16. ACIDS AND BASES. Solutions to Exercises

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1 16. ACIDS AND BASES Solutions to Exercises Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiplestep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off See labels below reaction: H 2 CO 3 (aq) CN (aq) HCN(aq) HCO 3 (aq) acid base acid base H 2 CO 3 is the proton donor (BronstedLowry acid) on the left, and HCN is the proton donor (BronstedLowry acid) on the right. The CN and HCO 3 ions are proton acceptors (BronstedLowry bases). HCN is the conjugate acid of CN Part (a) involves molecules with all single bonds; part (b) does not, so bonds are drawn in. a. B Lewis acid CH 3 :O: H Lewis base CH 3 B :O: H (continued)

2 564 CHAPTER 16 b. : O : 2 : O : C Lewis base : O : Lewis acid : O C 2 : O : : O : 16.3 The HC 2 H 3 O 2 is a stronger acid than H 2 S, and HS is a stronger base than the C 2 H 3 O 2 ion. The equilibrium favors the weaker acid and weaker base; therefore, the reactants are favored a. PH 3 b. HI c. H 2 SO 3 d. H 3 AsO 4 e. HSO A M solution of Ba(OH) 2, a strong base, ionizes completely to yield M Ba 2 ion and 2 x M, or M, OH ion. Use the K w equation to calculate the [H 3 O ]. [H 3 O ] K w [OH ] x 10 (0.250) 4.00 x x M 16.6 Use the K w equation to calculate the [H 3 O ]. [H 3 O ] K w [OH ] x x x x 10 9 M Since the [H 3 O ] concentration is less than 1.0 x 10 7, the solution is basic Calculate the negative log of the [H 3 O ]: ph log [H 3 O ] log (0.045) Calculate the poh of M OH, and then subtract from to find ph: poh log [OH ] log (0.025) ph

3 ACIDS AND BASES Because ph 3.16, by definition the log [H 3 O ] Enter this on the calculator and convert to the antilog (number) of [H 3 O ] antilog (3.16) x x 10 4 M Find the poh by subtracting the poh from Then enter 3.40 on the calculator to convert to the antilog (number) corresponding to poh [H 3 O ] antilog (3.40) x x 10 4 M Answers to Concept Checks 16.1 In any aqueous solution, you should consider the autoionization of water. And because we have a solution of a weak acid in water, you should also consider the equilibrium between this acid and water. Here are the two equilibria: H 2 O(l) H 2 O(l) HCHO 2 (aq) H 2 O(l) H 3 O (aq) OH (aq) CHO 2 (aq) H 3 O (aq) The species present in these equilibria are: H 2 O(l), H 3 O (aq), OH (aq), HCHO 2 (aq), and CHO 2 (aq) The stronger acid gives up its proton more readily, and, therefore, its conjugate base ion holds onto a proton less strongly. In other words, the stronger acid has the weaker conjugate base. Because formic acid is the stronger acid, the formate ion is the weaker base. Acetate ion is the stronger base Look at each solution, and determine whether it is acidic, basic, or neutral. In solution A, the numbers of H 3 O and OH ions are equal, so the solution is neutral. For solution B, the number of H 3 O ions is greater than the number of OH ions, so the solution is acidic. In solution C, the number of H 3 O ions is less than the number of OH ions, so the solution is basic. Therefore, ranking from most acidic to least acidic (most basic) is B > A > C.

4 566 CHAPTER In order to qualitatively answer this problem, it is essential that all the solutions have the same solute concentrations. Bases produce solutions of ph greater than seven while acids produce solutions of ph less than seven. NH 3 and NaOH are bases, and HCl, and HC 2 H 3 O 2 are acids. NaOH is a stronger base than NH 3, so the NaOH solution would have the highest ph, followed by the NH 3 solution. HC 2 H 3 O 2 is a much weaker acid than HCl so the HC 2 H 3 O 2 solution would have a higher ph than the HCl solution. Therefore, the ranking from highest to lowest ph for solutions with the same solute concentrations is: NaOH > NH 3 > HC 2 H 3 O 2 > HCl. Answers to Review Questions 16.1 You can classify these acids using the information in Section Also recall that all diatomic acids of Group VIIA halides are strong except for HF. a. Weak b. Weak c. Strong d. Strong e. Weak f. Weak 16.2 In Section 16.1, we are told that all neutralizations involving strong acids and bases evolve kj of heat per one mol H 3 O. Thus, the thermochemical evidence for the Arrhenius concept is based on the fact that when one mole of any strong acid (one mol H 3 O ) is neutralized by one mole of any strong base (one mol OH ), the heat of neutralization is always the same ( H kj/mol) A BronstedLowry acid is a molecule or ion that donates an H 3 O ion (proton donor) to a base in a protontransfer reaction. A BronstedLowry base is a molecule or ion that accepts an H 3 O ion (proton acceptor) from an acid in a protontransfer reaction. An example of an acidbase equation: HF(aq) NH 3 (aq) NH 4 (aq) F (aq) acid base acid base 16.4 The conjugate acid of a base is a species that differs from the base by only one H 3 O. Consider the base, HSO 3. Its conjugate acid would be H 2 SO 3 but not H 2 SO 4. H 2 SO 4 differs from HSO 3 by one H and one O You can write the equations by considering that H 2 PO 3 is both a BronstedLowry acid and a BronstedLowry base. The H 2 PO 3 acts as a BronstedLowry acid when it reacts with a base such as OH : H 2 PO 3 (aq) OH (aq) HPO 3 2 (aq) H 2 O(l) The H 2 PO 3 acts as a BronstedLowry base when it reacts with an acid such as HF: H 2 PO 3 (aq) HCl(aq) H 3 PO 3 (aq) Cl (aq)

