; assumptions good. HCO H CO Initial M M 0 Equil y y y If y is small, then [H [HCO, and a [H [CO [HCO y. y [CO M; assumptions good. The amount of H from the second dissociation is M or:

32 CHAPTER 1 ACIDS AND BASES % This result justifies our treating the equilibria separately. If the second dissociation contributed a significant amount of H, we would have to treat both equilibria simultaneously. The reaction that occurs when acid is added to a solution of HCO is: HCO (aq H (aq H CO (aq H O(l CO (g The bubbles are CO (g and are formed by the breakdown of unstable H CO molecules. We should write H O(l CO (aq or CO (aq for what we call carbonic acid. It is for convenience, however, that we write H CO (aq The dominant H producer is the strong acid H SO. A.0 M H SO solution produces.0 M HSO - and.0 M H. However, HSO is a weak acid that could also add H to the solution. HSO H SO Initial.0 M.0 M 0 mol/l HSO dissociates to reach equilibrium Change Equil..0.0 a [H [SO [HSO (.0.0(.0.0, M Because is 0.60% of.0, the assumption is valid by the % rule. The amount of additional H from HSO is M. The total amount of H present is: [H.0 ( M; ph log( Note: In this problem, H from HSO could have been ignored. However, this is not usually the case in more dilute solutions of H SO For H SO, the first dissociation occurs to completion. The hydrogen sulfate ion (HSO is a weak acid with a We will consider this equilibrium for additional H production: HSO H SO Initial M M 0 mol/l HSO dissociates to reach equilibrium Change Equil a 0.01 ( , 0.01; assumption is horrible (0% error.

33 8 CHAPTER 1 ACIDS AND BASES Using the quadratic formula: (0.01 (0.000, ( ± ( / ± 0.0,.0 10 M [H M; ph.10 Note: We had to consider both H SO and HSO for H production in this problem. Acid-Base Properties of Salts 111. One difficult aspect of acid-base chemistry is recognizing what types of species are present in solution, that is, whether a species is a strong acid, strong base, weak acid, weak base, or a neutral species. Below are some ideas and generalizations to keep in mind that will help in recognizing types of species present. a. Memorize the following strong acids: HCl, HBr, HI, HNO, HClO, and H SO b. Memorize the following strong bases: LiOH, NaOH, OH, RbOH, CsOH, Ca(OH, Sr(OH, and Ba(OH c. Weak acids have a a value of less than 1 but greater than w. Some weak acids are listed in Table 1. of the tet. Weak bases have a b value of less than 1 but greater than w. Some weak bases are listed in Table 1. of the tet. d. Conjugate bases of weak acids are weak bases; that is, all have a b value of less than 1 but greater than w. Some eamples of these are the conjugate bases of the weak acids listed in Table 1. of the tet. e. Conjugate acids of weak bases are weak acids; that is, all have a a value of less than 1 but greater than w. Some eamples of these are the conjugate acids of the weak bases listed in Table 1. of the tet. f. Alkali metal ions (Li, Na,, Rb, Cs and heavier alkaline earth metal ions (Ca, Sr, Ba have no acidic or basic properties in water. g. All conjugate bases of strong acids (Cl, Br -, I, NO, ClO, HSO have no basic properties in water ( b << w, and only HSO - has any acidic properties in water. Let s apply these ideas to this problem to see what type of species are present. The letters in parenthesis is(are the generalization(s above that identifies the species. OH: Strong base (b NO : Neutral; and NO have no acidic/basic properties (f and g. CN: CN is a weak base, b w / a, HCN 1.0 Ignore (f /6. NH Cl: NH is a weak acid, a (c and e. Ignore Cl (g. HCl: Strong acid (a (c and d.

