EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119227946
Author: Willard
Publisher: VST
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 16, Problem 75AE
Interpretation Introduction
Interpretation:
For the below reaction
Concept Introduction:
For equilibrium
Expression for equilibrium constant is written as follows:
Here,
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
0.0500 M HF (Ka= 7.20 x 10–4)
determine the equilibrium molar concentration of H3O+ and equilibrium molar concentration of HF (in 3 sig. figures).
Write the dissociation reaction and the corresponding K equilibrium expression for HCN
Explain why we can simplify the calculations for K when the value of K is very small (less
4.1
than -10*).
4.2
HCN will react very slightly with water with a K = 6.2 x 10-10.
HCN(aq) + H,O()
H,O* (aq) and CN" (aq)
A chemist starts with a 0.50 M HCN solution. What is the equilibrium concentration of
H;O*?. Then use the concentration of H3O* to calculate the pH of the solution with the equation
pH = -log[H3O*] (you can do this!!!). Put your final answers in the box. If you need more space
to show your work, you can use the backside of this
page.
[H;O'] at equilibrium
pH = -log(H,O]
pH =
% of HCN that
reacted
Chapter 16 Solutions
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
Ch. 16.1 - Prob. 16.1PCh. 16.2 - Prob. 16.2PCh. 16.3 - Prob. 16.3PCh. 16.3 - Prob. 16.4PCh. 16.3 - Prob. 16.5PCh. 16.3 - Prob. 16.6PCh. 16.4 - Prob. 16.7PCh. 16.4 - Prob. 16.8PCh. 16.5 - Prob. 16.9PCh. 16.5 - Prob. 16.10P
Ch. 16.6 - Prob. 16.11PCh. 16.6 - Prob. 16.12PCh. 16.7 - Prob. 16.13PCh. 16.7 - Prob. 16.14PCh. 16.7 - Prob. 16.15PCh. 16.8 - Prob. 16.16PCh. 16 - Prob. 1RQCh. 16 - Prob. 2RQCh. 16 - Prob. 3RQCh. 16 - Prob. 4RQCh. 16 - Prob. 5RQCh. 16 - Prob. 6RQCh. 16 - Prob. 7RQCh. 16 - Prob. 8RQCh. 16 - Prob. 9RQCh. 16 - Prob. 10RQCh. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 14RQCh. 16 - Prob. 15RQCh. 16 - Prob. 16RQCh. 16 - Prob. 17RQCh. 16 - Prob. 18RQCh. 16 - Prob. 19RQCh. 16 - Prob. 20RQCh. 16 - Prob. 21RQCh. 16 - Prob. 22RQCh. 16 - Prob. 23RQCh. 16 - Prob. 24RQCh. 16 - Prob. 25RQCh. 16 - Prob. 26RQCh. 16 - Prob. 27RQCh. 16 - Prob. 1PECh. 16 - Prob. 2PECh. 16 - Prob. 3PECh. 16 - Prob. 4PECh. 16 - Prob. 5PECh. 16 - Prob. 6PECh. 16 - Prob. 7PECh. 16 - Prob. 8PECh. 16 - Prob. 9PECh. 16 - Prob. 10PECh. 16 - Prob. 11PECh. 16 - Prob. 12PECh. 16 - Prob. 13PECh. 16 - Prob. 14PECh. 16 - Prob. 15PECh. 16 - Prob. 16PECh. 16 - Prob. 17PECh. 16 - Prob. 18PECh. 16 - Prob. 19PECh. 16 - Prob. 20PECh. 16 - Prob. 21PECh. 16 - Prob. 22PECh. 16 - Prob. 23PECh. 16 - Prob. 24PECh. 16 - Prob. 25PECh. 16 - Prob. 26PECh. 16 - Prob. 27PECh. 16 - Prob. 28PECh. 16 - Prob. 29PECh. 16 - Prob. 30PECh. 16 - Prob. 31PECh. 16 - Prob. 32PECh. 16 - Prob. 33PECh. 16 - Prob. 34PECh. 16 - Prob. 35PECh. 16 - Prob. 36PECh. 16 - Prob. 37PECh. 16 - Prob. 38PECh. 16 - Prob. 39PECh. 16 - Prob. 40PECh. 16 - Prob. 41PECh. 16 - Prob. 42PECh. 16 - Prob. 43PECh. 16 - Prob. 44PECh. 16 - Prob. 45PECh. 16 - Prob. 46PECh. 16 - Prob. 47PECh. 16 - Prob. 48PECh. 16 - Prob. 49AECh. 16 - Prob. 50AECh. 16 - Prob. 51AECh. 16 - Prob. 52AECh. 16 - Prob. 53AECh. 16 - Prob. 54AECh. 16 - Prob. 55AECh. 16 - Prob. 56AECh. 16 - Prob. 57AECh. 16 - Prob. 58AECh. 16 - Prob. 59AECh. 16 - Prob. 60AECh. 16 - Prob. 61AECh. 16 - Prob. 62AECh. 16 - Prob. 63AECh. 16 - Prob. 64AECh. 16 - Prob. 65AECh. 16 - Prob. 66AECh. 16 - Prob. 67AECh. 16 - Prob. 68AECh. 16 - Prob. 69AECh. 16 - Prob. 70AECh. 16 - Prob. 71AECh. 16 - Prob. 72AECh. 16 - Prob. 73AECh. 16 - Prob. 74AECh. 16 - Prob. 75AECh. 16 - Prob. 76AECh. 16 - Prob. 77AECh. 16 - Prob. 78AECh. 16 - Prob. 79AECh. 16 - Prob. 80AECh. 16 - Prob. 81AECh. 16 - Prob. 83AECh. 16 - Prob. 84AECh. 16 - Prob. 85AECh. 16 - Prob. 86CECh. 16 - Prob. 87CECh. 16 - Prob. 88CECh. 16 - Prob. 89CE
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
- Write an equation for an equilibrium system that would lead to the following expressions (ac) for K. (a) K=(Pco)2 (PH2)5(PC2H6)(PH2O)2 (b) K=(PNH3)4 (PO2)5(PNO)4 (PH2O)6 (c) K=[ ClO3 ]2 [ Mn2+ ]2(Pcl2)[ MNO4 ]2 [ H+ ]4 ; liquid water is a productarrow_forwardThe concentration of barium in a saturated solution of barium sulfate at a particular temperature is 1.2 g/mL. Calculate Ksp at this temperature.arrow_forwardCalculate the equilibrium molar concentrations of trichloroacetic acid in an aqueous solution that contains 284.9 mg of trichloroacetic acid,Cl3CCOOH (163.4 g/mol), in 15 mL (the acid is 73.0% ionized in water). Indicate unitsarrow_forward