5 ACIDS AND BASES The BronstedLowry concept enlarges on the Arrhenius concept in the following ways: (1) It expands the concept of a base to include any species that accepts protons, not just the OH ion or compounds containing the OH ion. (2) It enlarges the concepts of acids and bases to include ions as well as molecules. (3) It enables us to write acidbase reactions in nonaqueous solutions as well as in aqueous solutions, whereas the Arrhenius concept applies only to aqueous solutions. (4) It allows some species to be considered as acids or bases depending on the other reactant with which they are mixed According to the Lewis concept, an acid is an electronpair acceptor and a base is an electronpair donor. An example is Ag (aq) 2(:NH 3 ) Ag(NH 3 ) 2 (aq) acid base 16.8 Recall that, the weaker the acid, the stronger it holds on to its proton(s). Thus, if a reaction mixture consists of a stronger acid and base and a weaker acid and base, the weaker acid side will always be favored because the proton(s) will bond more strongly to the weaker acid The two factors that determine the strength of an acid are (1) the polarity of the bond to which the H atom is attached, and (2) the strength of the bond, or how tightly the proton is held by the atom to which it is bonded. An increase in the polarity of the bond makes it easier to remove the proton, increasing the strength of the acid. An increase in the strength of the bond makes it more difficult to remove the proton, decreasing the strength of the acid. The strength of the bond depends in turn on the size of the atom, so larger atoms have weaker bonds, whereas smaller atoms have stronger bonds The selfionization of water is the reaction of two water molecules in which a proton is transferred from one molecule to the other to form H 3 O and OH ions. At 25 C, the K w expression is K w [H 3 O ][OH ] 1.0 x The ph log [H 3 O ] of an aqueous solution. Measure ph by using electrodes and a ph meter, or by interpolating the ph from the color changes of a series of acidbase indicators A solution of ph 4 has a [H 3 O ] 1 x 10 4 M and is more acidic than a solution of ph 5, which has a [H 3 O ] 1 x 10 5 M For a neutral solution, [H 3 O ] [OH ]; thus, the [H 3 O ] of a neutral solution at 37 C is the square root of K w at 37 C: [H 3 O ] x x 10 7 M ph log (1.58 x 10 7 )

6 568 CHAPTER Because ph poh pk w at any temperature, ph poh log (2.5 x ) Answers to Conceptual Problems It is not necessary to have the species NH 4 OH in order to have OH in the solution. When ammonia reacts with water, hydroxide ion forms in the reaction. NH 3 (aq) H 2 O(l) NH 4 (aq) OH (aq) A reaction where HPO 4 2 acts as an acid is HPO 4 2 (aq) OH (aq) PO 4 3 (aq) H 2 O(l) A reaction where HPO 4 2 acts as a base is HPO 4 2 (aq) H 3 O (aq) H 2 PO 4 (aq) H 2 O(l) The hydroxide ion acts as a base and donates a pair of electrons on the O atom, forming a bond with CO 2 to give HCO Nitrogen has a greater electronegativity than carbon. You would expect the H O bond in the H O N group to be more polar (with the H atom having a positive partial charge) than the H O bond in the H O C group. Thus, based on their structure, you would expect HNO 2 to be the stronger acid When you lower the temperature of pure water, the value of K w decreases. In pure water, the hydronium ion concentration equals the hydroxide ion concentration, so K w [H 3 O ] 2. When K w decreases, the hydronium ion concentration decreases, and the corresponding ph increases Sodium hydroxide is a strong base, whereas ammonia is weak. As a strong base, NaOH exists in solution completely as ions, whereas NH 3 exists in solution as an equilibrium in which only part of the NH 3 has reacted to produce ions. Thus, a sodium hydroxide solution has a greater OH concentration than the same concentration solution of NH 3. At the same concentrations, the ph of the NaOH solution is greater (more basic) than that of the NH 3 solution.

