chapter 14 acids and bases. copyright © houghton mifflin company. all rights reserved.crs question,...

Chapter 14

Acids and BasesAcids and Bases

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 14–2

QUESTIONAniline, C6H5NH2, 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. C6H5NH2+

2. C6H5NH3+

3. C6H5NH–

4. C6H5NH+


ANSWERChoice 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


QUESTIONNitric acid, HNO3, is considered to be a strong acid whereas nitrous acid, HNO2, 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 theright and NO3

– 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 nitricacid.


ANSWERChoice 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


QUESTION


QUESTION (continued)Use information on this table to determine which of the following bases would have the weakest conjugate acid:

OC6H5–; C2H3O2

–; OCl–; NH3

1. OC6H5–

2. C2H3O2–

3. OCl– 4. NH3


ANSWERChoice 1 correctly identifies the base, among these four, with the weakest conjugate acid. The Ka’s in the table can be used to compare conjugate acid strength. The higher the Ka value, the stronger the acid.



QUESTIONIn 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?

1. 1.2 10–20 M; acidic2. 1.2 10–20 M; basic3. 8.3 10–9 M; basic4. 8.3 10–9 M; acidic


ANSWERChoice 4 correctly shows the OH– molarity and classifies the solution as acidic. Kw = [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.



QUESTIONAn 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 M2. pH = 5.46 ; pOH = 8.54; [OH–] = 2.9 10–9 M3. pH = 12.56 ; pOH =1.44 ; [OH–] = 3.6 10–2 M 4. pH = 8.54; pOH = 5.46; [OH–] = 2.9 10–9 M


ANSWERChoice 2 provides all three correct responses. The expression pH = –log[H+] can be used to find the pH then: 14.00 = pH + pOH can be used to obtain the pOH. Finally, [OH–] = 10–pOH.

Section 14.3: The pH Scale


QUESTIONWhich 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 astrong 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.


ANSWERChoice 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


QUESTIONButyric acid is a weak acid that can be found in spoiled butter. The compound has many uses in synthesizing other flavors. The Ka of HC4H7O2 at typical room temperatures is 1.5 10–5. What is the pH of a 0.20 M solution of the acid?

1. 5.522. 4.823. 2.764. –0.70


ANSWERChoice 3 is correct assuming that the amount of dissociation of this weak acid is negligible when compared to its molarity.

Ka = [H+][A–] / [HA – x] = x2/(0.20 M – x)

becomes 1.5 10–5 = x2/0.20; once x is found, representing the [H+], taking –log of that yields the pH.

Section 14.5: Calculating the pH of Weak Acid Solutions


QUESTIONA 0.35 M solution of an unknown acid is brought into a lab. The pH of the solution is found to be 2.67. From this data, what is the Ka value of the acid?

1. 6.1 10–3

2. 1.3 10–5

3. 7.5 10–4

4. 2.1 10–3


ANSWERChoice 2 shows the unknown acid’s Ka value. The pH could be used to find the [H+] concentration , 10–2.67, then the Ka expression only has one unknown:

Ka = [0.00214][0.00214]/(0.35–0.00214)



QUESTIONTwo 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 Ka if the pH of the HA solution were lower than the pH of the HB solution?

1. HA has the larger Ka.

2. HB has the larger Ka.

3. The Ka of HA = Ka of HB.4. My answer would actually just be a guess, does more

information need to be given?


ANSWERChoice 1 correctly identifies the acid with the larger Ka 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 Ka (product to reactant ratio) were larger.



QUESTIONAmine compounds often have pungent aromas. The compound ethylamine can be associated with a fish-like aroma. The Kb, at typical room conditions, of C2H5NH2 is 5.6 10–4. What is the pH of a 0.15 M solution of this weak base?

1. 9.922. 2.043. 11.964. 10.75


ANSWERChoice 3 provides the correct pH for the solution. The Kb expression can be used to solve for [OH–]. This can be used to either obtain the pOH (pH = 14.00 – pOH at 25°C) or [H+] = (Kw/[OH–]); and then pH can be calculated.

