Ch18part1

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Published on October 15, 2007

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Acid-Base Reactions:  Acid-Base Reactions 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, Florida Strong Acid and Strong Base HCl + NaOH  HCl + H2O --->H3O+ + Cl- NaOH(aq) ---> Na+(aq) + OH-(aq) H3O+ + Cl- + Na++ OH-  2 H2O + Cl- + Na+ Net ionic Equation doesn’t show spectator ions! Acid-Base Reactions:  Acid-Base Reactions 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, Florida H2O + H2O  H3O+ + OH- K = 1 x 10-14 H3O+ + OH-  2 H2O K = 1/ 1 x 10-14 = 1 x 1014 NEUTRALIZATION REACTION, pH = 7 Acid-Base Reactions:  Acid-Base Reactions 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, Florida Strong Acid and Weak Base NH3(aq) + H2O(liq)  NH4+(aq) + OH-(aq) HCl + NH3  H3O+ + OH-  2 H2O H3O+ + NH3  H2O + NH4+ Knet = Kb (1/Kw) = 1.8 x 109 Acid-Base Reactions:  Acid-Base Reactions 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, Florida STRONG Acid and Weak Base Mixing equal molar quantities gives an acidic solution. Weak Acid and STRONG Base Mixing equal molar quantities gives a basic solution. Acid-Base Reactions:  Acid-Base Reactions 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, Florida Weak Acid and Weak Base pH depends on Kb and Ka of conjugate base and acid. Stomach Acidity & Acid-Base Reactions:  Stomach Acidity & Acid-Base Reactions Example 18.1—Calculate pH:  Example 18.1—Calculate pH 100 mL of 0.10 M HCl with 50 mL of 0.20 M NH3 HCl + NH3  Cl- + NH4+ 0.100 L x 0.10 mol/L = 0.0100 mol HCl =0.0100 mol Cl- 0.050 L x 0.20 mol/L = 0.0100 mol NH3 =0.0100 mol NH4+ Example 18.1—Calculate pH:  Example 18.1—Calculate pH 100 mL of 0.10 M HCl with 50 mL of 0.20 M NH3 HCl + NH3  Cl- + NH4+ 0.0100 mol Cl- / 0.150 L solution = 0.0667 M Cl- 0.0100 mol NH4+ / 0.150 L solution = 0.0667 M NH4+ H2O + NH4+  H3O+ + NH3 Example 18.1—Calculate pH:  Example 18.1—Calculate pH Equilibria of NH4+ contributes to pH EQUIL Ka = x (x) / (0.0667 – x) Ka = 5.6 x 10-10 and x = 6.1 x 10-6 M H2O + NH4+  H3O+ + NH3 Ka = [H3O+ ] [NH3] / [NH4+] pH = 5.21 Acid-Base Reactions Section 18.4:  Acid-Base Reactions Section 18.4 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? HBz + NaOH ---> Na+ + Bz- + H2O C6H5CO2H = HBz Benzoate ion = Bz- Acid-Base Reactions Section 18.4:  Acid-Base Reactions Section 18.4 The product of the titration of benzoic acid, the benzoate ion, Bz-, is the conjugate base of a weak acid. The final solution is basic. Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 Acid-Base Reactions:  Acid-Base Reactions Strategy — find the conc. of the conjugate base B- in the solution AFTER the titration, then calculate pH. This is a two-step problem 1. stoichiometry of acid-base reaction 2. equilibrium calculation QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Acid-Base Reactions:  Acid-Base Reactions STOICHIOMETRY PORTION 1. Calc. moles of NaOH req’d (0.100 L HBz)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2. Calc. volume of NaOH req’d 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH req’d QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Acid-Base Reactions:  Acid-Base Reactions STOICHIOMETRY PORTION 25 mL of NaOH req’d 3. Moles of Bz- produced = moles HBz = 0.0025 mol 4. Calc. conc. of Bz- There are 0.0025 mol of Bz- in a TOTAL SOLUTION VOLUME of 125 mL [Bz-] = 0.0025 mol / 0.125 L = 0.020 M QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Acid-Base Reactions:  Acid-Base Reactions EQUILIBRIUM PORTION Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 [Bz-] [HBz] [OH-] initial change equilib QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Acid-Base Reactions:  Acid-Base Reactions EQUILIBRIUM PORTION Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 [Bz-] [HBz] [OH-] initial 0.020 0 0 change -x +x +x equilib 0.020 - x x x QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Acid-Base Reactions:  Acid-Base Reactions EQUILIBRIUM PORTION Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 [Bz-] [HBz] [OH-] equilib 0.020 - x x x Solving in the usual way, we find x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? Acid-Base Reactions:  Acid-Base Reactions HBz + H2O H3O+ + Bz- Ka = 6.3 x 10-5 At the half-way point, [HBz] = [Bz-], so [H3O+] = 1 (Ka ) = 6.3 x 10-5 pH = 4.20 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point? Henderson-Hasselbalch Equation:  Henderson-Hasselbalch Equation Take the negative log of both sides of this equation Henderson-Hasselbalch Equation:  Henderson-Hasselbalch Equation or This is called the Henderson-Hasselbalch equation. Henderson-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.

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