7 ACIDS AND BASES A strong monoprotic acid will dissolve completely in water to form hydronium ions (H 3 O ) and anions in equal number. None of the original monoprotic acid molecules will remain. This is represented by the picture in the middle. The picture on the left represents an undissociated acid. No ions are present in the solution. The picture on the right represents a weak acid where only some of the acid molecules have dissociated to form ions A weak acid will dissolve to form a solution that has some hydronium ions present, some anions present, and some undissociated acid molecules present. This is represented by beaker A. The solution in beaker B is a weak base, and the solution in beaker C is a strong acid. Solutions to Practice Problems Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiplestep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off The reaction with labels of acid or base written below is as follows: OH (aq) HF(l) F (aq) H 2 O(l) base acid base acid The reaction with labels of acid or base written below is as follows: HOCl(aq) H 2 O(l) H 3 O (aq) OCl (aq) acid base acid base a. PO 4 3 b. HS c. NO 2 d. HAsO a. HSO 4 b. PH 3 c. Se 2 d. OBr a. HClO b. AsH 4 c. H 3 PO 4 d. HTeO a. H 2 S b. NH 3 c. HBrO 2 d. N 2 H 5

8 570 CHAPTER Each equation is given below with the labels for acid or base: a. HSO 4 (aq) NH 3 (aq) SO 2 4 (aq) NH 4 (aq) acid base base acid The conjugate acidbase pairs are: HSO 4, SO 2 4, and NH 4, NH 3. b. HPO 4 2 (aq) NH 4 (aq) H 2 PO 4 (aq) NH 3 (aq) base acid acid base The conjugate acidbase pairs are: H 2 PO 4, HPO 4 2, and NH 4, NH 3. c. Al(H 2 O) 6 3 (aq) H 2 O(l) Al(H 2 O) 5 (OH) 2 (aq) H 3 O (aq) acid base base acid The conjugate acidbase pairs are: Al(H 2 O) 6 3, Al(H 2 O) 5 (OH) 2, and H 3 O, H 2 O. d. SO 3 2 (aq) NH 4 (aq) HSO 3 (aq) NH 3 (aq) base acid acid base The conjugate acidbase pairs are: HSO 3, SO 3 2, and NH 4, NH Each equation is given below with the labels for acid or base: a. H 2 PO 4 (aq) HCO 3 (aq) HPO 2 4 (aq) H 2 CO 3 (aq) acid base base acid The conjugate acidbase pairs are: H 2 PO 4, HPO 2 4, and H 2 CO 3, HCO 3. b. F (aq) HSO 4 (aq) HF(aq) SO 4 2 (aq) base acid acid base The conjugate acidbase pairs are: HF, F, and HSO 4, SO 4 2. c. HSO 4 (aq) H 2 O(l) SO 4 2 (aq) H 3 O (aq) acid base base acid The conjugate acidbase pairs are: HSO 4, SO 4 2, and H 3 O, H 2 O.

9 ACIDS AND BASES 571 d. H 2 S(aq) CN (aq) HS (aq) HCN(aq) acid base base acid The conjugate acidbase pairs are: H 2 S, HS, and HCN, CN The reaction is As : B Lewis base Lewis acid : F : As : B The reaction is :Cl: :Cl: Al :Cl: Lewis acid H : N : H H Lewis base :Cl: H :Cl: Al : N : H :Cl: H a. The completed equation is AlCl 3 Cl AlCl 4 The Lewis formula representation is : Cl Al : Cl : Cl : : Cl : : Cl : : Cl Al : Cl : Cl : AlCl 3 is the electronpair acceptor and is the acid. Cl is the electron pair donor and is the base. (continued)

10 572 CHAPTER 16 b. The completed equation is I I 2 I 3 The Lewis formula representation is : I : :. I.. I. : :. I.. I. I : I 2 is the electronpair acceptor and is the acid. The I ion is the electron pair donor and is the base a. The completed equation is GaBr 3 Br GaBr 4 The Lewis formula representation is : Br Ga : Br : Br : : Br : : Br : : Br Ga : Br : Br : GaBr 3 is the electronpair acceptor and is the acid. The Br ion is the electron pair donor and is the base. b. The completed equation is BF 3 F BF 4 The Lewis formula representation is : F B F : : F B F : BF 3 is the electronpair acceptor and is the acid. The F ion is the electron pair donor and is the base.