34 CHAPTER 1 ACIDS AND BASES 9 The most acidic solution will be the strong acid solution, with the weak acid solution less acidic. The most basic solution will be the strong base solution, with the weak base solution less basic. The NO solution will be neutral at ph Most acidic most basic: HCl > NH Cl > NO > CN > OH 11. See Eercise 111 for some generalizations on acid-base properties of salts. The letters in parenthesis is(are the generalization(s listed in Eercise 111 that identifies the species. CaBr : Neutral; Ca and Br have no acidic/basic properties (f and g. NO : NO is a weak base, b w / a, HNO 1 10 /( (c and d. Ignore (f. HClO : Strong acid (a HNO : Weak acid, a.0 10 (c HONH ClO : HONH is a weak acid, a w / b, HONH /( (c and e. Ignore ClO (g. Note that HNO has a larger a value than HONH, so HNO is a stronger weak acid than HONH. Using the information above (identity and the a or b values, the ordering is: Most acidic most basic: HClO > HNO > HONH ClO > CaBr > NO 11. From the a values, acetic acid is a stronger acid than hypochlorous acid. Conversely, the conjugate base of acetic acid, C H O, will be a weaker base than the conjugate base of hypochlorous acid, OCl. Thus the hypochlorite ion, OCl, is a stronger base than the acetate ion, C H O. In general, the stronger the acid, the weaker the conjugate base. This statement comes from the relationship w a b, which holds for all conjugate acid-base pairs. 11. Because NH is a weaker base (smaller b value than CH NH, the conjugate acid of NH will be a stronger acid than the conjugate acid of CH NH. Thus NH is a stronger acid than CH NH. 11. a. Cl is a soluble ionic compound that dissolves in water to produce (aq and Cl (aq. (like the other alkali metal cations has no acidic or basic properties. Cl is the conjugate base of the strong acid HCl. Cl has no basic (or acidic properties. Therefore, a solution of Cl will be neutral because neither of the ions has any acidic or basic properties. The 1.0 M Cl solution has [H [OH 7 M and ph poh b. F is also a soluble ionic compound that dissolves in water to produce (aq and F (aq. The difference between the Cl solution and the F solution is that F does have basic properties in water, unlike Cl. F is the conjugate base of the weak acid HF, and as is true for all conjugate bases of weak acids, F is a weak base in water. We must solve an equilibrium problem in order to determine the amount of OH this weak base produces in water.

35 0 CHAPTER 1 ACIDS AND BASES F H O HF OH b Initial 1.0 M 0 ~0 mol/l of F reacts with H O to reach equilibrium Change! Equil. 1.0 w a, HF b [HF[OH, 1. [F [OH M ; assumptions good poh.; ph ; [H M 116. C H NH Cl C H NH Cl ; C H NH is the conjugate acid of the weak base C H NH ( b As is true for all conjugate acids of weak bases, C H NH is a weak acid. Cl has no basic (or acidic properties. Ignore Cl. Solving the weak acid problem: C H NH C H NH H a w /.6 Initial 0. M 0 ~0 mol/l C H NH dissociates to reach equilibrium Change! Equil. 0.! a [CHNH[H [C H NH (assuming << 0. [H.1 [C H NH [H.1 [OH w /[H M; ph.68; assumptions good M; [C H NH 0. M; [Cl - 0. M 9 10 M 117. a. CH NH Cl CH NH Cl : CH NH is a weak acid. Cl is the conjugate base of a strong acid. Cl has no basic (or acidic properties. CH NH CH NH H a [CH [CH NH [H NH 1 w b CH NH CH NH H Initial 0.10 M 0 ~0 mol/l CH NH dissociates to reach equilibrium Change! Equil. 0.10!

36 CHAPTER 1 ACIDS AND BASES 1 a (assuming << 0.10 [H M; ph.8; assumptions good. b. NaCN Na CN : CN is a weak base. Na has no acidic (or basic properties. CN H O HCN OH b 1 w 10 a Initial 0.00 M 0 ~0 b 1.6 mol/l CN reacts with H O to reach equilibrium Change! Equil. 0.00! 10 b [HCN[OH [CN [OH M; poh.0; ph 10.9; assumptions good a. NO NO : NO is a weak base. Ignore. NO H O HNO OH b 1 w a Initial 0.1 M 0 ~0 Equil. 0.1! b [OH [HNO [NO [OH M; poh.77; ph 8.; assumptions good. b. NaOCl Na OCl : OCl is a weak base. Ignore Na. OCl H O HOCl OH b 1 w 8 a Initial 0. M 0 ~0 Equil. 0.! b [HOCl[OH [OCl [OH.6 10 M; poh.; ph 10.6; assumptions good.

37 CHAPTER 1 ACIDS AND BASES c. NH ClO NH ClO : NH is a weak acid. ClO is the conjugate base of a strong acid. ClO has no basic (or acidic properties. NH NH H 1 w a Initial 0.0 M 0 ~0 Equil. 0.0! b a [NH[H [NH [H M; ph.8; assumptions good All these salts contain Na, which has no acidic/basic properties, and a conjugate base of a weak acid (ecept for NaCl, where Cl is a neutral species. All conjugate bases of weak acids are weak bases since b values for these species are between w and 1. To identify the species, we will use the data given to determine the b value for the weak conjugate base. From the b value and data in Table 1. of the tet, we can identify the conjugate base present by calculating the a value for the weak acid. We will use A - as an abbreviation for the weak conjugate base. A H O HA OH Initial mol/1.00 L 0 ~0 mol/l A reacts with H O to reach equilibrium Change Equil b [HA[OH [A ; from the problem, ph 8.07: b poh ; [OH M ( ( b value for the conjugate base of a weak acid. The a value for the weak acid equals w / b : a From Table 1. of the tet, this a value is closest to HF. Therefore, the unknown salt is NaF. 10. BHCl BH Cl ; Cl is the conjugate base of the strong acid HCl, so Cl has no acidic/ basic properties. BH is a weak acid because it is the conjugate acid of a weak base B. Determining the a value for BH : 1