Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 12.4, Problem 12.6CE
(a)
Interpretation Introduction
Interpretation:
Is the equilibrium constant large or small for the reverse reaction has to be explained.
(b)
Interpretation Introduction
Interpretation:
The equilibrium constant for reverse reaction has to be calculated.
(c)
Interpretation Introduction
Interpretation:
The extent of reaction has to be explained, by using equilibrium constant value.
(d)
Interpretation Introduction
Interpretation:
It has to be predicted,
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
What is the equilibrium constant expression (Ka) for the acid dissociation of hypochlorous acid? HClO(aq)+H 2 O(l) H 3 O^ + (aq)+ClO-(aq) .
a) Ka=[H 3 O^ + ]+[ClO-]-[HClO]; b) Ka=([HClOl[H 2 O]/([H 3 O^ + ][ClO-]);
c) Ka=([H 3 O^ + ][ClO-])/([HClO][H 2 O]; d) Ka=(IH 3 O^ + ][ClO-])/[HClO]
A chemical system is set up by placing some solid ammonium chloride in an ammonia solution.
The equilibrium established can be represented as follows:
NH4*(aq) + H2O(e) 2 H30*(aq) + NH3(aq)
The pH of the solution is taken, then a small amount of NaOH(aq) is added and the pH is taken
again.
What can be said about the change in pH for the solution?
The pH significantly increases because a strong base has been added to the solution.
The pH significantly decreases because a strong base has been added to the solution.
There is very little change to the pH of the solution. If anything the pH of the solution
decreases slightly.
There is very little change to the pH of the solution. If anything the pH of the solution
increases slightly.
Aluminum hydroxide dissolves in strong acid by the following reaction.
Al(OH)3(s) + 3 H3O+(aq) Al3+ (aq) + 6 H2O(l)
(a) What is the equilibrium constant expression for the reaction? (b) Use the following equilibrium constants to calculate the value of K for the above reaction.
H3O+(aq) + OH −(aq) 2 H2O(l)
K1 = 5.6 x1015
Al(OH)3(s) Al3+(aq) + 3 OH −(aq)
K2 = 1.9 x10-33
Chapter 12 Solutions
Chemistry: The Molecular Science
Ch. 12.1 - The introduction to this chapter states that at a...Ch. 12.1 - Prob. 12.2CECh. 12.2 - After a mixture of cis-2-butene and trans-2-butene...Ch. 12.2 - Prob. 12.1PSPCh. 12.2 - Prob. 12.4ECh. 12.2 - When carbon dioxide dissolves in water it reacts...Ch. 12.2 - For each of these reactions, calculate KP from Kc....Ch. 12.3 - Prob. 12.3PSPCh. 12.4 - Suppose that solid AgCl and AgI are placed in 1.0...Ch. 12.4 - Prob. 12.6CE
Ch. 12.5 - For the equilibrium 2 SO2(g) + O2(g) 2 SO3(g) Kc...Ch. 12.5 - Prob. 12.7CECh. 12.5 - Prob. 12.6PSPCh. 12.5 - Prob. 12.7PSPCh. 12.6 - Prob. 12.8CECh. 12.6 - Prob. 12.9ECh. 12.6 - Prob. 12.10CECh. 12.6 - Prob. 12.8PSPCh. 12.7 - For the ammonia synthesis reaction
⇌
Does the...Ch. 12.8 - Prob. 12.13CECh. 12 - Prob. 1QRTCh. 12 - Prob. 2QRTCh. 12 - Prob. 3QRTCh. 12 - Decomposition of ammonium dichromate is shown in...Ch. 12 - For the equilibrium reaction in Question 4, write...Ch. 12 - Indicate whether each statement below is true or...Ch. 12 - Prob. 7QRTCh. 12 - Prob. 8QRTCh. 12 - Prob. 9QRTCh. 12 - Prob. 10QRTCh. 12 - The atmosphere consists of about 80% N2 and 20%...Ch. 12 - Prob. 12QRTCh. 12 - Prob. 13QRTCh. 12 - Prob. 14QRTCh. 12 - Prob. 15QRTCh. 12 - Prob. 16QRTCh. 12 - Prob. 17QRTCh. 12 - Prob. 18QRTCh. 12 - Prob. 19QRTCh. 12 - Prob. 20QRTCh. 12 - Prob. 21QRTCh. 12 - Prob. 22QRTCh. 12 - Prob. 23QRTCh. 12 - Prob. 24QRTCh. 12 - Prob. 25QRTCh. 12 - Prob. 26QRTCh. 12 - Prob. 27QRTCh. 12 - Prob. 28QRTCh. 12 - Prob. 29QRTCh. 12 - Prob. 30QRTCh. 12 - Given these data at a certain temperature,...Ch. 12 - The vapor pressure of water at 80. C is 0.467 atm....Ch. 12 - Prob. 33QRTCh. 12 - Prob. 34QRTCh. 12 - Prob. 35QRTCh. 12 - Prob. 36QRTCh. 