Chapter 14 Acids and Bases

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–2 QUESTION Aniline, C 6 H 5 NH 2, was isolated in the 1800s and began immediate use in the dye industry. What is the formula of the conjugate acid of this base? 1.C 6 H 5 NH C 6 H 5 NH C 6 H 5 NH – 4.C 6 H 5 NH +

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–3 ANSWER Choice 2 correctly represents the result of aniline accepting a H + ion as bases typically do. The conjugate acid of a base is represented as the base with the addition of a H +. Section 14.1: The Nature of Acids and Bases

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–4 QUESTION Nitric acid, HNO 3, is considered to be a strong acid whereas nitrous acid, HNO 2, is considered to be a weak acid. Which of the statements here is fully correct? 1.Nitric acid has an aqueous equilibrium that lies far to the right and NO 3 – is considered a weak conjugate base. 2.Nitric acid has a stronger conjugate base than nitrous acid. 3.The dissociation of nitrous acid compared to an equal concentration of nitric acid produces more H +. 4.The equilibrium of nitrous acid lies far to the left and the conjugate base is weaker than the conjugate base of nitric acid.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–5 ANSWER Choice 1 correctly compares equilibrium and conjugate base characteristics. The conjugate base of a strong acid is considered to be weak. The stronger the acid, the more reaction in water. Therefore, a weak acid’s equilibrium is favored to the left. Section 14.2: Acid Strength

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–6 QUESTION

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–7 QUESTION (continued) Use information on this table to determine which of the following bases would have the weakest conjugate acid: OC 6 H 5 – ; C 2 H 3 O 2 – ; OCl – ; NH 3 1.OC 6 H 5 – 2.C 2 H 3 O 2 – 3.OCl – 4.NH 3

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–8 ANSWER Choice 1 correctly identifies the base, among these four, with the weakest conjugate acid. The K a ’s in the table can be used to compare conjugate acid strength. The higher the K a value, the stronger the acid. Section 14.2: Acid Strength

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–9 QUESTION In a solution of water at a particular temperature the [H + ] may be 1.2  10 –6 M. What is the [OH – ] in the same solution? Is the solution acidic, basic, or neutral?  10 –20 M; acidic  10 –20 M; basic  10 –9 M; basic  10 –9 M; acidic

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–10 ANSWER Choice 4 correctly shows the OH – molarity and classifies the solution as acidic. K w = [H + ][OH – ] = 1.0  10 –14 at 25°C. The H + molarity is approximately 1,000 times greater than the OH – concentration. Solutions with higher H + concentrations than OH – are acidic. Section 14.2: Acid Strength

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–11 QUESTION An environmental chemist obtains a sample of rainwater near a large industrial city. The [H + ] was determined to be 3.5  10 –6 M. What is the pH, pOH, and [OH – ] of the solution? 1.pH = 5.46 ; pOH = 8.54; [OH – ] = 7.0  10 –6 M 2.pH = 5.46 ; pOH = 8.54; [OH – ] = 2.9  10 –9 M 3.pH = ; pOH =1.44 ; [OH – ] = 3.6  10 –2 M 4.pH = 8.54; pOH = 5.46; [OH – ] = 2.9  10 –9 M

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–12 ANSWER Choice 2 provides all three correct responses. The expression pH = –log[H + ] can be used to find the pH then: = pH + pOH can be used to obtain the pOH. Finally, [OH – ] = 10 –pOH. Section 14.3: The pH Scale

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–13 QUESTION Which of the following correctly compares strength of acids, pH, and concentrations? 1.A weak acid, at the same concentration of a strong acid, will have a lower pH. 2.A weak acid, at the same concentration of a strong acid, will have the same pH. 3.A weak acid, at a high enough concentration more than a strong acid, could have a lower pH than the strong acid. 4.A weak acid, at a concentration below a strong acid, could have a lower pH than a strong acid.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–14 ANSWER Choice 3 correctly predicts that it is possible to have a high enough concentration of the weak acid compared to a strong acid, and that the pH of the weaker acid would be lower (more acidic) than the more dilute stronger acid. Strength of an acid refers to its dissociation. The pH of a solution depends on the concentration, regardless of source, of the H + ion. Section 14.4: Calculating the pH of Strong Acid Solutions

