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
Question
Chapter 8, Problem 8.10VP
Interpretation Introduction
To find:
Which of the four graphs in Figure P8.10 of the textbook best represents the changes in conductivity that occur before and after the equivalence point in each of the following titrations?
- Sample, Cl- (aq);titrant, AgNO3 (aq)
- Sample, HCl(aq);titrant, LiOH(aq)
- Sample, CH3 COOH(aq);titrant, NaOH(aq).
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
At which point in the titration is the number of
moles of analyte and titrant the same?
EJA B
Co
D
Volume titrant added (mL)
A. O Point A
В. О Рoint B
С. О Рoint C
D. O Point D
Hd
A reg1.hu.edu.jo
العربية
ENGLISH
Question 4/12
ميز هذا السؤال
Titration of 0.2121 g of primary standard Na2C204 (133.999 g/mol) required 21.66 mL of KMNO4 (158.034 g/mol) to reach the end point.
Calculate the molar concentration of KMNO4? 2 MnO4¯+ 5 C204²-+ 16 H+ →
2 Mn2+ 10 CO2 +8 H2O
1. 00.0146
2. 00.0292
3. 00.00731
4. 00.1462
(f). Following are the titre values of two students when 25 mL of KIO3(aq) is titrated against Na2S2O3(aq).Titration no. Student 1 Student 21 2 3 24 24.9 25.7 25 24.1 24.0Which student is best in doing titration and why do you say so?(g). Taking the values of best student, calculate the concentration of the solution for which molarity is not known.(Note: first calculate moles of pipet solution in 25 mL, then moles of I2 and then moles of solution of unknown concentration and then its molarity by using average titre)
Chapter 8 Solutions
Chemistry: An Atoms-Focused Approach
Ch. 8 - Prob. 8.1VPCh. 8 - Prob. 8.2VPCh. 8 - Prob. 8.3VPCh. 8 - Prob. 8.4VPCh. 8 - Prob. 8.5VPCh. 8 - Prob. 8.6VPCh. 8 - Prob. 8.7VPCh. 8 - Prob. 8.8VPCh. 8 - Prob. 8.9VPCh. 8 - Prob. 8.10VP
Ch. 8 - Prob. 8.11QACh. 8 - Prob. 8.12QACh. 8 - Prob. 8.13QACh. 8 - Prob. 8.14QACh. 8 - Prob. 8.15QACh. 8 - Prob. 8.16QACh. 8 - Prob. 8.17QACh. 8 - Prob. 8.18QACh. 8 - Prob. 8.19QACh. 8 - Prob. 8.20QACh. 8 - Prob. 8.21QACh. 8 - Prob. 8.22QACh. 8 - Prob. 8.23QACh. 8 - Prob. 8.24QACh. 8 - Prob. 8.25QACh. 8 - Prob. 8.26QACh. 8 - Prob. 8.27QACh. 8 - Prob. 8.28QACh. 8 - Prob. 8.29QACh. 8 - Prob. 8.30QACh. 8 - Prob. 8.31QACh. 8 - Prob. 8.32QACh. 8 - Prob. 8.33QACh. 8 - Prob. 8.34QACh. 8 - Prob. 8.35QACh. 8 - Prob. 8.36QACh. 8 - Prob. 8.37QACh. 8 - Prob. 8.38QACh. 8 - Prob. 8.39QACh. 8 - Prob. 8.40QACh. 8 - Prob. 8.41QACh. 8 - Prob. 8.42QACh. 8 - Prob. 8.43QACh. 8 - Prob. 8.44QACh. 8 - Prob. 8.45QACh. 8 - Prob. 8.46QACh. 8 - Prob. 8.47QACh. 8 - Prob. 8.48QACh. 8 - Prob. 8.49QACh. 8 - Prob. 8.50QACh. 8 - Prob. 8.51QACh. 8 - Prob. 8.52QACh. 8 - Prob. 8.53QACh. 8 - Prob. 8.54QACh. 8 - Prob. 8.55QACh. 8 - Prob. 8.56QACh. 8 - Prob. 8.57QACh. 8 - Prob. 8.58QACh. 8 - Prob. 8.59QACh. 8 - Prob. 8.60QACh. 8 - Prob. 8.61QACh. 8 - Prob. 8.62QACh. 8 - Prob. 8.63QACh. 8 - Prob. 8.64QACh. 8 - Prob. 8.65QACh. 8 - Prob. 8.66QACh. 8 - Prob. 8.67QACh. 8 - Prob. 8.68QACh. 8 - Prob. 8.69QACh. 8 - Prob. 8.70QACh. 8 - Prob. 8.71QACh. 8 - Prob. 8.72QACh. 8 - Prob. 8.73QACh. 8 - Prob. 8.74QACh. 8 - Prob. 8.75QACh. 8 - Prob. 8.76QACh. 8 - Prob. 8.77QACh. 8 - Prob. 8.78QACh. 8 - Prob. 8.79QACh. 8 - Prob. 8.80QACh. 8 - Prob. 8.81QACh. 8 - Prob. 8.82QACh. 8 - Prob. 8.83QACh. 8 - Prob. 8.84QACh. 8 - Prob. 8.85QACh. 8 - Prob. 8.86QACh. 8 - Prob. 8.87QACh. 8 - Prob. 8.88QACh. 8 - Prob. 8.89QACh. 8 - Prob. 8.90QACh. 8 - Prob. 8.91QACh. 8 - Prob. 8.92QACh. 8 - Prob. 8.93QACh. 8 - Prob. 8.94QACh. 8 - Prob. 8.95QACh. 8 - Prob. 8.96QACh. 8 - Prob. 8.97QACh. 8 - Prob. 8.98QACh. 8 - Prob. 8.99QACh. 8 - Prob. 8.100QACh. 8 - Prob. 8.101QACh. 8 - Prob. 8.102QACh. 8 - Prob. 8.103QACh. 8 - Prob. 8.104QACh. 8 - Prob. 8.105QACh. 8 - Prob. 8.106QACh. 8 - Prob. 8.107QACh. 8 - Prob. 8.108QACh. 8 - Prob. 8.109QACh. 8 - Prob. 8.110QACh. 8 - Prob. 8.111QACh. 8 - Prob. 8.112QACh. 8 - Prob. 8.113QACh. 8 - Prob. 8.114QACh. 8 - Prob. 8.115QACh. 8 - Prob. 8.116QACh. 8 - Prob. 8.117QACh. 8 - Prob. 8.118QACh. 8 - Prob. 8.119QACh. 8 - Prob. 8.120QACh. 8 - Prob. 8.121QACh. 8 - Prob. 8.122QACh. 8 - Prob. 8.123QACh. 8 - Prob. 8.124QACh. 8 - Prob. 8.125QACh. 8 - Prob. 8.126QACh. 8 - Prob. 8.127QACh. 8 - Prob. 8.128QACh. 8 - Prob. 8.129QACh. 8 - Prob. 8.130QA
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
- Consider the nanoscale-level representations for Question 111 of the titration of the aqueous strong acid HA with aqueous NaOH, the titrant. Water molecules and Na+ ions are omitted for clarity. Which diagram corresponds to the situation: (a) After a very small volume of titrant has been added to the initial HA solution? (b) Halfway to the equivalence point? (c) When enough titrant has been added to take the solution just past the equivalence point? (d) At the equivalence point? Nanoscale representations for Question 111.arrow_forwardConsider the following two acids: In two separate experiments the pH was measured during the titration of 5.00 mmol of each acid with 0.200 M NaOH. Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at 25.00 mL added NaOH, and in the other experiment the stoichiometric point was at 50.00 mL NaOH. Explain these results. (See Exercise 113.)arrow_forwardA saturated solution of copper(II) iodate in pure water has a copper-ion concentration of 2.7 103 M. a What is the molar solubility of copper iodate in a 0.35 M potassium iodate solution? b What is the molar solubility of copper iodate in a 0.35 M copper nitrate solution? c Should there be a difference in the answers to parts a and b? Why?arrow_forward
- Consider the nanoscale-level representations for Question 110 of the titration of the aqueous weak acid HX with aqueous NaOH, the titrant. Water molecules and Na+ ions are omitted for clarity. Which diagram corresponds to the situation: After a very small volume of titrant has been added to the initial HX solution? When enough titrant has been added to take the solution just past the equivalence point? Halfway to the equivalence point? At the equivalence point? Nanoscale representations for Question 110.arrow_forwardPotassium hydrogen phthalate, known as KHP (molar mass = 204.22 g/mol), can be obtained in high purity and is used to determine the concentration of solutions of strong bases by the reaction HP(aq)+OH(aq)H2O(l)+P2(aq) If a typical titration experiment begins with approximately 0.5 g KHP and has a final volume of about 100 mL, what is an appropriate indicator to use? The pKa for HP is 5.51.arrow_forwardA saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?arrow_forward
