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suppose there is 1.00L of an aqueous buffer containing 600.0 moral & acetic acid (pk. 4.76) and 40.0 ml of acetate. Calculate the pott of this buffer. pH = pka + log [#] and [HA]- acetic acid 60 moal [A-] acetate = 40 mmal pka = 4.76 PH= 4.76 log (488) = 4,58² # PH What volume of 3.50 M NaOH would be required to increase the PH to 4.93?

suppose there is 1.00L of an aqueous buffer containing 600.0 moral & acetic acid (pk. 4.76) and 40.0 ml of acetate. Calculate the pott of this buffer. pH = pka + log [#] and [HA]- acetic acid 60 moal [A-] acetate = 40 mmal pka = 4.76 PH= 4.76 log (488) = 4,58² # PH What volume of 3.50 M NaOH would be required to increase the PH to 4.93?

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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Suppose there is look of an aqueous buffer containing 600.0 moral & acetic acid (pk. - 4.76) and 40.0 mmol of acetate. Calculate the pt of this buffer. pH = pka + log [#] acid [HA]- acetic acid = 60 mmol [A-] acetate = 40 mmal pka = 4.76 PH² 4.76 log (128)-14.58 B What volume of 3.50 M NaOH would be required to increase. the pH to 4.93? Required ph= 4.93 I e E 0.17 = 10 Substitute into Henderson pH = pka + log [CH₂COO] [CH3COOL+] 4.93 = 4.76 + log (40+ 3. -4.76 -4.76 60- 1.47 TO CH3COOH + OH → CH ₂ COO + H₂O 3.50x 40 1 -3.50x (60-3.50x) D 40 -3.50x 0.1740+3.50x 60-2x X = 20.35 +3.50X 40+ 8.50x1 60-2x PH 40+ 3.50 x 40-2x) -3.50 x +3.50x/ (40+3,50x کو
Transcribed Image Text:Suppose there is look of an aqueous buffer containing 600.0 moral & acetic acid (pk. - 4.76) and 40.0 mmol of acetate. Calculate the pt of this buffer. pH = pka + log [#] acid [HA]- acetic acid = 60 mmol [A-] acetate = 40 mmal pka = 4.76 PH² 4.76 log (128)-14.58 B What volume of 3.50 M NaOH would be required to increase. the pH to 4.93? Required ph= 4.93 I e E 0.17 = 10 Substitute into Henderson pH = pka + log [CH₂COO] [CH3COOL+] 4.93 = 4.76 + log (40+ 3. -4.76 -4.76 60- 1.47 TO CH3COOH + OH → CH ₂ COO + H₂O 3.50x 40 1 -3.50x (60-3.50x) D 40 -3.50x 0.1740+3.50x 60-2x X = 20.35 +3.50X 40+ 8.50x1 60-2x PH 40+ 3.50 x 40-2x) -3.50 x +3.50x/ (40+3,50x کو
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