12 - Carbon dioxide reacts with carbon to give carbon...Ch. 12 - Prob. 38QRTCh. 12 - Prob. 39QRTCh. 12 - Prob. 40QRTCh. 12 - Nitrosyl chloride, NOC1, decomposes to NO and Cl2...Ch. 12 - Suppose 0.086 mol Br2 is placed in a 1.26-L flask....Ch. 12 - Prob. 43QRTCh. 12 - Prob. 44QRTCh. 12 - Prob. 45QRTCh. 12 - Using the data of Table 12.1, predict which of...Ch. 12 - Prob. 47QRTCh. 12 - The equilibrium constants for dissolving silver...Ch. 12 - Prob. 49QRTCh. 12 - Prob. 50QRTCh. 12 - At room temperature, the equilibrium constant Kc...Ch. 12 - Prob. 52QRTCh. 12 - Consider the equilibrium N2(g)+O2(g)2NO(g) At 2300...Ch. 12 - The equilibrium constant, Kc, for the reaction...Ch. 12 - Prob. 55QRTCh. 12 - Prob. 56QRTCh. 12 - Prob. 57QRTCh. 12 - At 503 K the equilibrium constant Kc for the...Ch. 12 - Prob. 59QRTCh. 12 - Prob. 60QRTCh. 12 - Prob. 61QRTCh. 12 - Prob. 62QRTCh. 12 - Prob. 63QRTCh. 12 - Prob. 64QRTCh. 12 - Prob. 65QRTCh. 12 - Prob. 66QRTCh. 12 - Prob. 67QRTCh. 12 - Hydrogen, bromine, and HBr in the gas phase are in...Ch. 12 - Prob. 69QRTCh. 12 - Prob. 70QRTCh. 12 - Prob. 71QRTCh. 12 - Prob. 72QRTCh. 12 - Prob. 73QRTCh. 12 - Prob. 74QRTCh. 12 - Consider the system
4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) +...Ch. 12 - Prob. 76QRTCh. 12 - Predict whether the equilibrium for the...Ch. 12 - Prob. 78QRTCh. 12 - Prob. 79QRTCh. 12 - Prob. 80QRTCh. 12 - Prob. 81QRTCh. 12 - Prob. 82QRTCh. 12 - Prob. 83QRTCh. 12 - Prob. 84QRTCh. 12 - Prob. 85QRTCh. 12 - Prob. 86QRTCh. 12 - Prob. 87QRTCh. 12 - Consider the decomposition of ammonium hydrogen...Ch. 12 - Prob. 89QRTCh. 12 - Prob. 90QRTCh. 12 - Prob. 91QRTCh. 12 - Prob. 92QRTCh. 12 - Prob. 93QRTCh. 12 - Prob. 94QRTCh. 12 - Prob. 95QRTCh. 12 - Prob. 96QRTCh. 12 - Prob. 97QRTCh. 12 - Prob. 98QRTCh. 12 - Prob. 99QRTCh. 12 - Prob. 100QRTCh. 12 - Two molecules of A react to form one molecule of...Ch. 12 - Prob. 102QRTCh. 12 - In Table 12.1 (←Sec. 12-3a) the equilibrium...Ch. 12 - Prob. 104QRTCh. 12 - Prob. 105QRTCh. 12 - Prob. 106QRTCh. 12 - Prob. 107QRTCh. 12 - Which of the diagrams for Questions 107 and 108...Ch. 12 - Draw a nanoscale (particulate) level diagram for...Ch. 12 -
The diagram represents an equilibrium mixture for...Ch. 12 - The equilibrium constant, Kc, is 1.05 at 350 K for...Ch. 12 - For the reaction in Question 111, which diagram...Ch. 12 - Prob. 113QRTCh. 12 - Prob. 114QRTCh. 12 - Prob. 115QRTCh. 12 - For the equilibrium...Ch. 12 - Prob. 117QRTCh. 12 - Prob. 119QRTCh. 12 - Prob. 120QRTCh. 12 - When a mixture of hydrogen and bromine is...Ch. 12 - Prob. 122QRTCh. 12 - Prob. 123QRTCh. 12 - Prob. 124QRTCh. 12 - Prob. 125QRTCh. 12 - Prob. 12.ACPCh. 12 - Prob. 12.BCPCh. 12 - Prob. 12.CCPCh. 12 - Prob. 12.DCPCh. 12 - Prob. 12.ECPCh. 12 - Prob. 12.FCP
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
- Complete each of these reactions by filling in the blanks. Predict whether each reaction is product-favored or reactant-favored, and explain your reasoning. (a) _________ (aq) + Br(aq) NH3(aq) + HBr(aq) (b) CH3COOH(aq) + CN(aq) ________ (aq) + HCN(aq) (c) ________ (aq)+H2O () NH3(aq) + OH(aq)arrow_forwardFollow the directions of Question 19 for the following acids: (a) hypochlorous acid (b) formic acid, HCHO2 (c) acetic acid, HC2H3O2 (d) hydrobromic acid (e) sulfurous acidarrow_forwardMost naturally occurring acids are weak acids. Lactic acid is one example. CH3CH(OH)CO2H(s)+H2O(l)H3O+(aq)+CH3CH(OH)CO2(aq) If you place some lactic acid in water, it will ionize to a small extent, and an equilibrium will be established. Suggest some experiments to prow that this is a weak acid and that the establishment of equilibrium is a reversible process.arrow_forward