11 ACIDS AND BASES a. Each water molecule donates a pair of electrons to copper(ii), making the water molecule a Lewis base and the Cu 2 ion a Lewis acid. b. The AsH 3 donates a pair of electrons to the boron atom in BBr 3, making AsH 3 a Lewis base and the BBr 3 molecule a Lewis acid a. Each OH ion donates a pair of electrons to Al in Al(OH) 3, making OH a Lewis base and Al(OH) 3 a Lewis acid. b. Each CN ion donates a pair of electrons to silver ion, making CN ion a Lewis base and the Ag ion a Lewis acid The equation is H 2 S HOCH 2 CH 2 NH 2 HOCH 2 CH 2 NH 3 HS. The H 2 S is a Lewis acid, and HOCH 2 CH 2 NH 2 is a Lewis base. The hydrogen ion from H 2 S accepts a pair of electrons from the N atom in HOCH 2 CH 2 NH The equation is CaO SO 2 CaSO 3. The CaO is a Lewis base, and SO 2 is a Lewis acid. The oxide ion from CaO donates a pair of electrons to the sulfur of SO 2, forming the SO 3 2 ion in CaSO The reaction is HSO 4 ClO HClO SO 4 2. According to Table 16.2, HClO is a weaker acid than HSO 4. Because the equilibrium for this type of reaction favors formation of the weaker acid (or weaker base), the reaction occurs to a significant extent The reaction is HCN SO 4 CN HSO 4. According to Table 16.2, HCN is a weaker acid than HSO 4. Because the equilibrium for this type of reaction favors formation of the weaker acid (or weaker base), the reaction occurs in the opposite direction a. NH 4 is a weaker acid than H 3 PO 4, so the lefthand species are favored at equilibrium. b. HCN is a weaker acid than H 2 S, so the lefthand species are favored at equilibrium. c. H 2 O is a weaker acid than HCO 3, so the righthand species are favored at equilibrium. d. H 2 O is a weaker acid than Al(H 2 O) 6 3, so the righthand species are favored at equilibrium.

12 574 CHAPTER a. HCO 3 is a weaker acid than NH 4, so the righthand species are favored at equilibrium. b. H 2 S is a weaker acid than H 2 CO 3, so the lefthand species are favored at equilibrium. c. H 2 O is a weaker acid than HCN, so the lefthand species are favored at equilibrium. d. HCN is a weaker acid than H 2 CO 3, so the righthand species are favored at equilibrium Trichloroacetic acid is the stronger acid because, in general, the equilibrium favors the formation of the weaker acid, which is formic acid in this case The BF 4 ion is the weaker base because the equilibrium favors the formation of the weaker base a. H 2 S is stronger because acid strength decreases with increasing anion charge for polyprotic acid species. b. H 2 SO 3 is stronger because, for a series of oxoacids, acid strength increases with increasing electronegativity. c. HBr is stronger because Br is more electronegative than Se. Within a period, acid strength increases as electronegativity increases. d. HIO 4 is stronger because acid strength increases with the number of oxygen atoms bonded to the central atom. e. H 2 S is stronger because, within a group, acid strength increases with the increasing size of the central atom in binary acids a. HNO 2 < HNO 3 : Acid strength increases with the number of oxygen atoms bonded to the central atom. b. HCO 3 < H 2 CO 3 : Acid strength decreases with increasing anion charge for polyprotic acid species. c. H 2 S < H 2 Te: Within a group, acid strength increases with increasing size of the central atom for binary acids. d. H 2 S < HCl: Within a period, acid strength increases with increasing electronegativity of the central atom for binary acids. e. H 3 AsO 4 < H 3 PO 4 : For a series of oxoacids, acid strength increases with increasing electronegativity.

13 ACIDS AND BASES a. [H 3 O ] 1.2 M [OH ] Kw [H3O ] x x x M b. [OH ] 0.32 M [H 3 O ] K w [OH ] x x x M c. [OH ] 2 x (0.085 M) M [H 3 O ] K w [OH ] x x x M d. [H 3 O ] 0.38 M [OH ] K w 3 [H O ] x x x M a. [OH ] 1.65 M [H 3 O ] K w [OH ] 1.00 x x x M b. [OH ] 2 x (0.35 M) 0.70 M [H 3 O ] K w [OH ] x x x M c. [H 3 O ] M [OH ] K w 3 [H O ] x x x M d. [H 3 O ] 0.58 M [OH ] K w 3 [H O ] x x x M The [H 3 O ] M (HCl is a strong acid); using K w, the [OH ] 2.0 x M The [H 3 O ] M (HNO 3 is a strong acid); using K w, the [OH ] 5.0 x M.

14 576 CHAPTER Because the Ba(OH) 2 forms two OH per formula unit, the [OH ] 2 x M. [H 3 O ] K w [OH ] x x x M Because Mg(OH) 2 forms two OH per formula unit, the [OH ] 2 x 3.2 x x 10 4 M. [H 3 O ] K w [OH ] (1.0 x ) (6.4 x 10 4 ) 1.56 x x M a. 5 x 10 6 M H 3 O > 1.0 x 10 7, so the solution is acidic. b. Use K w to determine [H 3 O ] [H 3 O ] K w [OH ] 1.00 x 10 5 x x x 10 6 M Since 2 x 10 6 M > 1.0 x 10 7, the solution is acidic. c. When [OH ] 1.0 x 10 7 M, [H 3 O ] 1.0 x 10 7 M, and the solution is neutral. d. 2 x 10 9 M H 3 O < 1.0 x 10 7, so the solution is basic a. Use K w to determine [H 3 O ] [H 3 O ] K w [OH ] 1.00 x 10 2 x x x 10 4 M Since 5 x 10 4 M > 1.0 x 10 7, the solution is acidic. b. 2 x 10 6 M > 1.0 x 10 7, so the solution is acidic. c. Use K w to determine [H 3 O ] [H 3 O ] K w [OH ] 1.00 x 10 6 x x x 10 5 M Since 2 x 10 5 M > 1.0 x 10 7, the solution is acidic. d. 6 x 10 3 M > 1.0 x 10 7 M, so the solution is acidic.

