AP Chemistry: Chapter 16 Quiz

1. What is the conjugate acid of NH₃?

NH4+

NH3+

NH2+

NH3

NH4OH

The conjugate acid of NH3 is NH4+. In a chemical reaction, NH3 can act as a base by accepting a proton (H+) to form NH4+. The addition of the proton converts NH3 into its conjugate acid, NH4+.

Explanation

The conjugate acid of NH3 is NH4+. In a chemical reaction, NH3 can act as a base by accepting a proton (H+) to form NH4+. The addition of the proton converts NH3 into its conjugate acid, NH4+.

2. Which one of the following is a Bronsted-Lowry acid?

HNO2

(CH3)3NH+

HF

CH3COOH

All of the above

All of the given compounds, HNO2, (CH3)3NH+, HF, and CH3COOH, can act as Bronsted-Lowry acids because they are capable of donating a proton (H+) to another species. In the case of HNO2, it can donate a proton to form NO2-. (CH3)3NH+ can donate a proton to form (CH3)3N. HF can donate a proton to form F-. CH3COOH can donate a proton to form CH3COO-. Therefore, all of the above compounds can act as Bronsted-Lowry acids.

Explanation

All of the given compounds, HNO2, (CH3)3NH+, HF, and CH3COOH, can act as Bronsted-Lowry acids because they are capable of donating a proton (H+) to another species. In the case of HNO2, it can donate a proton to form NO2-. (CH3)3NH+ can donate a proton to form (CH3)3N. HF can donate a proton to form F-. CH3COOH can donate a proton to form CH3COO-. Therefore, all of the above compounds can act as Bronsted-Lowry acids.

3. What is the pH of an aqueous solution at 25C in which [H+] is 0.0025 M?

2.60

2.25

-3.40

3.40

-2.60

The pH of a solution is a measure of its acidity or alkalinity. It is calculated using the formula pH = -log[H+]. In this case, the concentration of H+ ions is given as 0.0025 M. Taking the negative logarithm of this value gives a pH of 2.60.

Explanation

The pH of a solution is a measure of its acidity or alkalinity. It is calculated using the formula pH = -log[H+]. In this case, the concentration of H+ ions is given as 0.0025 M. Taking the negative logarithm of this value gives a pH of 2.60.

4. Which one of the following is the weakest acid?

HClO (Ka = 3.0 x 10-8)

HCN (Ka = 4.9 x 10-10)

Acetic Acid (Ka = 1.8 x 10-5)

HNO2 (Ka = 4.5 x 10-4)

HF (Ka = 6.8 x 10-4)

The Ka value represents the acid dissociation constant, which indicates the strength of an acid. A lower Ka value indicates a weaker acid. Among the given options, HCN has the lowest Ka value of 4.9 x 10-10, making it the weakest acid.

Explanation

The Ka value represents the acid dissociation constant, which indicates the strength of an acid. A lower Ka value indicates a weaker acid. Among the given options, HCN has the lowest Ka value of 4.9 x 10-10, making it the weakest acid.

5. Calculate the molarity of hydroxide ion in an aqueous solution that has a pOH of 5.33.

8.67

2.1 x 10-9

5.3 x 10-14

8.7 x 10-14

4.7 x 10-6

The molarity of hydroxide ion in an aqueous solution can be calculated using the equation: pOH = -log[OH-]. Rearranging the equation, we get [OH-] = 10^(-pOH). Plugging in the given pOH value of 5.33 into the equation, we get [OH-] = 10^(-5.33) = 4.7 x 10^(-6). Therefore, the molarity of hydroxide ion in the solution is 4.7 x 10^(-6).

Explanation

The molarity of hydroxide ion in an aqueous solution can be calculated using the equation: pOH = -log[OH-]. Rearranging the equation, we get [OH-] = 10^(-pOH). Plugging in the given pOH value of 5.33 into the equation, we get [OH-] = 10^(-5.33) = 4.7 x 10^(-6). Therefore, the molarity of hydroxide ion in the solution is 4.7 x 10^(-6).

6. A- is a weak base. Which equilbrium corresponds to the equilibrium constant Ka for HA?

A- + H3O+ HA + H2O

A- + OH- HOA-2

A- + H2O HA + OH-

HA + H2O H2A+ + A-

HA + H2O H3O+ + A-

The equilibrium that corresponds to the equilibrium constant Ka for HA is HA + H2O H3O+ + A-. This is because the Ka value represents the acidity constant of an acid, which is the ratio of the concentration of the dissociated form (H3O+ and A-) to the concentration of the undissociated form (HA). Therefore, the equation that includes both the dissociated and undissociated forms of HA is the one that corresponds to Ka.

