Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Write the acid dissociation reaction for formic acid, HCO2H.
Consider the following acidic equilibrium: H₂CO₃(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + H₃O⁺(aq). If you add NaHCO₃ to this solution, which of the following will occur?
A) The reaction quotient will decrease.
B) The reaction will shift in the reverse direction.
C) The equilibrium constant will increase.
D) No changes to the equilibrium positions will take place.
Write the acid dissociation constant expression for nitrous acid, HNO2.
Chapter 15 Solutions
Chemistry: An Atoms-Focused Approach
Ch. 15 - Prob. 15.1VPCh. 15 - Prob. 15.2VPCh. 15 - Prob. 15.3VPCh. 15 - Prob. 15.4VPCh. 15 - Prob. 15.5VPCh. 15 - Prob. 15.6VPCh. 15 - Prob. 15.7VPCh. 15 - Prob. 15.8VPCh. 15 - Prob. 15.9VPCh. 15 - Prob. 15.10VP
Ch. 15 - Prob. 15.11QACh. 15 - Prob. 15.12QACh. 15 - Prob. 15.13QACh. 15 - Prob. 15.14QACh. 15 - Prob. 15.15QACh. 15 - Prob. 15.16QACh. 15 - Prob. 15.17QACh. 15 - Prob. 15.18QACh. 15 - Prob. 15.19QACh. 15 - Prob. 15.20QACh. 15 - Prob. 15.21QACh. 15 - Prob. 15.22QACh. 15 - Prob. 15.23QACh. 15 - Prob. 15.24QACh. 15 - Prob. 15.25QACh. 15 - Prob. 15.26QACh. 15 - Prob. 15.27QACh. 15 - Prob. 15.28QACh. 15 - Prob. 15.29QACh. 15 - Prob. 15.30QACh. 15 - Prob. 15.31QACh. 15 - Prob. 15.32QACh. 15 - Prob. 15.34QACh. 15 - Prob. 15.35QACh. 15 - Prob. 15.36QACh. 15 - Prob. 15.37QACh. 15 - Prob. 15.38QACh. 15 - Prob. 15.39QACh. 15 - Prob. 15.40QACh. 15 - Prob. 15.41QACh. 15 - Prob. 15.42QACh. 15 - Prob. 15.43QACh. 15 - Prob. 15.44QACh. 15 - Prob. 15.45QACh. 15 - Prob. 15.46QACh. 15 - Prob. 15.47QACh. 15 - Prob. 15.48QACh. 15 - Prob. 15.49QACh. 15 - Prob. 15.50QACh. 15 - Prob. 15.51QACh. 15 - Prob. 15.52QACh. 15 - Prob. 15.53QACh. 15 - Prob. 15.54QACh. 15 - Prob. 15.55QACh. 15 - Prob. 15.56QACh. 15 - Prob. 15.57QACh. 15 - Prob. 15.58QACh. 15 - Prob. 15.59QACh. 15 - Prob. 15.61QACh. 15 - Prob. 15.62QACh. 15 - Prob. 15.63QACh. 15 - Prob. 15.64QACh. 15 - Prob. 15.65QACh. 15 - Prob. 15.66QACh. 15 - Prob. 15.67QACh. 15 - Prob. 15.68QACh. 15 - Prob. 15.69QACh. 15 - Prob. 15.70QACh. 15 - Prob. 15.71QACh. 15 - Prob. 15.72QACh. 15 - Prob. 15.73QACh. 15 - Prob. 15.74QACh. 15 - Prob. 15.75QACh. 15 - Prob. 15.76QACh. 15 - Prob. 15.77QACh. 15 - Prob. 15.78QACh. 15 - Prob. 15.79QACh. 15 - Prob. 15.80QACh. 15 - Prob. 15.81QACh. 15 - Prob. 15.82QACh. 15 - Prob. 15.83QACh. 15 - Prob. 15.84QACh. 15 - Prob. 15.85QACh. 15 - Prob. 15.86QACh. 15 - Prob. 15.87QACh. 15 - Prob. 15.88QACh. 15 - Prob. 15.89QACh. 15 - Prob. 15.90QACh. 15 - Prob. 15.91QACh. 15 - Prob. 15.92QACh. 15 - Prob. 15.93QACh. 15 - Prob. 15.94QACh. 15 - Prob. 15.95QACh. 15 - Prob. 15.96QACh. 15 - Prob. 15.97QACh. 15 - Prob. 15.98QACh. 15 - Prob. 15.99QACh. 15 - Prob. 15.100QACh. 15 - Prob. 15.101QACh. 15 - Prob. 15.102QACh. 15 - Prob. 15.103QACh. 15 - Prob. 15.104QACh. 15 - Prob. 15.105QACh. 15 - Prob. 15.106QACh. 15 - Prob. 15.107QACh. 15 - Prob. 15.108QACh. 15 - Prob. 15.109QACh. 15 - Prob. 15.110QACh. 15 - Prob. 15.111QACh. 15 - Prob. 15.112QACh. 15 - Prob. 15.113QACh. 15 - Prob. 15.114QACh. 15 - Prob. 15.115QACh. 15 - Prob. 15.116QACh. 15 - Prob. 15.117QACh. 15 - Prob. 15.118QACh. 15 - Prob. 15.119QACh. 15 - Prob. 15.120QACh. 15 - Prob. 15.121QACh. 15 - Prob. 15.122QACh. 15 - Prob. 15.123QACh. 15 - Prob. 