15 16.55 The [H 3 O ] calculated below is > 1.0 x 10 7 M, so the solution is acidic. ACIDS AND BASES 577 [H 3 O ] K w [OH ] 1.00 x x x x 10 6 M The [H 3 O ] calculated below is < 1.0 x 10 7 M, so the solution is basic. [H 3 O ] K w [OH ] 1.00 x x x x M a. ph 4.6, acidic solution b. ph 7.0, neutral solution c. ph 1.6, acidic solution d. ph 10.5, basic solution a. ph 12.9, basic solution b. ph 8.9, basic solution c. ph 5.1, acidic solution d. ph 1.6, acidic solution a. Acidic (3.5 < 7.0) b. Neutral ( ) c. Basic (9.0 > 7.0) d. Acidic (5.5 < 7.0) a. Basic (12.0 > 7.0) b. Neutral ( ) c. Acidic (4.0 < 7.0) d. Acidic (5.7 < 7.0) Record the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ]. a. log (1.0 x 10 8 ) b. log (5.0 x ) c. log (7.5 x 10 3 ) d. log (6.35 x 10 9 ) Record the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ]. a. log (1.0 x 10 4 ) b. log (3.2 x ) c. log (2.3 x 10 5 ) d. log (2.91 x ) Record the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ]. log (7.5 x 10 3 )

16 578 CHAPTER Record the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ]. log (5.0 x 10 3 ) a. poh log (5.25 x 10 9 ) ; ph b. poh log (8.3 x 10 3 ) ; ph c. poh log (3.6 x ) ; ph d. poh log (2.1 x 10 8 ) ; ph a. poh log (4.83 x ) ; ph b. poh log (3.2 x 10 5 ) 4.494; ph c. poh log (2.7 x ) 9.568; ph d. poh log (5.0 x 10 4 ) 3.301; ph First, convert the [OH ] to [H 3 O ] using the K w equation. Then, find the ph, recording the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ]. [H 3 O ] K w [OH ] (1.0 x ) (0.0040) 2.50 x M ph log (2.50 x ) First, convert the [OH ] to [H 3 O ] using the K w equation. Then, find the ph, recording the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ]. [H 3 O ] K w [OH ] (1.0 x ) (0.050) 2.00 x M ph log (2.00 x ) From the definition, ph log [H 3 O ] and ph log [H 3 O ], so enter the negative value of the ph on the calculator, and use the inverse and log keys (or 10 x ) key to find the antilog of ph: log [H 3 O ] ph 5.12 [H 3 O ] antilog (5.12) x x 10 6 M

17 ACIDS AND BASES From the definition, ph log [H 3 O ] and ph log [H 3 O ], so enter the negative value of the ph on the calculator, and use the inverse and log keys (or 10 x ) key to find the antilog of ph: log [H 3 O ] ph 3.85 [H 3 O ] antilog (3.85) x x 10 4 M From the definition, ph log [H 3 O ] and ph log [H 3 O ], so enter the negative value of the ph on the calculator, and use the inverse and log keys (or 10 x ) key to find the antilog of ph. Then, use the K w equation to calculate [OH ] from [H 3 O ]. log [H 3 O ] ph [H 3 O ] antilog (11.63) x M [OH ] K w [H 3 O ] (1.0 x ) (2.34 x ) 4.27 x x 10 3 M From the definition, ph log [H 3 O ] and ph log [H 3 O ], so enter the negative value of the ph on the calculator, and use the inverse and log keys (or 10 x ) key to find the antilog of ph. Then, use the K w equation to calculate [OH ] from [H 3 O ]. log [H 3 O ] ph 9.61 [H 3 O ] antilog (9.61) x M [OH ] K w [H 3 O ] (1.0 x ) (2.45 x ) 4.08 x x 10 5 M First, calculate the molarity of the OH ion from the mass of NaOH. Then, convert the [OH ] to [H 3 O ] using the K w equation. Then, find the ph, recording the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ] g NaOH 1.00 L x 1 mol NaOH g NaOH mol NaOH 1.00 L M OH [H 3 O ] K w [OH ] (1.0 x ) (0.1450) x M ph log [H 3 O ] log (6.896 x )