- For conjugate acidbase pairs, how are Ka and Kb related? Consider the reaction of acetic acid in water CH3CO2H(aq)+H2O(l)CH3CO2(aq)+H3O+(aq) where Ka = 1.8 105 a. Which two bases are competing for the proton? b. Which is the stronger base? c. In light of your answer to part b. why do we classify the acetate ion (CH3CO2) as a weak base? Use an appropriate reaction to justify your answer. In general, as base strength increases, conjugate acid strength decreases. Explain why the conjugate acid of the weak base NH3 is a weak acid. To summarize, the conjugate base of a weak acid is a weak base and the conjugate acid of a weak base is a weak acid (weak gives you weak). Assuming Ka for a monoprotic strong acid is 1 106, calculate Kb for the conjugate base of this strong acid. Why do conjugate bases of strong acids have no basic properties in water? List the conjugate bases of the six common strong acids. To tie it all together, some instructors have students think of Li+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+ as the conjugate acids of the strong bases LiOH, KOH. RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. Although not technically correct, the conjugate acid strength of these cations is similar to the conjugate base strength of the strong acids. That is, these cations have no acidic properties in water; similarly, the conjugate bases of strong acids have no basic properties (strong gives you worthless). Fill in the blanks with the correct response. The conjugate base of a weak acid is a_____base. The conjugate acid of a weak base is a_____acid. The conjugate base of a strong acid is a_____base. The conjugate acid of a strong base is a_____ acid. (Hint: Weak gives you weak and strong gives you worthless.)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_forwardWrite the equilibrium constant expression for this reaction: H;PO,(aq) - 3 H (aq)+PO (aq) ?arrow_forward
- 8. (a) HA(aq) is a weak acid with a dissociation constant, Ka, of 8.8 x 10−12. What is the pH of a 0.022 M solution of A−(aq)? The temperature is 25 ◦C. (b) For the reaction A(g) =A(l), the equilibrium constant is 0.666 at 25.0 ◦C and 0.222 at 75.0 ◦C. Making the approximation that the entropy and enthalpy changes of this reaction do not change with temperature, at what temperature will the equilibrium constant be equal to 0.777?arrow_forwardWrite the equilibrium constant expression, K, for the following reaction taking place in dilute aqueous solution.ClO- (aq) + H2O (l)HClO (aq) + OH- (aq)arrow_forwardWrite the equilibrium constant expression for this reaction: CH;Cl(aq)+OH (aq) → CH;OH(aq)+Cl (aq)arrow_forward
- The active ingredient of bleach such as Clorox is sodium hypochlorite (NaClO). Its conjugate acid, hypochlorous acid (HClO), has a Ka of 3.0 × 10^–8. (a) The undiluted bleach contains roughly 1 M NaClO. Calculate the pH of 1 M NaClO solution. (b) Some applications require extremely diluted bleach solution, such as swimming pools. Suppose the solution in (a) is diluted by 10,000 -fold. Calculate the pH of the diluted solution, and demonstrate that you can still neglect the autoionization of water in your calculation. (c) Suppose the solution in (a) is diluted by 1 million-fold, briefly explain how your approach will be different. Write the equation with [H3O+] as the unknown, but you do not need to solve it.arrow_forwardCalculate the pH of 0.10 M (COOH)2 (aq), oxalic acid. Ka1 = 5.9 × 10–2 ; Ka2 = 6.4 × 10–5.arrow_forwardGiven 0.01 M solutions of each of the following acids, which solution would have the lowest pH? -117 Hypoiodous acid (HOI), K = 2.3 x 10 Hypobromous acid (HOBr), K = 2.5 x 10 Lactic acid (HC₂H₂O₂), K = 1.3 x 10 Chlorous acid (HClO₂), K = 1.1 x 10²arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY