Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter U4, Problem C16.3RE
Interpretation Introduction
Interpretation:
The reason due to which a titration procedure helps in identifying the concentration of hydrogen ion in a sample solution should be stated.
Concept Introduction:
The reaction of an acid with a base results in the formation of salt and water. This type of reaction is known as neutralization reaction.
Expert Solution & Answer
Answer to Problem C16.3RE
With the help of the titration, the equivalence point can be found out and from that the concentration of hydrogen ions can be identified.
Explanation of Solution
Titration is the process in which the unknown concentration of an acid with known volume is titrated with known concentration of a base. The measurement of the hydrogen ion concentration can be used to find out the location of equivalence point that is equal to volume of base used by an acid.
In the titration procedure, an acid is titrated with a base in which a neutralization reaction takes place. The volume used by a base is equal to the volume used by an acid if the reaction ratio is 1:1. The remaining amount of acid is calculated by subtracting of reacted amount by initial amount. The number of mole of acid is divided by the total volume (base + acid) and the concentration of the acid would be observed. This is equal to the concentration of the hydrogen ions.
Conclusion
With the help of the titration, the equivalence point can be found out and from that the concentration of hydrogen ions can be identified.
Chapter U4 Solutions
Living by Chemistry
Ch. U4.68 - Prob. 1TAICh. U4.68 - Prob. 1ECh. U4.68 - Prob. 2ECh. U4.68 - Prob. 3ECh. U4.68 - Prob. 4ECh. U4.68 - Prob. 5ECh. U4.68 - Prob. 6ECh. U4.69 - Prob. 1TAICh. U4.69 - Prob. 1ECh. U4.69 - Prob. 2E
Ch. U4.69 - Prob. 3ECh. U4.69 - Prob. 4ECh. U4.69 - Prob. 5ECh. U4.70 - Prob. 1TAICh. U4.70 - Prob. 1ECh. U4.70 - Prob. 2ECh. U4.70 - Prob. 3ECh. U4.70 - Prob. 4ECh. U4.70 - Prob. 6ECh. U4.71 - Prob. 1TAICh. U4.71 - Prob. 1ECh. U4.71 - Prob. 2ECh. U4.71 - Prob. 3ECh. U4.71 - Prob. 4ECh. U4.71 - Prob. 5ECh. U4.71 - Prob. 6ECh. U4.71 - Prob. 7ECh. U4.72 - Prob. 1TAICh. U4.72 - Prob. 1ECh. U4.72 - Prob. 2ECh. U4.72 - Prob. 3ECh. U4.72 - Prob. 4ECh. U4.73 - Prob. 1TAICh. U4.73 - Prob. 1ECh. U4.73 - Prob. 2ECh. U4.73 - Prob. 3ECh. U4.73 - Prob. 5ECh. U4.73 - Prob. 6ECh. U4.73 - Prob. 7ECh. U4.74 - Prob. 1TAICh. U4.74 - Prob. 1ECh. U4.74 - Prob. 2ECh. U4.74 - Prob. 4ECh. U4.74 - Prob. 5ECh. U4.75 - Prob. 1TAICh. U4.75 - Prob. 1ECh. U4.75 - Prob. 2ECh. U4.75 - Prob. 4ECh. U4.75 - Prob. 5ECh. U4.75 - Prob. 6ECh. U4.75 - Prob. 7ECh. U4.75 - Prob. 8ECh. U4.75 - Prob. 9ECh. U4.75 - Prob. 10ECh. U4.75 - Prob. 11ECh. U4.76 - Prob. 1TAICh. U4.76 - Prob. 1ECh. U4.76 - Prob. 2ECh. U4.76 - Prob. 3ECh. U4.76 - Prob. 4ECh. U4.76 - Prob. 5ECh. U4.76 - Prob. 6ECh. U4.76 - Prob. 7ECh. U4.76 - Prob. 8ECh. U4.77 - Prob. 1TAICh. U4.77 - Prob. 1ECh. U4.77 - Prob. 2ECh. U4.77 - Prob. 3ECh. U4.77 - Prob. 4ECh. U4.77 - Prob. 5ECh. U4.77 - Prob. 6ECh. U4.77 - Prob. 7ECh. U4.77 - Prob. 8ECh. U4.78 - Prob. 1TAICh. U4.78 - Prob. 1ECh. U4.78 - Prob. 2ECh. U4.78 - Prob. 3ECh. U4.78 - Prob. 4ECh. U4.78 - Prob. 5ECh. U4.78 - Prob. 6ECh. U4.78 - Prob. 7ECh. U4.78 - Prob. 8ECh. U4.79 - Prob. 1TAICh. U4.79 - Prob. 1ECh. U4.79 - Prob. 2ECh. U4.79 - Prob. 3ECh. U4.79 - Prob. 4ECh. U4.80 - Prob. 1TAICh. U4.80 - Prob. 1ECh. U4.80 - Prob. 2ECh. U4.80 - Prob. 3ECh. U4.80 - Prob. 4ECh. U4.80 - Prob. 5ECh. U4.80 - Prob. 6ECh. U4.80 - Prob. 7ECh. U4.80 - Prob. 8ECh. U4.80 - Prob. 9ECh. U4.80 - Prob. 10ECh. U4.81 - Prob. 1TAICh. U4.81 - Prob. 1ECh. U4.81 - Prob. 2ECh. U4.81 - Prob. 3ECh. U4.81 - Prob. 4ECh. U4.81 - Prob. 5ECh. U4.81 - Prob. 6ECh. U4.81 - Prob. 7ECh. U4.81 - Prob. 8ECh. U4.81 - Prob. 9ECh. U4.82 - Prob. 1TAICh. U4.82 - Prob. 1ECh. U4.82 - Prob. 2ECh. U4.82 - Prob. 3ECh. U4.82 - Prob. 4ECh. U4.82 - Prob. 5ECh. U4.82 - Prob. 6ECh. U4.82 - Prob. 7ECh. U4.82 - Prob. 8ECh. U4.83 - Prob. 1TAICh. U4.83 - Prob. 1ECh. U4.83 - Prob. 2ECh. U4.83 - Prob. 3ECh. U4.83 - Prob. 4ECh. U4.83 - Prob. 5ECh. U4.83 - Prob. 6ECh. U4.83 - Prob. 7ECh. U4.84 - Prob. 1TAICh. U4.84 - Prob. 1ECh. U4.84 - Prob. 2ECh. U4.84 - Prob. 3ECh. U4.84 - Prob. 5ECh. U4.84 - Prob. 6ECh. U4.84 - Prob. 7ECh. U4.85 - Prob. 1TAICh. U4.85 - Prob. 1ECh. U4.85 - Prob. 2ECh. U4.85 - Prob. 3ECh. U4.85 - Prob. 4ECh. U4.85 - Prob. 5ECh. U4.85 - Prob. 6ECh. U4.85 - Prob. 7ECh. U4.85 - Prob. 8ECh. U4.86 - Prob. 1TAICh. U4.86 - Prob. 1ECh. U4.86 - Prob. 2ECh. U4.86 - Prob. 3ECh. U4.86 - Prob. 4ECh. U4.86 - Prob. 6ECh. U4.86 - Prob. 7ECh. U4.86 - Prob. 8ECh. U4.87 - Prob. 1TAICh. U4.87 - Prob. 1ECh. U4.87 - Prob. 2ECh. U4.87 - Prob. 3ECh. U4.87 - Prob. 4ECh. U4.87 - Prob. 5ECh. U4.87 - Prob. 6ECh. U4.87 - Prob. 7ECh. U4.87 - Prob. 8ECh. U4.88 - Prob. 1TAICh. U4.88 - Prob. 1ECh. U4.88 - Prob. 2ECh. U4.88 - Prob. 4ECh. U4.88 - Prob. 5ECh. U4.88 - Prob. 6ECh. U4.88 - Prob. 7ECh. U4.88 - Prob. 8ECh. U4.89 - Prob. 1TAICh. U4.89 - Prob. 1ECh. U4.89 - Prob. 2ECh. U4.89 - Prob. 3ECh. U4.89 - Prob. 4ECh. U4.89 - Prob. 5ECh. U4.89 - Prob. 6ECh. U4.90 - Prob. 1ECh. U4.90 - Prob. 2ECh. U4.90 - Prob. 3ECh. U4.90 - Prob. 4ECh. U4.90 - Prob. 5ECh. U4.90 - Prob. 6ECh. U4.90 - Prob. 7ECh. U4.91 - Prob. 1ECh. U4.91 - Prob. 2ECh. U4.91 - Prob. 3ECh. U4.91 - Prob. 5ECh. U4.91 - Prob. 6ECh. U4.92 - Prob. 1TAICh. U4.92 - Prob. 1ECh. U4.92 - Prob. 2ECh. U4.92 - Prob. 3ECh. U4.92 - Prob. 4ECh. U4.93 - Prob. 1TAICh. U4.93 - Prob. 1ECh. U4.93 - Prob. 2ECh. U4.93 - Prob. 4ECh. U4.93 - Prob. 5ECh. U4.93 - Prob. 6ECh. U4 - Prob. C13.3RECh. U4 - Prob. C13.4RECh. U4 - Prob. C14.1RECh. U4 - Prob. C14.2RECh. U4 - Prob. C14.3RECh. U4 - Prob. C14.5RECh. U4 - Prob. C14.6RECh. U4 - Prob. C15.1RECh. U4 - Prob. C15.2RECh. U4 - Prob. C15.3RECh. U4 - Prob. C15.4RECh. U4 - Prob. C15.5RECh. U4 - Prob. C15.6RECh. U4 - Prob. C15.7RECh. U4 - Prob. C15.8RECh. U4 - Prob. C16.1RECh. U4 - Prob. C16.2RECh. U4 - Prob. C16.3RECh. U4 - Prob. C16.4RECh. U4 - Prob. C17.1RECh. U4 - Prob. C17.2RECh. U4 - Prob. C17.3RECh. U4 - Prob. 1RECh. U4 - Prob. 4RECh. U4 - Prob. 5RECh. U4 - Prob. 6RECh. U4 - Prob. 7RECh. U4 - Prob. 8RECh. U4 - Prob. 9RECh. U4 - Prob. 10RECh. U4 - Prob. 11RECh. U4 - Prob. 12RECh. U4 - Prob. 1STPCh. U4 - Prob. 2STPCh. U4 - Prob. 3STPCh. U4 - Prob. 4STPCh. U4 - Prob. 5STPCh. U4 - Prob. 6STPCh. U4 - Prob. 7STPCh. U4 - Prob. 8STPCh. U4 - Prob. 9STPCh. U4 - Prob. 10STPCh. U4 - Prob. 11STPCh. U4 - Prob. 12STPCh. U4 - Prob. 13STPCh. U4 - Prob. 14STPCh. U4 - Prob. 15STPCh. U4 - Prob. 16STPCh. U4 - Prob. 17STPCh. U4 - Prob. 18STPCh. U4 - Prob. 19STPCh. U4 - Prob. 20STP