18 580 CHAPTER First, calculate the molarity of the Ba(OH) 2 from the mass of Ba(OH) 2, and multiply the molarity of Ba(OH) 2 by two to obtain [OH ]. Then, convert the [OH ] to [H 3 O ] using the K w equation. Then, find the ph, recording the same number of places after the decimal point in the ph as the number of significant figures in the [H 3 O ] g Ba(OH) L 1 mol Ba(OH) g Ba(OH) x mol Ba(OH) L [OH ] 2 mol OH /1 mol Ba(OH) 2 2 x M Ba(OH) M [H 3 O ] K w [OH ] (1.0 x ) ( ) x M ph log [H 3 O ] log (1.264 x ) Figure shows that the methylred indicator is yellow at ph values above about 5.5 (slightly past the midpoint of the range for methyl red). Bromthymol blue is yellow at ph values up to about 6.5 (slightly below the midpoint of the range for bromthymol blue). Therefore, the ph of the solution is between 5.5 and 6.5, and the solution is acidic Thymol blue is yellow at ph values above about 2.5. Bromphenol blue is yellow at ph values below about 3.5. Therefore, the ph of the aspirin solution must be in the range of 2.5 to 3.5. The solution is, of course, acidic. Solutions to General Problems a. BaO is a base; BaO H 2 O Ba 2 2OH b. H 2 S is an acid; H 2 S H 2 O H 3 O HS c. CH 3 NH 2 is a base; CH 3 NH 2 H 2 O CH 3 NH 3 OH d. SO 2 is an acid; SO 2 2H 2 O H 3 O HSO a. P 4 O 10 is an acid; P 4 O 10 10H 2 O 4H 3 O 4H 2 PO 4 b. K 2 O is a base; K 2 O H 2 O 2K 2OH c. N 2 H 4 is a base; N 2 H 4 H 2 O N 2 H 5 OH d. H 2 Se is an acid; H 2 Se H 2 O H 3 O HSe

19 ACIDS AND BASES a. H 2 O 2 (aq) S 2 (aq) HO 2 (aq) HS (aq) b. HCO 3 (aq) OH (aq) CO 2 3 (aq) H 2 O(l) c. NH 4 (aq) CN (aq) NH 3 (aq) HCN(aq) d. H 2 PO 4 (aq) OH (aq) HPO 2 4 (aq) H 2 O(l) a. H 2 O(l) HCl(aq) H 3 O (aq) Cl (aq) b. HCO 3 (aq) HF(aq) H 2 CO 3 (aq) F (aq) c. NH 3 (aq) HBrO(aq) NH 4 (aq) BrO (aq) d. H 2 PO 4 (aq) H 2 SO 3 (aq) H 3 PO 4 (aq) HSO 3 (aq) a. The ClO ion is a Bronsted base, and water is a Bronsted acid. The complete chemical equation is ClO (aq) H 2 O(l) HClO(aq) OH (aq). The equilibrium does not favor the products because ClO is a weaker base than OH. In Lewis language, a proton from H 2 O acts as a Lewis acid by accepting a pair of electrons on the oxygen of ClO. H : Cl : O : : Cl : O : H b. The NH 2 ion is a Bronsted base, and NH 4 is a Bronsted acid. The complete chemical equation is NH 4 NH 2 2NH 3. The equilibrium favors the products because the reactants form the solvent, a weakly ionized molecule. In Lewis language, the proton from NH 4 acts as a Lewis acid by accepting a pair of electrons on the nitrogen of NH 2. H H : N : H H : N : H H 2 H : N : H H a. The HS ion is a Bronsted base, and water is a Bronsted acid. The complete chemical equation is HS (aq) H 2 O(l) H 2 S(aq) OH (aq). The equilibrium does not favor the products because the HS ion is a weaker base than OH. In Lewis language, a proton from H 2 O acts as a Lewis acid by accepting a pair of electrons on the sulfur of HS. H : O : H : S : H H : S : H OH (continued)

20 582 CHAPTER 16 b. The complete chemical equation is Fe 3 (aq) CN (aq) Fe(CN) 2. The reaction cannot be described in Bronsted language because no proton transfer occurs. In Lewis language, Fe 3 acts as a Lewis acid by sharing a pair of electrons on the carbon of CN. Fe 3 [: C N : ] 2 [ : Fe : C N : ] Table 16.2 shows that HNO 2 is a stronger acid than HF. Because an acidbase reaction normally goes in the direction of the weaker acid, the reaction is more likely to go in the direction written: HNO 2 F HF NO The HS ion is acting as a base and will form its conjugate acid, H 2 S. Table 16.2 shows that H 2 S is a stronger acid than HCN. Because an acidbase reaction normally goes in the direction of the weaker acid, the reaction is more likely to go in the opposite direction: H 2 S CN HS HCN The order is H 2 S < H 2 Se < HBr. H 2 Se is stronger than H 2 S because, within a group, acid strength increases with increasing size of the central atom in binary acids. HBr is a strong acid, whereas the others are weak acids The order is HBrO < HBrO 2 < HClO 2. HClO 2 is stronger than HBrO 2 because, for oxoacids, acid strength increases with increasing electronegativity. HBrO 2 is stronger than HBrO because acid strength increases with increasing oxidation number of Br The KOH is a strong base and is fully ionized in solution, so you can use its formula and molar concentration to determine the [OH ] of the solution. Therefore, the 0.25 M KOH contains 0.25 M OH. The [H 3 O ] is obtained from the K w expression: K w 1.0 x [H 3 O ] x 0.25 M OH [H3 O ] x x x 10 M