15.124QACh. 15 - Prob. 15.125QACh. 15 - Prob. 15.126QACh. 15 - Prob. 15.127QACh. 15 - Prob. 15.128QACh. 15 - Prob. 15.129QACh. 15 - Prob. 15.130QACh. 15 - Prob. 15.131QACh. 15 - Prob. 15.132QACh. 15 - Prob. 15.133QACh. 15 - Prob. 15.134QACh. 15 - Prob. 15.135QACh. 15 - Prob. 15.136QACh. 15 - Prob. 15.137QACh. 15 - Prob. 15.138QACh. 15 - Prob. 15.139QACh. 15 - Prob. 15.140QACh. 15 - Prob. 15.141QACh. 15 - Prob. 15.142QACh. 15 - Prob. 15.143QACh. 15 - Prob. 15.144QACh. 15 - Prob. 15.145QACh. 15 - Prob. 15.146QACh. 15 - Prob. 15.147QACh. 15 - Prob. 15.148QACh. 15 - Prob. 15.149QACh. 15 - Prob. 15.150QACh. 15 - Prob. 15.151QACh. 15 - Prob. 15.152QA
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- The reaction just described is reversible. Deprotonation of the conjugate acid of an organic base by water provides another example of simultaneous making and breaking of sigma bonds. Thus, in the deprotonation of anilinium ion by water, the base is water, which has unshared electrons on the ________ atom. The acid is ________ ion. A pair of ________ electrons on the oxygen atom of water is pushed toward the ________ atom. Simultaneously, the pair of ________ electrons between the hydrogen and ________ atom of the anilinium ion is pushed toward the ________ atom. Thus, the oxygen- ________ sigma bond is made and a hydrogen- ________ sigma bond is broken. The nitrogen atom, which possessed a positive charge, is now ________, and the oxygen atom, which was neutral, now possesses a formal ________ charge.arrow_forwardWrite the acid ionization constant expression for the ionization of each of the following monoprotic acids. a. HCN (hydrocyanic acid) b. HC6H7O6 (ascorbic acid)arrow_forwardWrite the chemical equation and the expression for the equilibrium constant, and calculate Kb for the reaction of each of the following ions as a base. (a) sulfate ion (b) citrate ionarrow_forward
- Write the acid dissociation constant expression (K,) for the dissociation of HCN.arrow_forwardFormic acid dissociates reversibly according to the following equation. \ HCOOH(aq) ⇌ HCOO-(aq) + H+(aq) Write the law of mass action for this equilibrium.K = Is the equilibrium constant in (a) a Ka? Briefly explain. If a solution is initially 0.20 mol/L HCOOH and the equilibrium constant in (a) is 1.8 × 10-4, what is the concentration of hydrogen ions at equilibrium? Show your work. Use the value obtained in (c) to find the solution’s pH and pOH at 25 °C. If you were unable to complete (c), you may assume a 0.0030 mol/L concentration of hydrogen ions. Would the addition of sodium formate (NaHCOO) to this solution increase pH? Explain.arrow_forwardRefer to the following ionization of hypochlorous acid: HOCI (aq) + H₂O (1) H3O+ (aq) + OCI (aq) a. The acid-dissociation constant, Ka, for hypochlorous acid, HOCI, is 3.5 X 10-8. Calculate the equilibrium concentrations of H3O+, OCI, and HOCI if the initial concentration of HOCI is 0.125 M. b. Calculate the pH and pOH of the above solution once equilibrium is established.arrow_forward