- A monoprotic organic acid that has a molar mass of 176.1 g/mol is synthesized. Unfortunately, the acid produced is not completely pure. In addition, it is not soluble in water. A chemist weighs a 1.8451-g sample of the impure acid and adds it to 100.0 mL of 0.1050 M NaOH. The acid is soluble in the NaOH solution and reacts to consume most of the NaOH. The amount of excess NaOH is determined by titration: It takes 3.28 mL of 0.0970 M HCl to neutralize the excess NaOH. What is the purity of the original acid, in percent?arrow_forwardAssume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.00 102 mL of solution and then titrate the solution with 0.108 M NaOH. C6H5CO2H(aq) + OH(aq) C6H5CO2(aq) + H2O() (a) What was the pH of the original benzoic add solution? (b) What are the concentrations of all of the following ions at the equivalence point: Na+, H3O+, OH, and C6H5CO2? (c) What is the pH of the solution at the equivalence point?arrow_forwardA volume of 50 mL of 1.8 M NH3 is mixed with an equal volume of a solution containing 0.95 g of MgCl2. What mass of NH4Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)2?arrow_forward
- The weak base ethanolamine. HOCH2CH2NH2, can be titrated with HCl. HOCH2CH2NH2(aq)+H3O+(aq)HOCH2CH2NH3+(aq)+H2O(l) Assume you have 25.0 mL of a 0.010 M solution of ethanolamine and titrate it with 0.0095 M HCl. (Kb for ethanolamine is 3.2 107.) (a) What is the pH of the ethanolamine solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 would be the best choice to detect the equivalence point? (e) Calculate the pH of the solution after adding 5.00, 10.0, 20.0, and 30.0 mL of the acid. (f) Combine the information in parts (a), (b), (c), and (e), and plot an approximate titration curve.arrow_forwardDescribe how the amount of sodium hydroxide in a mixture can be determined by titration with hydrochloric acid of known molarity.arrow_forward88c. [Tutorial: Titration stoichiometry] This question will walk you through the steps of calculating the volume of titrant required to reach the endpoint based on the volume and concentration of analyte. Consider the standardization of NaOH by titration with C₆H₈O₇ according to the following chemical reaction: 3 NaOH(aq) + C₆H₈O₇(aq) ⇌ Na₃C₆H₅O₇(aq) + 3 H₂O(l) Step 3: Calculate the volume of titrant required to reach the endpoint. Calculate the volume in mL of a 0.200 M NaOH solution needed to neutralize 345 mL of 0.0333 M C₆H₈O₇ standard solution. 88d. Based on the fact that NaOH is a strong base and C6H807 is a weak acid, how would you describe the pH of the solution at the endpoint, where moles of base = 3 times the moles of acid?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
General 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 Learning
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
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher: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
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Precipitation Reactions: Crash Course Chemistry #9; Author: Crash Course;https://www.youtube.com/watch?v=IIu16dy3ThI;License: Standard YouTube License, CC-BY