21 ACIDS AND BASES The Sr(OH) 2 is a strong base and is fully ionized in solution, so you can use its formula and molar concentration to determine the [OH ] of the solution. Therefore, the 0.35 M Sr(OH) 2 contains 2 x M OH. The [H 3 O ] is obtained from the K w expression: K w 1.0 x [H 3 O ] x 0.70 M OH 14 [H 3 O 1.0 x ] 1.42 x x 10 M Enter the H 3 O concentration of 1.5 x 10 3 into the calculator, press the log key, and press the sign key to change the negative log to a positive log. This follows the negative log definition of ph. The number of decimal places of the ph should equal the significant figures in the H 3 O. ph log [H 3 O ] log (1.5 x 10 3 ) Enter the H 3 O concentration of 2.5 x 10 2 into the calculator, press the log key, and press the sign key to change the negative log to a positive log. This follows the negative log definition of ph. The number of decimal places of the ph should equal the significant figures in the H 3 O. ph log [H 3 O ] log (2.5 x 10 2 ) Find the poh from the ph using ph poh Then, calculate the [OH ] from the poh by entering the poh into the calculator, pressing the sign key to change the positive log to a negative log, and finding the antilog. On some calculators, the antilog is found by using the inverse of the log; on other calculators, the antilog is found using the 10 x key. The number of significant figures in the [OH ] should equal the number of decimal places in the poh. poh [OH ] antilog (10.85) x x M Find the poh from the ph using ph poh Then, calculate the [OH ] from the poh by entering the poh into the calculator, pressing the sign key to change the positive log to a negative log, and finding the antilog. On some calculators, the antilog is found by using the inverse of the log; on other calculators, the antilog is found using the 10 x key. The number of significant figures in the [OH ] should equal the number of decimal places in the poh. poh [OH ] antilog (9.95) x x M

22 584 CHAPTER a. H 2 SO 4 (aq) 2NaHCO 3 (aq) Na 2 SO 4 (aq) 2CO 2 (g) 2H 2 O(l) (molecular) H 3 O (aq) HCO 3 (aq) CO 2 (g) 2H 2 O(l) (net ionic) b. The total moles of H 3 O from the H 2 SO 4 is mol H 3 O mol H 2SO4 1L mol H 3 O x L x 2molHO 3 1molHSO 2 4 The moles of H 3 O that reacted with the NaOH are given by mol H 3 O mol NaOH x L mol 1L The moles of NaHCO 3 present in the original sample are equal to the moles of H 3 O that reacted with the HCO 3, which is given by Total moles H 3 O moles H 3 O reacted with the NaOH moles HCO mol mol mol NaHCO 3 c. The mass of NaHCO 3 (molar mass g/mol) present in the original sample is mol NaHCO 3 x g NaHCO3 1 mol NaHCO g Thus, the percent NaHCO 3 in the original sample is given by Percent NaHCO g g x 100% percent The percent KCl in the original sample is Percent KCl percent a. CO 2 3 (aq) 2 H 3 O (aq) 3 H 2 O (l) CO 2 (g) b. The total moles of H 3 O from the HCl is mol H 3 O mol HCl x L mol 1L (continued)

23 The moles of H 3 O that reacted with the NaOH is given by mol H 3 O mol NaOH x L mol 1L ACIDS AND BASES 585 The moles of Na 2 CO 3 present in the original sample are equal to the moles of H 3 O that reacted with the CO 3 2, which is given by Total moles H 3 O moles H 3 O reacted with the NaOH moles CO mol mol mol Na 2 CO 3 c. The mass of Na 2 CO 3 (molar mass g/mol) present in the original sample is mol Na 2 CO 3 x g Na2CO3 1 mol Na2CO g Thus, the percent Na 2 CO 3 in the original sample is given by Percent Na 2 CO g g x 100% percent The percent NaCl in the original sample is Percent NaCl percent HCO 3 (aq) H 2 O(l) H 3 O (aq) CO 3 2 (aq) HCO 3 (aq) H 2 O(l) H 2 CO 3 (aq) OH (aq) HCO 3 (aq) Na (aq) OH (aq) HCO 3 (aq) H (aq) Cl (aq) Na (aq) CO 3 2 (aq) H 2 O(l) H 2 O(l) CO 2 (g) Cl (aq) H 2 PO 4 is amphiprotic or amphoteric. H 2 PO 4 (aq) H 2 O(l) H 2 PO 4 (aq) H 2 O(l) H 3 O (aq) HPO 4 2 (aq) H 3 PO 4 (aq) OH (aq) H 2 PO 4 (aq) K (aq) OH (aq) H 2 PO 4 (aq) H (aq) NO 3 (aq) K (aq) HPO 4 2 (aq) H 2 O(l) H 3 PO 4 (aq) NO 3 (aq)