- You are on one of Saturn's many moons and discover a substance, Y2O that undergoes autoionization just like water on Earth. Y2O + Y2O ⇌ Y3O+ + OY– Through a series of experiments, you determine the equilibrium constant (Keq) for this autoionization reaction at various temperatures. The value of Keq at 28.7ºC is 7.6 x 10-14. What is the pY of the pure substance at this temperature? Report your answer to the hundreths placearrow_forwardExplain why we can simplify the calculations for K when the value of K is very small (less than ~104). 4.1 4.2 HCN will react very slightly with water with aK = 6.2 x 101º. HCN(aq) + H,O() = H;O* (aq) and CN¯ (aq) A chemist starts with a 0.50 M HCN solution. What is the equilibrium concentration of H,O*?. Then use the concentration of H30* to calculate the pH of the solution with the equation pH = -log[H;O*] (you can do this!!!). Put your final answers in the box. If you need more space to show your work, ou can use the backside of this page. %3D [H,O*] at equilibrium M. pH = -log[H;O*] pH = % of HCN that reactedarrow_forwardState the standard (reference) concentration for NH3 appearing in the equilibrium of the following reaction: NH3 (g) + H3O+ (aq) ⇒ NH4+ (aq) + H2O (l) Concentration is 1.2e-2 Marrow_forward
- A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) + H₂O(1) ⇒ H₂O* (aq) + A¯(aq) The equilibrium concentrations of the reactants and products are [HA] = 0.100 M, [H3O+] = 3.00 × 10−4 M, and [A-] = 3.00 × 10-4 M. Calculate the Ką value for the acid HA. Ka =arrow_forwardermine the pH of a solution by constructing a BCA table, constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. Complete Parts 1-4 before submitting your answer. 2 NEXT > Two solutions are mixed: 40.0 mL of 0.500 M NH, and 25.0 mL of 0.300 M HCI. Fill in the table with the appropriate value for each involved species to determine the moles of reactant and product after the reaction of the acid and base.. NH3(aq) NH, (aq) (be),H Before (mol) Change (mol) After (mol) 5 RESET 40.0 0.500 25.0 0.300 0.300 -x 0.0075 -0.0075 0.0100 -0.0100 0.0125 -0.0125 0.0200 -0.0200 Based on the result of the acid-base reaction (Part 1), set up the ICE table in order to determine the unknown concentrations of all reactants and products. NH3(aq) H,O(1) OH (aq) NH, (aq) Initial (M) Change (M) Equilibrium (M) 5 RESET 0.500 0.300 0.192 0.313 0.115 0.500 -x 0.300 -x 0.192 +* 0.192 -x 0.500 + x 0.300 +x 0.313 + x 0.313 -x 0.115 +x 0.115-xarrow_forwardDetermine the pOH of a weak acid (HA) solution by constructing an ICE table and using this information to determine the pOH of the solution. Complete Parts 1-2 before submitting your answer. 1 A 0.200 M aqueous solution of HA is 3.0% dissociated at equilibrium. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) HA(aq) + H₂O(l) 2 H3O+ (aq) 2 NEXT > + A-(aq)arrow_forward
- Given the two reactions 3. PbCl 2 (aq)⇌Pb 2+ (aq)+2Cl − (aq), K 3 = 1.80×10−10, and4. AgCl(aq)⇌Ag + (aq)+Cl − (aq), K 4 = 1.13×10−4,what is the equilibrium constant K final for the following reaction?PbCl 2 (aq)+2Ag + (aq)⇌2AgCl(aq)+Pb 2+ (aq)arrow_forwardA weak base has K, = 2.4 x 10-4 at 25°C. By adding a strong acid to solution, there is an acid-base reaction. Calculate the pH at which [B] = [BHt].arrow_forwardIn a lab experiment, a student manages to dissolve 0.00250 moles of a mysterious compound AB into 1 L of water. The mysterious compound dissociates using the following equation: AB(S) ⇌ A(aq) + B(aq) What is the value of the Ksp for this compound? (Please provide your answer to 3 significant figures in scientific notation such as 1.23*10^-5.)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Chemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
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
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Types of Matter: Elements, Compounds and Mixtures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=dggHWvFJ8Xs;License: Standard YouTube License, CC-BY