Activity 9Buffer SolutionsThis activity practices with two skills: (1) Determining the pH of a buffer solution, and (2)figuring out changes in pH when strong acid or strong base is added to a volume of the buffersolution. Finally, you will compare these pH changes to what would happen if the sameamount of strong acid or base is added to water.(1) What is the pH of a solution that is 0.371 M carbonic acid (H2CO3) and 0.350 M sodiumbicarbonate (NaHCO;)? Ka for carbonic acid is 4.3 x 10-7. Figure this out two ways:A) Starting with the acid dissociation equation and solving the ICE problem, figure out the pH.[A*]B) Use the Henderson-Hasselbalch Equation instead: pH = pK, + log[HA]Remember that pKa is defined as –log(Ka). You should get the same answer.13 Activity 9, continued2A) What happens to the pH of this buffer solution when strong acid is added. If 1.0 mL of 6.00M HCl is added to 200. mL of the buffer solution in problem (1), what is the new pH?Remember that first you have to do a dilution problem.Then, ask yourself: What does the H* react with?Change the concentrations of H2CO3 and HCO3 as appropriate, to do this problem.It is much easier to do this using H-H than the other way, but they are both good methods.2B) What happens if 1.0 mL of 6.00 M HCl is added to 200. mL WATER instead of to the buffer?2C) Next, figure out what happens to the pH if 1.0 mL of 6.00 M NAOH (strong base) is added to200. mL of the original buffer solution. What does the OH react with? Change the concentrationsof H2CO3 and HCO; as appropriate, and finish the problem.2D) Compare this to what the change in pH would be if you added 1.00 mL of 6.00M NaOH to200. mL of WATER.

1.) a.) In the given buffer solution , there is acid H2CO3 and sodium salt of its conjugate base ,…

(1) What is the pH of a solution that is 0.371 M carbonic acid (H2CO;) and 0.350 M sodium bicarbonate (NaHCO;)? Ka for carbonic acid is 4.3 x 107. Figure this out two ways: A) Starting with the acid dissociation equation and solving the ICE problem, figure out the pH. B) Use the Henderson-Hasselbalch Equation instead: pH = pK, + log| [HA], Remember that pK, is defined as –log(K.). You should get the same answer.

(1) What is the pH of a solution that is 0.371 M carbonic acid (H2CO;) and 0.350 M sodium bicarbonate (NaHCO;)? Ka for carbonic acid is 4.3 x 107. Figure this out two ways: A) Starting with the acid dissociation equation and solving the ICE problem, figure out the pH. B) Use the Henderson-Hasselbalch Equation instead: pH = pK, + log| [HA], Remember that pK, is defined as –log(K.). You should get the same answer.

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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Question

Activity 9
Buffer Solutions
This activity practices with two skills: (1) Determining the pH of a buffer solution, and (2)
figuring out changes in pH when strong acid or strong base is added to a volume of the buffer
solution. Finally, you will compare these pH changes to what would happen if the same
amount of strong acid or base is added to water.
(1) What is the pH of a solution that is 0.371 M carbonic acid (H2CO3) and 0.350 M sodium
bicarbonate (NaHCO;)? Ka for carbonic acid is 4.3 x 10-7. Figure this out two ways:
A) Starting with the acid dissociation equation and solving the ICE problem, figure out the pH.
[A*]
B) Use the Henderson-Hasselbalch Equation instead: pH = pK, + log
[HA]
Remember that pKa is defined as –log(Ka). You should get the same answer.
13

Activity 9, continued
2A) What happens to the pH of this buffer solution when strong acid is added. If 1.0 mL of 6.00
M HCl is added to 200. mL of the buffer solution in problem (1), what is the new pH?
Remember that first you have to do a dilution problem.
Then, ask yourself: What does the H* react with?
Change the concentrations of H2CO3 and HCO3 as appropriate, to do this problem.
It is much easier to do this using H-H than the other way, but they are both good methods.
2B) What happens if 1.0 mL of 6.00 M HCl is added to 200. mL WATER instead of to the buffer?
2C) Next, figure out what happens to the pH if 1.0 mL of 6.00 M NAOH (strong base) is added to
200. mL of the original buffer solution. What does the OH react with? Change the concentrations
of H2CO3 and HCO; as appropriate, and finish the problem.
2D) Compare this to what the change in pH would be if you added 1.00 mL of 6.00M NaOH to
200. mL of WATER.

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