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This is a Dry Lab since we cannot meet on Campus. After calibrating a pH meter you would have measured the pH of four different Basic Aqueous Solutions and examined the data for trends. You would have calculated the expected pH of each of the strong base solutions by assuming that for every NaOH formula unit that dissolves, one OH- ion is released. Thus, we can assume that the OH- Molarity equals the NaOH Molarity. Using the OH- Molarity and the water ionization equilibrium expression (Kw= 1.0 X 10 to the -14 power = H+ Molarity X OH- Molarity), we can calculate the H+ Molarity of the solution. Using the H+ Molarity, we can calculate the solution's pH. Thus a 1.0 X 10 to the 4 power NaOH solution has a pH of 10. The calculated pH value is arrived at as follows. By rearranging the water ionization equilibrium expression and plugging in the OH- Molarity, we can solve for H+ Molarity. H+ M = 1.0 X 10 to the -14 power divided by 1.0 X 10 to the -4 power= 1.0 X 10 to the 10 power. pH= -log (1.0 X 10 to the -10 power)= 10 I Calculate the expected pH of these three NaOH solutions, complete the table, answer the questions and email it to me at jpanek@mc3.edu. NaOH Aqueous Solution 0.10 M 0.010 M ▼ 0.0010 M Calculated expected pH Experimental pH (typical) 12.85 11.95 Answer these Questions. #1 What happens to the pH as OH- Molarity increases ? 11.05 tion hocomo

This is a Dry Lab since we cannot meet on Campus. After calibrating a pH meter you would have measured the pH of four different Basic Aqueous Solutions and examined the data for trends. You would have calculated the expected pH of each of the strong base solutions by assuming that for every NaOH formula unit that dissolves, one OH- ion is released. Thus, we can assume that the OH- Molarity equals the NaOH Molarity. Using the OH- Molarity and the water ionization equilibrium expression (Kw= 1.0 X 10 to the -14 power = H+ Molarity X OH- Molarity), we can calculate the H+ Molarity of the solution. Using the H+ Molarity, we can calculate the solution's pH. Thus a 1.0 X 10 to the 4 power NaOH solution has a pH of 10. The calculated pH value is arrived at as follows. By rearranging the water ionization equilibrium expression and plugging in the OH- Molarity, we can solve for H+ Molarity. H+ M = 1.0 X 10 to the -14 power divided by 1.0 X 10 to the -4 power= 1.0 X 10 to the 10 power. pH= -log (1.0 X 10 to the -10 power)= 10 I Calculate the expected pH of these three NaOH solutions, complete the table, answer the questions and email it to me at jpanek@mc3.edu. NaOH Aqueous Solution 0.10 M 0.010 M ▼ 0.0010 M Calculated expected pH Experimental pH (typical) 12.85 11.95 Answer these Questions. #1 What happens to the pH as OH- Molarity increases ? 11.05 tion hocomo

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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First read the text sections 10.5 and 10.6 on pH. This is a Dry Lab since we cannot meet on Campus. After calibrating a pH meter you would have measured the pH of four different Basic Aqueous Solutions and examined the data for trends. You would have calculated the expected pH of each of the strong base solutions by assuming that for every NaOH formula unit that dissolves, one OH- ion is released. Thus, we can assume that the OH- Molarity equals the NaOH Molarity. Using the OH- Molarity and the water ionization equilibrium expression (Kw= 1.0 X 10 to the -14 power = H+ Molarity X OH- Molarity), we can calculate the H+ Molarity of the solution. Using the H+ Molarity, we can calculate the solution's pH. Thus a 1.0 X 10 to the 4 power NaOH solution has a pH of 10. The calculated pH value is arrived at as follows. By rearranging the water ionization equilibrium expression and plugging in the OH- Molarity, we can solve for H+ Molarity. H+ M = 1.0 X 10 to the -14 power divided by 1.0 X 10 to the -4 power= 1.0 X 10 to the 10 power. pH= -log (1.0 X 10 to the -10 power)= 10 I Calculate the expected pH of these three NaOH solutions, complete the table, answer the questions and email it to me at jpanek@mc3.edu. NaOH Aqueous Solution 0.10 M Calculated expected pH Experimental pH (typical) 0.010 M more acidic or more basic? 0.0010 M 12.85 11.95 11.05 Answer these Questions. #1 What happens to the pH as OH- Molarity increases ? #2 As OH- Molarity increases, does the solution become
Transcribed Image Text:First read the text sections 10.5 and 10.6 on pH. This is a Dry Lab since we cannot meet on Campus. After calibrating a pH meter you would have measured the pH of four different Basic Aqueous Solutions and examined the data for trends. You would have calculated the expected pH of each of the strong base solutions by assuming that for every NaOH formula unit that dissolves, one OH- ion is released. Thus, we can assume that the OH- Molarity equals the NaOH Molarity. Using the OH- Molarity and the water ionization equilibrium expression (Kw= 1.0 X 10 to the -14 power = H+ Molarity X OH- Molarity), we can calculate the H+ Molarity of the solution. Using the H+ Molarity, we can calculate the solution's pH. Thus a 1.0 X 10 to the 4 power NaOH solution has a pH of 10. The calculated pH value is arrived at as follows. By rearranging the water ionization equilibrium expression and plugging in the OH- Molarity, we can solve for H+ Molarity. H+ M = 1.0 X 10 to the -14 power divided by 1.0 X 10 to the -4 power= 1.0 X 10 to the 10 power. pH= -log (1.0 X 10 to the -10 power)= 10 I Calculate the expected pH of these three NaOH solutions, complete the table, answer the questions and email it to me at jpanek@mc3.edu. NaOH Aqueous Solution 0.10 M Calculated expected pH Experimental pH (typical) 0.010 M more acidic or more basic? 0.0010 M 12.85 11.95 11.05 Answer these Questions. #1 What happens to the pH as OH- Molarity increases ? #2 As OH- Molarity increases, does the solution become
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