logging in or signing up Ch18part2 Berenger Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINTLite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 1107 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 04, 2008 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript The Common Ion EffectSection 18.2: QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Here we are adding an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the ____________. The pH will go _____________. After all, NH4+ is an acid! The Common Ion Effect Section 18.2 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, FloridaSlide2: Let us first calculate the pH of a 0.25 M NH3 solution. [NH3] [NH4+] [OH-] initial 0.25 0 0 change -x +x +x equilib 0.25 - x x x QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq)Slide3: QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Assuming x is << 0.25, we have [OH-] = x = 0.0021 M This gives pOH = 2.67 and so pH = 11.33 for 0.25 M NH3Slide4: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Calc pH = 11.33 for 0.25 M NH3 We expect that the pH will decline on adding NH4Cl. The Common Ion EffectSection 18.2: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) [NH3] [NH4+] [OH-] initial 0.25 0.10 0 change -x +x +x equilib 0.25 - x 0.10 + x x The Common Ion Effect Section 18.2The Common Ion EffectSection 18.2: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Because equilibrium shifts left, x is MUCH less than 0.0021 M, the value without NH4Cl. The Common Ion Effect Section 18.2The Common Ion EffectSection 18.2: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) [OH-] = x = 4.5 x 10-5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion. The Common Ion Effect Section 18.2Buffer Solutions: Buffer Solutions HCl is added to pure water. HCl is added to a solution of a weak acid H2PO4- and its conjugate base HPO42-.Buffer Solutions: The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the common ion effect. Buffer Composition Weak Acid + Conj. Base HOAc + OAc- H2PO4- + HPO42- Weak Base + Conj. Acid NH3 + NH4+ Buffer SolutionsBuffer Solutions: ACID USES UP ADDED OH- OAc- + H2O HOAc + OH- Kb = 5.6 x 10-10 Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb = 1.8 x 109 Kreverse is VERY LARGE, so HOAc completely snatches up OH- ! Buffer SolutionsBuffer Solutions: CONJ. BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 Therefore, the reverse reaction of the WEAK BASE with added H+ has Kreverse = 1/ Ka = 5.6 x 104 Kreverse is VERY LARGE, so OAc- completely snatches up H+ ! Buffer SolutionsBuffer Solutions: Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] initial change equilib Buffer SolutionsBuffer Solutions: [HOAc] [OAc-] [H3O+] initial 0.700 0.600 x change -x +x +x equilib 0.700 - x 0.600 + x x Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 Buffer SolutionsBuffer Solutions: [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have [H3O+] = 2.1 x 10-5 and pH = 4.68 Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 Buffer SolutionsBuffer Solutions: Notice that the H+ or OH- concs. depend on K and the ratio of acid and base concs. Buffer SolutionsBuffer Solutions: Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? Ka = 1.8 x 10-5 pH = 4.74 + log [0.600M/0.700M] pH = 4.74 –0.07 = 4.67Henderson-Hasselbalch Equation: Henderson-Hasselbalch Equation This shows that the pH is determined largely by the pKa of the acid and then adjusted by the ratio of acid and conjugate base. Adding an Acid to a Buffer: Adding an Acid to a Buffer Problem: What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (a) Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of water M1•V1 = M2 • V2 M2 = 1.00 x 10-3 M pH = 3.00Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer) The reaction occurs completely because K is very large. What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H3O+] [OAc-] [HOAc] Before rxn Change After rxn What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H3O+] [OAc-] [HOAc] Before rxn 0.00100 0.600 0.700 Change -0.00100 -0.00100 +0.00100 After rxn 0 0.599 0.701 What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [H3O+] [OAc-] Before rxn 0.701 0.599 0 Change -x +x +x After rxn 0.710-x 0.599+x x What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [H3O+] [OAc-] After rxn 0.710-x 0.599+x x Because [H3O+] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599. What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed on adding HCl to the buffer! What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Preparing a Buffer: Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10-5 M It is best to choose an acid such that [H3O+] is about equal to Ka (or pH pKa). —then you get the exact [H3O+] by adjusting the ratio of acid to conjugate base. Preparing a Buffer Solution: Preparing a Buffer Solution Buffer prepared from HCO3- weak acid CO32- conjugate base HCO3- + H2O ¸ H3O+ + CO32-Preparing a Buffer: Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M POSSIBLE ACIDS Ka HSO4- / SO42- 1.2 x 10-2 HOAc / OAc- 1.8 x 10-5 HCN / CN- 4.0 x 10-10 Best choice is acetic acid / acetate.Preparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Preparing a BufferPreparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Preparing a BufferPreparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Solve for [HOAc]/[OAc-] ratio = 2.78/ 1 Preparing a BufferPreparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Solve for [HOAc]/[OAc-] ratio = 2.78/ 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30. Preparing a BufferPreparing a Buffer: A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. Preparing a Buffer You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