Explanation

The equilibrium that corresponds to the equilibrium constant Ka for HA is HA + H2O H3O+ + A-. This is because the Ka value represents the acidity constant of an acid, which is the ratio of the concentration of the dissociated form (H3O+ and A-) to the concentration of the undissociated form (HA). Therefore, the equation that includes both the dissociated and undissociated forms of HA is the one that corresponds to Ka.

7. The pH of a 0.15 M aqueous solution of NaZ (the sodium salt of HZ) is 10.7. What is the Ka for HZ?

3.3 x 10-8

1.6 x 10-6

6.0 x 10-9

1.3 x 10-12

8.9 x 10-4

The pH of a solution is a measure of its acidity. A pH of 10.7 indicates that the solution is basic. In the given question, NaZ is the sodium salt of HZ, which means it is a conjugate base of the weak acid HZ. The Ka value represents the acid dissociation constant, which indicates the extent to which an acid dissociates in water. Since the solution is basic, it means that the concentration of the conjugate base (NaZ) is higher than the concentration of the weak acid (HZ). The Ka value for HZ will be relatively small, indicating that it is a weak acid. Among the given options, 6.0 x 10-9 is the closest value to represent a weak acid.

Explanation

The pH of a solution is a measure of its acidity. A pH of 10.7 indicates that the solution is basic. In the given question, NaZ is the sodium salt of HZ, which means it is a conjugate base of the weak acid HZ. The Ka value represents the acid dissociation constant, which indicates the extent to which an acid dissociates in water. Since the solution is basic, it means that the concentration of the conjugate base (NaZ) is higher than the concentration of the weak acid (HZ). The Ka value for HZ will be relatively small, indicating that it is a weak acid. Among the given options, 6.0 x 10-9 is the closest value to represent a weak acid.

8. In the reaction below, NH3 is acting as a(n) ______________ base, but not as a(n) ________ base.

Arrhenius, Lewis

Arrhenius, Bronsted-Lowry

Lewis, Arrhenius

Bronsted-Lowry, Lewis

Lewis, Bronsted-Lowry

NH3 is acting as a Lewis base because it donates a pair of electrons to form a coordinate bond with a metal ion or a Lewis acid. It is not acting as an Arrhenius base because it does not produce hydroxide ions (OH-) when dissolved in water.

Explanation

NH3 is acting as a Lewis base because it donates a pair of electrons to form a coordinate bond with a metal ion or a Lewis acid. It is not acting as an Arrhenius base because it does not produce hydroxide ions (OH-) when dissolved in water.

9. What is the conjugate base of OH-?

O-2

H2O

O-

O2

H3O+

The conjugate base of OH- is O-2 because when OH- loses a proton (H+), it forms O-2. This is because OH- is a hydroxide ion, which is a strong base. In the process of losing a proton, it becomes negatively charged, resulting in O-2.

Explanation

The conjugate base of OH- is O-2 because when OH- loses a proton (H+), it forms O-2. This is because OH- is a hydroxide ion, which is a strong base. In the process of losing a proton, it becomes negatively charged, resulting in O-2.

10. The base dissociation constant of ethylamine (C2H5NH2) is 6.4 x 10-4. What is the [H+] in a 1.6x10-12 M solution of ethylamine?

3.5 x 10-12

11.46

2.9 x 10-3

3.2 x 10-3

3.1 x 10-12

The base dissociation constant (Kb) is a measure of the strength of a base. It is the equilibrium constant for the reaction of the base with water to form the conjugate acid and hydroxide ions. In this case, ethylamine (C2H5NH2) is a base, and its Kb value is given as 6.4 x 10-4. To find the [H+] in a solution of ethylamine, we need to calculate the concentration of hydroxide ions (OH-) using the Kb value. Since water is amphoteric, the concentration of [OH-] will be equal to the concentration of [H+]. Therefore, the [H+] in the solution of ethylamine is 3.5 x 10-12 M.

Explanation

The base dissociation constant (Kb) is a measure of the strength of a base. It is the equilibrium constant for the reaction of the base with water to form the conjugate acid and hydroxide ions. In this case, ethylamine (C2H5NH2) is a base, and its Kb value is given as 6.4 x 10-4. To find the [H+] in a solution of ethylamine, we need to calculate the concentration of hydroxide ions (OH-) using the Kb value. Since water is amphoteric, the concentration of [OH-] will be equal to the concentration of [H+]. Therefore, the [H+] in the solution of ethylamine is 3.5 x 10-12 M.