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What is the pH after the addition of 1.0 mL of 0.10 M HCI to 20 mL of the solution above?

What is the pH after the addition of 1.0 mL of 0.10 M HCI to 20 mL of the solution above?

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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What is the pH after the addition of 1.0 mL of 0.10 M HCl to 20 mL of the solution above? Given: VHCI =1.0 mL CHC] = 0.10 Concentration of Sulution Albove - let HA=Acid->HA + THAT = (25.0mL) (0.10o) – t20.0 mL0.10ml) (25.0+20.0) mL [HAJ = 0.011| nal [A-] = Of M. (20-0ML5(0.10) I (25.0+ 20.0) mL mol [A-] = 0.0444 l or mol [A] 20.0 mL (0.0444 ) (20.0+ 0.1)m 1.0mL (0.101) L(20.0+1.0) mL pH =6.4478 Answer: 94
Transcribed Image Text:What is the pH after the addition of 1.0 mL of 0.10 M HCl to 20 mL of the solution above? Given: VHCI =1.0 mL CHC] = 0.10 Concentration of Sulution Albove - let HA=Acid->HA + THAT = (25.0mL) (0.10o) – t20.0 mL0.10ml) (25.0+20.0) mL [HAJ = 0.011| nal [A-] = Of M. (20-0ML5(0.10) I (25.0+ 20.0) mL mol [A-] = 0.0444 l or mol [A] 20.0 mL (0.0444 ) (20.0+ 0.1)m 1.0mL (0.101) L(20.0+1.0) mL pH =6.4478 Answer: 94
Calculate the amount of solid NHẠC1 to add to 60.0 mL of 0.100 M ammonia to form a mixture that has the desired pH. Assume no volume change in the solution. Hint: Use the same buffer ratio that you used in (a). pH = -log ka + log EVA4, - wknowm LNH,] INHT]-> Uu known [(0.100)] [x] 1.0=-109(5.81016)+ log mel X = 0.1724 = [NH4C1] Since volune of solution= Constaut: 1724 IL %3D mol NHYCI =0.01o34 mol Molar Mass NHYCI = 0.01034 (33. 4919) Mass of NtyCI mol Mass NHyCl = 0.55346 close Volume of NH3 Grams of NH4CI 60.00 mL Di5534. 97
Transcribed Image Text:Calculate the amount of solid NHẠC1 to add to 60.0 mL of 0.100 M ammonia to form a mixture that has the desired pH. Assume no volume change in the solution. Hint: Use the same buffer ratio that you used in (a). pH = -log ka + log EVA4, - wknowm LNH,] INHT]-> Uu known [(0.100)] [x] 1.0=-109(5.81016)+ log mel X = 0.1724 = [NH4C1] Since volune of solution= Constaut: 1724 IL %3D mol NHYCI =0.01o34 mol Molar Mass NHYCI = 0.01034 (33. 4919) Mass of NtyCI mol Mass NHyCl = 0.55346 close Volume of NH3 Grams of NH4CI 60.00 mL Di5534. 97
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