Concept explainers
Classify each of the following species as a Brønsted acid or base, or both: (a) H2O, (b) OH−, (c) H3O+, (d) NH3, (e)

Interpretation: Given set of species has to be classified as Bronsted acid or base, or both.
Concept Introduction: Bronsted's definition is based on the chemical reaction that occurs when both acids and bases are added with each other. In Bronsted's theory acid donates proton, while base accepts proton from acid resulting in the formation of water.
Example: Consider the following reaction.
Hydrogen chloride donates a proton, and hence it is a Bronsted acid. Ammonia accepts a proton, and hence it is a Bronsted base.
Bronsted base accepts a proton to give a protonated species known as conjugate acid and Bronsted acid loses a proton deprotonated species is known as conjugate base. When a proton is removed the resulting species will have a negative charge and when a proton is added the resulting species will have a positive charge.
Answer to Problem 15.3QP
The species (a) is both Bronsted acid and Bronsted base.
The species (b) is Bronsted base.
The species (c) is Bronsted acid.
The species (d) is Bronsted base.
The species (e) is Bronsted acid.
The species (f) is Bronsted base.
The species (g) is Bronsted base.
The species (h) is Bronsted base.
The species (i) is Bronsted acid.
The species (j) is Bronsted acid.
Explanation of Solution
(a)
To classify:
To identify the species as Bronsted acid.
Water molecule loses a proton to form a conjugate base as shown above. Therefore, water can act as Bronsted acid.
To identify the species as Bronsted base.
Water molecule accepts a proton to form hydronium ion. Therefore, water can act as Bronsted base.
From this we can conclude that water can act as both Bronsted acid and Bronsted base.
(b)
To classify:
To identify the species as Bronsted acid.
Hydroxide ion cannot lose a proton to form a conjugate base. Therefore, hydroxide ion cannot act as Bronsted acid.
To identify the species as Bronsted base.
Hydroxide ion accepts a proton to form water molecule. Therefore, hydroxide ion can act as Bronsted base.
From this we can conclude that hydroxide ion can only act as Bronsted base.
(c)
To classify:
To identify the species as Bronsted acid.
The hydronium ion can lose a proton to form a conjugate base as shown above. Therefore, hydronium ion can act as Bronsted acid.
To identify the species as Bronsted base.
Hydronium ion cannot accept proton to form a conjugate acid.
From this we can conclude that hydronium ion can act only as Bronsted acid.
(d)
To classify:
To identify the species as Bronsted acid.
Ammonia cannot lose a proton to form a conjugate base. Therefore, ammonia cannot act as Bronsted acid.
To identify the species as Bronsted base.
Ammonia accepts a proton to form ammonium ion. Therefore, ammonia ion can act as Bronsted base.
From this we can conclude that ammonia can act only as Bronsted base.
(e)
To classify:
To identify the species as Bronsted acid.
The ammonium ion can lose a proton to form a conjugate base as shown above. Therefore ammonium ion can act as Bronsted acid.
To identify the species as Bronsted base.
Ammonium ion cannot accept proton to form a conjugate acid.
From this we can conclude that ammonium ion can act only as Bronsted acid.
(f)
To classify:
To identify the species as Bronsted acid.
To identify the species as Bronsted base.
From this we can conclude that
(g)
To classify:
To identify the species as Bronsted acid.
To identify the species as Bronsted base.
From this we can conclude that
(h)
To classify:
To identify the species as Bronsted acid.
Explanation:
To identify the species as Bronsted base.
From this we can conclude that
(i)
To classify:
To identify the species as Bronsted acid.
The
To identify the species as Bronsted base.
From this we can conclude that
(j)
To classify:
To identify the species as Bronsted acid.
The
To identify the species as Bronsted base.
From this we can conclude that
The given set of species are classified as Bronsted acid or base, or both.
Want to see more full solutions like this?
Chapter 15 Solutions
CHEMISTRY-ALEKS 360 ACCESS
- true or false, given that a 20.00 mL sample of NaOH took 24.15 mL of 0.141 M HCI to reach the endpoint in a titration, the concentration of the NaOH is 1.17 M.arrow_forwardin the bromothymol blue experiment, pKa was measured. A closely related compound has a Ka of 2.10 x 10-5. What is the pKa?a) 7.1b) 4.7c) 2.0arrow_forwardcalculate the equilibrium concentration of H2 given that K= 0.017 at a constant temperature for this reaction. The inital concentration of HBr is 0.050 M.2HBr(g) ↔ H2(g) + Br2(g)a) 4.48 x 10-2 M b) 5.17 x 10-3 Mc) 1.03 x 10-2 Md) 1.70 x 10-2 Marrow_forward
- true or falsegiven these two equilibria with their equilibrium constants:H2(g) + CI2(l) ↔ 2HCI(g) K= 0.006 CI2(l) ↔ CI2(g) K= 0.30The equilibrium contstant for the following reaction is 1.8H2(g) + CI2 ↔ 2HCI(g)arrow_forwardI2(g) + CI2(g) ↔ 2ICIK for this reaction is 81.9. Find the equilibrium concentration of I2 if the inital concentration of I2 and CI2 are 0.010 Marrow_forwardtrue or false,the equilibrium constant for this reaction is 0.50.PCI5(g) ↔ PCI3(g) + CI2(g)Based on the above, the equilibrium constant for the following reaction is 0.25.2PCI5(g) ↔. 2PCI3(g) + 2CI2(g)arrow_forward
- true or false, using the following equilibrium, if carbon dioxide is added the equilibrium will shift toward the productsC(s) + CO2(g) ↔ 2CO(g)arrow_forward2S2O2/3- (aq) + I2 (aq) ---> S4O2/6- (aq) +2I- (aq) Experiment I2 (M) S2O3- (M) Initital Rate (M/s) 1 0.01 0.01 0.0004 2 0.01 0.02 0.0004 3 0.02 0.01 0.0008 Calculate the overall order for this reaction using the table data a) 3b) 0c) 2d) 1arrow_forwardthe decomposition of N2O5 is the first order with a half-life of 1.98 minutes. If the inital concentration of N2O5 is 0.200 M, what is the concentration after 6 minutes?a) 0.612 Mb) 0.035 Mc) 0.024 Md) 0.100 Marrow_forward
- 20.00 mL of 0.150 M HCI is titrated with 0.075 M NaOH. What volume of NaOH is needed?a) 50 mLb) 20 mLc) 40 mLd) 26.66 mLarrow_forward20.00 mL of 0.150 M NaOH is titrated with 37.75 mL of HCI. What is the molarity of the HCI?a) 0.150 Mb) 0.079 Mc) 0.025 Md) 0.050 Marrow_forwardin the following reaction, the OH- acts as which of these?NO2- (aq) + H2O (l) ⇌ OH- (aq) + HNO2 (aq)a) not a weak acidb) basec) acidarrow_forward
- Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningWorld of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning