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–15 QUESTION Butyric acid is a weak acid that can be found in spoiled butter. The compound has many uses in synthesizing other flavors. The K a of HC 4 H 7 O 2 at typical room temperatures is 1.5  10 –5. What is the pH of a 0.20 M solution of the acid? –0.70

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–16 ANSWER Choice 3 is correct assuming that the amount of dissociation of this weak acid is negligible when compared to its molarity. K a = [H + ][A – ] / [HA – x] = x 2 /(0.20 M – x) becomes 1.5  10 –5 = x 2 /0.20; once x is found, representing the [H + ], taking –log of that yields the pH. Section 14.5: Calculating the pH of Weak Acid Solutions

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–17 QUESTION A 0.35 M solution of an unknown acid is brought into a lab. The pH of the solution is found to be From this data, what is the K a value of the acid?  10 –  10 –  10 –  10 –3

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–18 ANSWER Choice 2 shows the unknown acid’s K a value. The pH could be used to find the [H + ] concentration, 10 –2.67, then the K a expression only has one unknown: K a = [ ][ ]/(0.35– ) Section 14.5: Calculating the pH of Weak Acid Solutions

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–19 QUESTION Two weak acids, HA and HB are placed in separate solutions so that their molarities are the same. Which, if either, would have the larger value for K a if the pH of the HA solution were lower than the pH of the HB solution? 1.HA has the larger K a. 2.HB has the larger K a. 3.The K a of HA = K a of HB. 4.My answer would actually just be a guess, does more information need to be given?

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–20 ANSWER Choice 1 correctly identifies the acid with the larger K a value. When equal concentrations are supplied there is an equal opportunity to dissociate into H +. However, the lower pH in the HA solution indicates a higher concentration of H + present in the solution. Therefore, HA must have a higher level of dissociation. This would be expected if the K a (product to reactant ratio) were larger. Section 14.5: Calculating the pH of Weak Acid Solutions

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–21 QUESTION Amine compounds often have pungent aromas. The compound ethylamine can be associated with a fish-like aroma. The K b, at typical room conditions, of C 2 H 5 NH 2 is 5.6  10 –4. What is the pH of a 0.15 M solution of this weak base?

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–22 ANSWER Choice 3 provides the correct pH for the solution. The K b expression can be used to solve for [OH – ]. This can be used to either obtain the pOH (pH = – pOH at 25°C) or [H + ] = (K w /[OH – ]); and then pH can be calculated. Section 14.6: Bases

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–23 QUESTION Pyridine is a very weak base with a strong, sharp odor. If a 0.57 M solution were measured to have a % dissociation, what is the value of the K b of pyridine?  10 –  10 –  10 –  10 –5

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–24 ANSWER Choice 1 displays the correct K b value. The initial molarity (0.57 M) is relatively unchanged by the % dissociation, so in the K b expression 0.57 M could be used for the equilibrium concentration. Of the original concentration only a fraction reacts. So is the molarity of OH – and pyridine’s conjugate acid. Now, K b can be calculated as K b = ( ) 2 /0.57 Section 14.6: Bases

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–25 QUESTION Ascorbic acid, also known as vitamin C, has two hydrogen atoms that ionize from the acid. K a 1 = 7.9  10 –5 ; K a 2 = 1.6  10 –12. What is the pH, and C 6 H 6 O 6 2– concentration of a 0.10 M solution of H 2 C 6 H 6 O 6 ? ; [C 6 H 6 O 6 2– ] = M ; [C 6 H 6 O 6 2– ] = 1.6  10 –12 M ; [C 6 H 6 O 6 2– ] = 1.6  10 –12 M ; [C 6 H 6 O 6 2– ] = M

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–26 ANSWER Choice 2 shows both correct answers. In a diprotic acid with two small, widely separated K a values the pH of a solution can be obtained from using the first K a and the molarity. The concentration of the dianion can be closely approximated by assuming very little dissociation of the second acidic hydrogen, so that K a 2 is very close to the molarity. Section 14.7: Polyprotic Acids