Additional Science Textbook Solutions
Find more solutions based on key concepts
27. Consider the reaction.
Express the rate of the reaction in terms of the change in concentration of each of...
Chemistry: Structure and Properties (2nd Edition)
Starting with 10 bacterial cells per milliliter in a sufficient amount of complete culture medium with a 1-hour...
Microbiology with Diseases by Body System (5th Edition)
1. Light can pass easily through water and through air, but light will reflect from the surface of a lake. What...
College Physics: A Strategic Approach (3rd Edition)
How can the freezing of water crack boulders?
Campbell Biology in Focus (2nd Edition)
A source of electromagnetic radiation produces infrared light. Which of the following could be the wavelength ...
Chemistry: The Central Science (14th Edition)
10.71 Identify each of the following as an acid or a base: (10.1)
H2SO4
RbOH
Ca(OH)2
HI
...
Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)
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
- Q4: Rank the relative nucleophilicity of halide ions in water solution and DMF solution, respectively. F CI Br | Q5: Determine which of the substrates will and will not react with NaSCH3 in an SN2 reaction to have a reasonable yield of product. NH2 Br Br Br OH Brarrow_forwardQ7: Rank the following groups in order of basicity, nucleophilicity, and leaving group ability. a) H₂O, OH, CH3COOT b) NH3, H₂O, H₂Sarrow_forwardQ8: Rank the following compounds in order of increasing reactivity in a nucleophilic substitution reaction with CN as the nucleophile. Br A B NH2 LL F C D OH CI LLI E Q9: Complete the missing entities for following reactions (e.g., major product(s), reactants, and/or solvents) for the SN2 reactions to occur efficiently. Include curved-arrow mechanism for reactions a) to d). a) H "Cl D + -OCH 3 Page 3 of 5arrow_forward
- Q10: (a) Propose a synthesis of C from A. (b) Propose a synthesis of C from B. Br Br ...\SCH 3 A B Carrow_forward9: Complete the missing entities for following reactions (e.g., major product(s), reactants, and/or solvents) for the SN2 reactions to occur efficiently. Include curved-arrow mechanism for reactions a) to d).arrow_forwardComplete the missing entities for following reactions (e.g., major product(s), reactants, and/or solvents) for the SN2 reactions to occur efficiently. Include curved-arrow mechanism for reactions a) to d).arrow_forward
- QUESTION 3: Provide the synthetic steps that convert the starting material into the product (no mechanism required). HO OH NH CH3 multiple steps 요요 H3Carrow_forwardQ6: Predict the effect of the changes given on the rate of the reaction below. CH3OH CH3Cl + NaOCH3 → CH3OCH3 + NaCl a) Change the substrate from CH3CI to CH31: b) Change the nucleophile from NaOCH 3 to NaSCH3: c) Change the substrate from CH3CI to (CH3)2CHCI: d) Change the solvent from CH3OH to DMSO.arrow_forwardQ3: Arrange each group of compounds from fastest SN2 reaction rate to slowest SN2 reaction rate. a) CI Cl فيكم H3C-Cl A B C D Br Br b) A B C Br H3C-Br Darrow_forward
- Q2: Group these solvents into either protic solvents or aprotic solvents. Acetonitrile (CH3CN), H₂O, Acetic acid (CH3COOH), Acetone (CH3COCH3), CH3CH2OH, DMSO (CH3SOCH3), DMF (HCON(CH3)2), CH3OHarrow_forwardSuppose the rate of evaporation in a hot, dry region is 1.76 meters per year, and the seawater there has a salinity of 35 ‰. Assuming a 93% yield, how much salt (NaCl) can be harvested each year from 1 km2 of solar evaporation ponds that use this seawater as a source?arrow_forwardhelparrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
General Chemistry | Acids & Bases; Author: Ninja Nerd;https://www.youtube.com/watch?v=AOr_5tbgfQ0;License: Standard YouTube License, CC-BY