Section 14.6: Bases


QUESTIONPyridine is a very weak base with a strong, sharp odor. If a 0.57 M solution were measured to have a 0.0055% dissociation, what is the value of the Kb of pyridine?

1. 1.7 10–9

2. 1.7 10–5

3. 5.3 10–5

4. 3.1 10–5


ANSWERChoice 1 displays the correct Kb value. The initial molarity (0.57 M) is relatively unchanged by the 0.0055% dissociation, so in the Kb expression 0.57 M could be used for the equilibrium concentration. Of the original concentration only a 0.000055 fraction reacts. So 0.000031 is the molarity of OH– and pyridine’s conjugate acid. Now, Kb can be calculated as Kb = (0.000031)2/0.57

Section 14.6: Bases


QUESTIONAscorbic acid, also known as vitamin C, has two hydrogen atoms that ionize from the acid. Ka1 = 7.9 10–5; Ka2 = 1.6 10–12. What is the pH, and C6H6O6

2– concentration of a 0.10 M solution of H2C6H6O6?

1. 2.55; [C6H6O62–] = 0.050 M

2. 2.55; [C6H6O62–] = 1.6 10–12 M

3. 1.00; [C6H6O62–] = 1.6 10–12 M

4. 5.10; [C6H6O62–] = 0.050 M


ANSWERChoice 2 shows both correct answers. In a diprotic acid with two small, widely separated Ka values the pH of a solution can be obtained from using the first Ka and the molarity. The concentration of the dianion can be closely approximated by assuming very little dissociation of the second acidic hydrogen, so that Ka2 is very close to the molarity.

Section 14.7: Polyprotic Acids


QUESTIONThe 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; NH4NO3; Ca(C2H3O2)2; AlCl3

Additional information: Kb for NH3 = 1.8 10–5; Ka for HC2H3O2 = 1.8 10–5; Ka for Al(H2O)3+ = 1.4 10–5.

1. NaCl; NH4NO3; Ca(C2H3O2)2; AlCl3

2. AlCl3; NaCl; NH4NO3; Ca(C2H3O2)2

3. AlCl3; NH4NO3; NaCl; Ca(C2H3O2)2

4. NH4NO3; AlCl3; NaCl; Ca(C2H3O2)2


ANSWERChoice 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 Al3+ can produce H+ as can NH4

+. However, the Ka of the aluminum’s reaction is larger than the Ka for NH4

+. 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


QUESTIONThe 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 0.110 M solution of the salt at 25°C?

1. 7.9562. 13.0413. 6.0444. 12.086


ANSWERChoice 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 Ka) it can regain H+ from water thus producing OH– ions. The Kb for this reaction is obtained through Kw/Ka. Then the [OH–] can be obtained from the Kb and molarity. This is then converted to pH after using the Kw for water.

Section 14.8: Acid–Base Properties of Salts


QUESTIONTo 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?


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.


ANSWERChoice 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


QUESTIONSome CO2 gas is bubbled into water. Which of the following compounds would be suitable to now neutralize the CO2 infused solution?

1. K2O2. NaCl3. NH4Br4. AlCl3


ANSWERChoice 1 is the only compound that would begin to neutralize an acid solution. Nonmetal oxides, such as CO2, produce acid solutions in water. So a compound with basic properties would be needed to neutralize the solution. Of the choices, K2O (a metal oxide) is the only one with basic properties.

Section 14.10: Acid–Base Properties of Oxides


QUESTIONHydrated 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 pairof electrons from water.

2. The metal ion is the Lewis base because it donates a pairof electrons to water.

3. Water is the Lewis base because it accepts a pair ofelectrons from the metal ion.

4. Water is the Lewis base because it donates a pair ofelectrons to the metal ion.


ANSWERChoice 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

chapter 14 acids and bases. copyright © houghton mifflin company. all rights reserved.crs question,...

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