- The acid-dissociation constant for hypochlorous acid (HClO) is 3.0 × 10-8. Calculatethe concentrations of H3O+, ClO- and HClO at equilibrium if the initial concentration of HClO is 0.0075 Marrow_forwardThe preparaons of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HCN is a weak acid. acids: 1 mol of NaOH is added to 1.0 L of a 0.7 M HCN bases: solution. other: 0.57 mol of HCl is added to acids: 1.0 L of a solution that is bases: I 1.5M in both HCN and NaCN. other: I Explanation Check 2021 McGraw.HiN Education. All Rights Reserved Terms of Use Privacy I Accessibility acer 8. O Oarrow_forwardAcetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions: HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone protons, H+ (aq), do not exist!). (d) When starting with completely un-dissociated acetic acid, is it accurate to assume that [HOAc]0 = [HOAc]eq? Why or why not? (e) A highly concentrated acetic acid solution contains 15.0M acetic acid at equilibrium. What are the equilibrium concentrations of the hydronium and acetate ions in this solution? (f) Creating the concentrated acetic acid solution by dissolving liquid HOAc in water raises the temperature of the water by about 5°C from room temperature. At 50°C, do you expect the solution to contain more or less acetate ion OAc– than what you calculated in (c)? Why?arrow_forward
- Acetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions: HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone protons, H+ (aq), do not exist!). (a) Write the equilibrium constant expression for the dissociation of acetic acid. (b) Vinegar sold commercially is typically 0.8 − 1.0 M acetic acid. A 1.00 M solution of acetic acid is measured by its pH to have an equilibrium concentration of 4.19×10−3 M for both acetate ions and hydronium ions at room temperature. Assuming [HOAc]0 = 1.00M, what is the equilibrium concentration of undissociated acetic acid [HOAc]eq to the correct number of significant figures? (c) What is the value of the equilibrium constant Keq for the dissociation according to the concentrations from part (b)? (d) When starting with completely un-dissociated…arrow_forwardHydrazoic acid, HN3, has an acid dissociation constant of 2.5 x 10-5. Calculate the equilibrium concentrations of all substances if the initial concentration of HN3 is 0.0750 M. Determine the pH of the solution. Would a 0.0750 M solution of HBr have a higher or lower pH than the 0.0750 M HN3 solution? Explain why.arrow_forwardThe pH scale for acidity is defined by pH log H| where H"| is the concentration of hydrogen ions measured in moles per liter (M). (A) The pH of Drano is 13.3. Calculate the concentration of hydrogen ions in moles per liter (M). Note: You may have to type in a lot of zeros. For example, an answer could be something like this 0.000000000000123. Or you can type your answer in scientific notation, such as 1.23*10-13 [H*] = (B) The pH of lemon juice is 2.0. Calculate the concentration of hydrogen ions in moles Niter (M). per [H*] =arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co
Introductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage Learning
General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning
Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Introductory Chemistry: An Active Learning Approa...
Chemistry
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Cengage Learning
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
General Chemistry | Acids & Bases; Author: Ninja Nerd;https://www.youtube.com/watch?v=AOr_5tbgfQ0;License: Standard YouTube License, CC-BY