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B. Calculate the pH after 17.50 mL of NaOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz • First assume the acid-base neutralization reaction is ireversible and goes to completion: HCH3O2 (ag) + OH" (ag) → H2O () + C2H3O2 (ag) Moles before reaction Change in moles Moles remaining after reaction

B. Calculate the pH after 17.50 mL of NaOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz • First assume the acid-base neutralization reaction is ireversible and goes to completion: HCH3O2 (ag) + OH" (ag) → H2O () + C2H3O2 (ag) Moles before reaction Change in moles Moles remaining after reaction

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.106QP: Codeine, C23H21NO3, is an alkaloid (Kb = 6 2 109) used as a painkiller and cough suppressant. A...
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Original concentration of Acetic Acid was 0.1033 M and NaOH was 0.1104 M

there is part a done, need to do part B

A. Calculate the pH of 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: HC2H3O2 (aa) 2 H* (aa) + C2H3O2 (aa) I Initial Concentration (M) 0.1033 Change in Concentration (M) +x +x -X- Equilibrium Concentration (M) 0.1033 - x E Equilibrium Concentration, Calculated (M) 0.1020 1.36x10-3 1.36x10-3 1. Initial concentration [HC2H3O2]0=0.1033 M 2. Ka symbolic expression: [H*][ C2H3O2 ] [HC2H3O2] (x)(x) 3. Ka numerical equation with x: 1.8 x 10-S = 0.1033-x 4. Quadratic Formula Method a. Quadratic equation: x2+ 1.8 x 10-S – 1.86 x 10-6 = 0 -1.8 x 10A-5t/(-1.8x 10)-4(1)(1.86 x 10A-6) b.Quadratic formula: x = 2(1) 5. Assumption (5% rule) Method (1.8x10)(0.1033) a. Equation for x: x = = 1.36x10-°M b.% ionization = (1.36x10^-3M)(0.1033M)x100 = 1.3% 6. Equilibrium concentrations a. [HC2H3O2lm =0.1033 - 1.36x103 = 0.1020M b.[H* Jeg =1.36x10-3 M c.[C2H3O2] =1.36x103 M 7. pH = - log(1.36x10-3 ) = 2.8
Transcribed Image Text:A. Calculate the pH of 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: HC2H3O2 (aa) 2 H* (aa) + C2H3O2 (aa) I Initial Concentration (M) 0.1033 Change in Concentration (M) +x +x -X- Equilibrium Concentration (M) 0.1033 - x E Equilibrium Concentration, Calculated (M) 0.1020 1.36x10-3 1.36x10-3 1. Initial concentration [HC2H3O2]0=0.1033 M 2. Ka symbolic expression: [H*][ C2H3O2 ] [HC2H3O2] (x)(x) 3. Ka numerical equation with x: 1.8 x 10-S = 0.1033-x 4. Quadratic Formula Method a. Quadratic equation: x2+ 1.8 x 10-S – 1.86 x 10-6 = 0 -1.8 x 10A-5t/(-1.8x 10)-4(1)(1.86 x 10A-6) b.Quadratic formula: x = 2(1) 5. Assumption (5% rule) Method (1.8x10)(0.1033) a. Equation for x: x = = 1.36x10-°M b.% ionization = (1.36x10^-3M)(0.1033M)x100 = 1.3% 6. Equilibrium concentrations a. [HC2H3O2lm =0.1033 - 1.36x103 = 0.1020M b.[H* Jeg =1.36x10-3 M c.[C2H3O2] =1.36x103 M 7. pH = - log(1.36x10-3 ) = 2.8
B. Calculate the pH after 17.50 mL of NaOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: First assume the acid-base neutralization reaction is irreversible and goes to completion: HC2H3O2 (aa) + ОH (ag) →H2O () + C2H3O2 (aa) Moles before reaction Change in moles Moles remaining after reaction Then equilibrium is established HCH3O2 (aa) 2H* (ag) + C2H3O2 (aa) I Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) E Equilibrium Concentration, Calculated (M)
Transcribed Image Text:B. Calculate the pH after 17.50 mL of NaOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: First assume the acid-base neutralization reaction is irreversible and goes to completion: HC2H3O2 (aa) + ОH (ag) →H2O () + C2H3O2 (aa) Moles before reaction Change in moles Moles remaining after reaction Then equilibrium is established HCH3O2 (aa) 2H* (ag) + C2H3O2 (aa) I Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) E Equilibrium Concentration, Calculated (M)
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