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Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. 4. You are given 60 mL of 0.50M phosphate buffer, pH = 6.83, to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: Concentration of HPO,: 0.304 M Molar quantity of HPO,: 18.2 mmol Concentration of H,PO: 0.196 M Molar quantity of H,PO = 11.8 mmol You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H,O added as HCI, and the final molar quantity of HPO, and H,PO, at equilibrium. a. 2- b. What is the new HPO,/H,PO,¯ ratio, and the new pH of the solution? The pK, of H,PO, is 6.64. Use the Henderson-Hasselbalch equation to calculate the new pH. (Note: You can use the molar ratio rather than the concentration ratio because both species are in the same volume.) 2- Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NAOH, Using steps similar to those above, calculate the new pH of the solution. C.

Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. 4. You are given 60 mL of 0.50M phosphate buffer, pH = 6.83, to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: Concentration of HPO,: 0.304 M Molar quantity of HPO,: 18.2 mmol Concentration of H,PO: 0.196 M Molar quantity of H,PO = 11.8 mmol You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H,O added as HCI, and the final molar quantity of HPO, and H,PO, at equilibrium. a. 2- b. What is the new HPO,/H,PO,¯ ratio, and the new pH of the solution? The pK, of H,PO, is 6.64. Use the Henderson-Hasselbalch equation to calculate the new pH. (Note: You can use the molar ratio rather than the concentration ratio because both species are in the same volume.) 2- Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NAOH, Using steps similar to those above, calculate the new pH of the solution. C.

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter16: Reactions Between Acids And Bases
Section: Chapter Questions
Problem 16.78QE
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Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. 4. You are given 60 mL of 0.50M phosphate buffer, pH = 6.83, to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: Concentration of HPO,: 0.304 M Molar quantity of HPO,: 18.2 mmol Concentration of H,PO: 0.196 M Molar quantity of H,PO = 11.8 mmol You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H,O added as HCI, and the final molar quantity of HPO, and H,PO, at equilibrium. a. 2- b. What is the new HPO,/H,PO,¯ ratio, and the new pH of the solution? The pK, of H,PO, is 6.64. Use the Henderson-Hasselbalch equation to calculate the new pH. (Note: You can use the molar ratio rather than the concentration ratio because both species are in the same volume.) 2- Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NAOH, Using steps similar to those above, calculate the new pH of the solution. C.
Transcribed Image Text:Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. 4. You are given 60 mL of 0.50M phosphate buffer, pH = 6.83, to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: Concentration of HPO,: 0.304 M Molar quantity of HPO,: 18.2 mmol Concentration of H,PO: 0.196 M Molar quantity of H,PO = 11.8 mmol You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H,O added as HCI, and the final molar quantity of HPO, and H,PO, at equilibrium. a. 2- b. What is the new HPO,/H,PO,¯ ratio, and the new pH of the solution? The pK, of H,PO, is 6.64. Use the Henderson-Hasselbalch equation to calculate the new pH. (Note: You can use the molar ratio rather than the concentration ratio because both species are in the same volume.) 2- Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NAOH, Using steps similar to those above, calculate the new pH of the solution. C.
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