24 586 CHAPTER CaH 2 (s) 2H 2 O(l) Ca(OH) 2 (s) H 2 (g) The hydride ion is a stronger base because it took an H from water, leaving the OH ion. Every time a strong base is added to water, it will react with the water leaving the OH as the product, so a strong base cannot exist in water Ca 3 N 2 (s) 6H 2 O(l) 3Ca(OH) 2 (s) 2NH 3 (g) NaNH 2 (s) H 2 O(l) NaOH(aq) NH 3 (g) N 3 is a stronger base than NH 2 because it has more negative charge, so it will have a greater attraction for H. Also, we could consider that NH 2 is a N 3 that has already reacted with two H 's. Stronger bases than OH will produce OH in aqueous solution a. 2HF(l) H 2 F F b. NaF will be a base because F is a conjugate base of HF. c. HClO 4 HF H 2 F ClO 4 The conjugate acid is H 2 F a. 2NH 3 (l) NH 4 NH 2 b. NH 2 is the conjugate base of NH 3, so NaNH 2 will be a base. c. NaNH 2 NH 4 Cl 2NH 3 NaCl The reaction of ammonia with water is given by NH 3 (aq) H 2 O(l) NH 4 (aq) OH (aq) The initial concentration of NH 3 (molar mass g/mol) is Molarity 1molNH 4.25 g NH 3 3 x g NH L M Since the NH 3 is 0.42 percent reacted, the concentration of OH is [OH ] M x M poh log [OH ] log ( ) ph 14 poh

25 ACIDS AND BASES The reaction of C 2 H 5 NH 2 with water is given by C 2 H 5 NH 2 (aq) H 2 O(l) C 2 H 5 NH 3 (aq) OH (aq) The initial concentration of C 2 H 5 NH 2 (molar mass g/mol) is given by Molarity 1molCHNH 0.675gCHNH x 45.09gCHNH L M Since the C 2 H 5 NH 2 is 0.98 percent reacted, the concentration of OH is [OH ] M x M poh log [OH ] log ( ) ph 14 poh Solutions to CumulativeSkills Problems For (HO) m YO n acids, acid strength increases with n regardless of the number of OH's. The structure of H 3 PO 4 is (HO) 3 PO; because H 3 PO 3 and H 3 PO 4 have about the same acidity, H 3 PO 3 must also have n 1; thus, m 2. This leaves one H, which must bond to phosphorus, giving a structure of (HO) 2 (O)PH. Assuming that only two H's react with NaOH, the mass of NaOH that reacts with 1.00 g of H 3 PO 3 (PA) is calculated as follows: 1.00 g PA x 1 mol PA g PA x 2 mol NaOH 1 mol PA x g NaOH 1 mol NaOH g NaOH For (HO) m YO n acids, acid strength increases with n regardless of the number of OH's. The structure of H 3 PO 4 is (HO) 3 PO; because H 3 PO 2 and H 3 PO 4 have the same acidity, H 3 PO 2 must also have n 1; thus, m 1. This leaves two H's, which must bond to phosphorus, giving a structure of (HO)(O)PH 2. Assuming that only one H reacts with NaOH, the mass of NaOH that reacts with 1.00 g of H 3 PO 2 (HA) is calculated as follows: 1.00 g HA x 1 mol PA g PA x 1 mol NaOH 1 mol HA x g NaOH 1 mol NaOH g NaOH

26 588 CHAPTER BF 3 acts as a Lewis acid, accepting an electron pair from NH 3 : BF 3 :NH 3 F 3 B:NH 3 The NH 3 acts as a Lewis base in donating an electron pair to BF 3. When 10.0 g of each are mixed, the BF 3 is the limiting reagent because it has the higher formula weight. The mass of BF 3 :NH 3 formed is 10.0 g BF 3 x 1 mol BF g BF3 x 1 mol BF 3:NH3 1 mol BF3 x g BF 3:NH3 1 mol BF 3:NH g BF 3 :NH BF 3 is the Lewis acid and accepts an electron pair from ether: BF 3 :OR 2 F 3 B:OR 2 The ether (:OR 2 ) acts as a Lewis base in donating an electron pair to BF 3. When 10.0 g of BF 3 and 20.0 g of ether are mixed, the BF 3 is the limiting reagent because the formula weights are nearly equal. The mass of BF 3 :OR 2 formed is 10.0 g BF 3 x 1 mol BF g BF3 x 1 mol BF 3:OR2 1 mol BF3 x g BF 3:OR2 1 mol BF 3:OR g BF 3 :OR 2

27 Filename: chapter16.doc Directory: D:\hmco\chemistry\general\ebbing\general_chem\8e\instructors\ solutions_manual Template: C:\Documents and Settings\willsoj\Application Data\Microsoft\Templates\Normal.dot Title: 15 Subject: Author: David Bookin Keywords: Comments: Creation Date: 7/2/2003 3:43 PM Change Number: 27 Last Saved On: 11/30/2003 7:22 AM Last Saved By: David Bookin Total Editing Time: 155 Minutes Last Printed On: 1/9/2004 4:11 PM As of Last Complete Printing Number of Pages: 26 Number of Words: 5,838 (approx.) Number of Characters: 33,277 (approx.)