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–27 QUESTION The following salts were all placed in separate solutions at the same temperature so that their concentrations were all equal. Arrange them in order from lowest pH to highest pH. NaCl; NH 4 NO 3 ; Ca(C 2 H 3 O 2 ) 2 ; AlCl 3 Additional information: K b for NH 3 = 1.8  10 –5 ; K a for HC 2 H 3 O 2 = 1.8  10 –5 ; K a for Al(H 2 O) 3+ = 1.4  10 –5. 1.NaCl; NH 4 NO 3 ; Ca(C 2 H 3 O 2 ) 2 ; AlCl 3 2.AlCl 3 ; NaCl; NH 4 NO 3 ; Ca(C 2 H 3 O 2 ) 2 3.AlCl 3 ; NH 4 NO 3 ; NaCl; Ca(C 2 H 3 O 2 ) 2 4.NH 4 NO 3 ; AlCl 3 ; NaCl; Ca(C 2 H 3 O 2 ) 2

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–28 ANSWER Choice 3 correctly ranks the salt solutions from lowest pH (most acidic solution) to highest pH. The ranking is based on production of H + from the salt ions interacting with water. Highly charged small metal ions such as Al 3+ can produce H + as can NH 4 +. However, the K a of the aluminum’s reaction is larger than the K a for NH 4 +. NaCl is neutral and the acetate ion undergoes a reaction that produces OH –, so it has a high pH. Section 14.8: Acid–Base Properties of Salts

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–29 QUESTION The Ka value for monochloroacetic acid is 1.35  10 –3 at 25°C. Neutralizing the acid with KOH would produce the salt potassium chloroacetate. What would be the pH of a M solution of the salt at 25°C?

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–30 ANSWER Choice 1 is the correct pH for this basic salt. The salt, in water, will dissociate and provide the opportunity for the cation and anion to interact with water. Since the anion is from a weak acid (note the K a ) it can regain H + from water thus producing OH – ions. The K b for this reaction is obtained through K w /K a. Then the [OH – ] can be obtained from the K b and molarity. This is then converted to pH after using the K w for water. Section 14.8: Acid–Base Properties of Salts

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–31 QUESTION To fit our traditional definition of an acid, a molecule must be able to produce H + ions in water. Some molecules can do this while others cannot. For example, the C–H bond does not allow H to easily leave. H–Cl however, does allow H to easily leave the Cl. H–F also allows H + to leave in water, but not as well as HCl. Which statement here offers the best explanation for these observations?

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–32 QUESTION (continued) 1.The C–H bond is strong, but not very polar. The H–Cl bond is strong and polar. The H–F bond is unusually strong and yet very polar. 2.The C–H bond is very strong, and polar. The H–Cl bond is very strong and not polar. The H–F bond is weak and polar. 3.The C–H bond is not strong, or very polar. The H–Cl bond is very strong and very polar. The H–F bond is weak and polar. 4.A weaker bond produces a more polar bond, hence a stronger acid.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–33 ANSWER Choice 1 correctly relates bond strength and polarity to the observations described in the question. Generally weaker bonds such as O–H would be likely to release H +, but polarity plays a key role in attracting H + to water. C–H is also non- polar so it does not release H +. H–F has a more polar bond than H–Cl so it could be expected to release H + to polar water, but the H–F is unusually strong so only a small percentage of HF molecules may behave this way. Section 14.9: The Effect of Structure on Acid–Base Properties

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–34 QUESTION Some CO 2 gas is bubbled into water. Which of the following compounds would be suitable to now neutralize the CO 2 infused solution? 1.K 2 O 2.NaCl 3.NH 4 Br 4.AlCl 3

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–35 ANSWER Choice 1 is the only compound that would begin to neutralize an acid solution. Nonmetal oxides, such as CO 2, produce acid solutions in water. So a compound with basic properties would be needed to neutralize the solution. Of the choices, K 2 O (a metal oxide) is the only one with basic properties. Section 14.10: Acid–Base Properties of Oxides

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–36 QUESTION Hydrated metal ions are stabilized by Lewis acid-base bonding. Which statement correctly identifies both the Lewis classification and the rationale for the decision when a metal ion bonds to a water molecule? 1.The metal ion is the Lewis base because it accepts a pair of electrons from water. 2.The metal ion is the Lewis base because it donates a pair of electrons to water. 3.Water is the Lewis base because it accepts a pair of electrons from the metal ion. 4.Water is the Lewis base because it donates a pair of electrons to the metal ion.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 14–37 ANSWER Choice 4 provides the correct classification for water as a Lewis base and provides the correct rationale. Lewis bases are electron donors. Metal ions have lost electrons; thus water has the opportunity, through its lone electron pairs, to donate electrons to the metal ion. Section 14.11: The Lewis Acid–Base Model