Ch18part2 Berenger Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINTLite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 1107 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 04, 2008 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript The Common Ion EffectSection 18.2: QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Here we are adding an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the ____________. The pH will go _____________. After all, NH4+ is an acid! The Common Ion Effect Section 18.2 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, FloridaSlide2: Let us first calculate the pH of a 0.25 M NH3 solution. [NH3] [NH4+] [OH-] initial 0.25 0 0 change -x +x +x equilib 0.25 - x x x QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq)Slide3: QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Assuming x is << 0.25, we have [OH-] = x = 0.0021 M This gives pOH = 2.67 and so pH = 11.33 for 0.25 M NH3Slide4: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Calc pH = 11.33 for 0.25 M NH3 We expect that the pH will decline on adding NH4Cl. The Common Ion EffectSection 18.2: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) [NH3] [NH4+] [OH-] initial 0.25 0.10 0 change -x +x +x equilib 0.25 - x 0.10 + x x The Common Ion Effect Section 18.2The Common Ion EffectSection 18.2: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Because equilibrium shifts left, x is MUCH less than 0.0021 M, the value without NH4Cl. The Common Ion Effect Section 18.2The Common Ion EffectSection 18.2: Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) [OH-] = x = 4.5 x 10-5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion. The Common Ion Effect Section 18.2Buffer Solutions: Buffer Solutions HCl is added to pure water. HCl is added to a solution of a weak acid H2PO4- and its conjugate base HPO42-.Buffer Solutions: The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the common ion effect. Buffer Composition Weak Acid + Conj. Base HOAc + OAc- H2PO4- + HPO42- Weak Base + Conj. Acid NH3 + NH4+ Buffer SolutionsBuffer Solutions: ACID USES UP ADDED OH- OAc- + H2O HOAc + OH- Kb = 5.6 x 10-10 Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb = 1.8 x 109 Kreverse is VERY LARGE, so HOAc completely snatches up OH- ! Buffer SolutionsBuffer Solutions: CONJ. BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 Therefore, the reverse reaction of the WEAK BASE with added H+ has Kreverse = 1/ Ka = 5.6 x 104 Kreverse is VERY LARGE, so OAc- completely snatches up H+ ! Buffer SolutionsBuffer Solutions: Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] initial change equilib Buffer SolutionsBuffer Solutions: [HOAc] [OAc-] [H3O+] initial 0.700 0.600 x change -x +x +x equilib 0.700 - x 0.600 + x x Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 Buffer SolutionsBuffer Solutions: [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have [H3O+] = 2.1 x 10-5 and pH = 4.68 Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 Buffer SolutionsBuffer Solutions: Notice that the H+ or OH- concs. depend on K and the ratio of acid and base concs. Buffer SolutionsBuffer Solutions: Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? Ka = 1.8 x 10-5 pH = 4.74 + log [0.600M/0.700M] pH = 4.74 –0.07 = 4.67Henderson-Hasselbalch Equation: Henderson-Hasselbalch Equation This shows that the pH is determined largely by the pKa of the acid and then adjusted by the ratio of acid and conjugate base. Adding an Acid to a Buffer: Adding an Acid to a Buffer Problem: What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (a) Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of water M1•V1 = M2 • V2 M2 = 1.00 x 10-3 M pH = 3.00Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer) The reaction occurs completely because K is very large. What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H3O+] [OAc-] [HOAc] Before rxn Change After rxn What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H3O+] [OAc-] [HOAc] Before rxn 0.00100 0.600 0.700 Change -0.00100 -0.00100 +0.00100 After rxn 0 0.599 0.701 What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [H3O+] [OAc-] Before rxn 0.701 0.599 0 Change -x +x +x After rxn 0.710-x 0.599+x x What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [H3O+] [OAc-] After rxn 0.710-x 0.599+x x Because [H3O+] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599. What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Adding an Acid to a Buffer: Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed on adding HCl to the buffer! What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)Preparing a Buffer: Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10-5 M It is best to choose an acid such that [H3O+] is about equal to Ka (or pH pKa). —then you get the exact [H3O+] by adjusting the ratio of acid to conjugate base. Preparing a Buffer Solution: Preparing a Buffer Solution Buffer prepared from HCO3- weak acid CO32- conjugate base HCO3- + H2O ¸ H3O+ + CO32-Preparing a Buffer: Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M POSSIBLE ACIDS Ka HSO4- / SO42- 1.2 x 10-2 HOAc / OAc- 1.8 x 10-5 HCN / CN- 4.0 x 10-10 Best choice is acetic acid / acetate.Preparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Preparing a BufferPreparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Preparing a BufferPreparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Solve for [HOAc]/[OAc-] ratio = 2.78/ 1 Preparing a BufferPreparing a Buffer: You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Solve for [HOAc]/[OAc-] ratio = 2.78/ 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30. Preparing a BufferPreparing a Buffer: A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. Preparing a Buffer