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Acid-Base Reactions

Acid-Base Reactions

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Acid-Base Reactions

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  1. Acid-Base Reactions Conjugates do not react!! CHAPTER 17 Part 1

  2. Stomach Acidity &Acid-Base Reactions

  3. ACIDS-BASE REACTIONS • For any acid-base reaction where only one hydrogen ion is transferred, the equilibrium constant for the reaction can be calculated and is often called Knet. • For the general reaction: • Knet is always Ka(reactant acid) / Ka(product acid) • When Knet >> 1, products are favored. • When Knet << 1, reactants are favored. See 16.5 for manipulating K

  4. ACIDS-BASE REACTIONS • There are four classifications or types of reactions: strong acid with strong base, strong acid with weak base, weak acid with strong base, and weak acid with weak base. • NOTE: For all four reaction types the limiting reactant problem is carried out first. Once this is accomplished, one must determine which reactants and products remain and write an appropriate equilibrium equation for the remaining mixture.

  5. STRONG ACID WITH STRONG BASE The net reaction is: The product, water, is neutral.

  6. STRONG ACID WITH WEAK BASE The net reaction is: The product is HB+ and the solution is acidic.

  7. WEAK ACID WITH STRONG BASE The net reaction is: The product is A- and the solution is basic.

  8. WEAK ACID WITH WEAK BASE The net reaction is: Notice that Knet may even be less than one. This will occur when Ka HB+ > Ka HA.

  9. Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na+ + Bz- + H2O Ka = 6.3 × 10-5 Kb = 1.6 × 10-10 C6H5CO2H = HBz Benzoate ion = Bz-

  10. Bz- + H2O HBz + OH- + + Kb = 1.6 x 10-10 Acid-Base Reactions 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.

  11. Acid-Base Reactions 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? Strategy —find the concentration of the conjugate base Bz- in the solution AFTER the titration, then calculate pH. This is a two-step problem: 1st. stoichiometry of acid-base reaction 2nd. equilibrium calculation

  12. Acid-Base Reactions 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? STOICHIOMETRY PORTION 1. Calculate moles of NaOH required. (0.100 L HBz)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2. Calculate volume of NaOH required. 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH required

  13. Acid-Base Reactions 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? STOICHIOMETRY PORTION 25 mL of NaOH required 3. Moles of Bz- produced = moles HBz = 0.0025 mol Bz- 4. Calculate concentration 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

  14. Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point (mol HBz = mol NaOH). What is the pH of the final solution? Note: HBz and NaOH are used up! HBz + NaOH ---> Na+ + Bz- + H2O Ka = 6.3 × 10-5 Kb = 1.6 × 10-10 C6H5CO2H = HBz Benzoate ion = Bz-

  15. Acid-Base Reactions 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? EQUILIBRIUM PORTION Bz- + H2O HBz + OH- Kb = 1.6 x 10-10 [Bz-][HBz] [OH-] initial change equilib 0.020 0 0 -x +x +x 0.020 - x x x

  16. 2 x -10 K 1.6 x 10 = = b 0.020 - x Acid-Base Reactions 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? 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

  17. Acid-Base Reactions 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? HBz+ H2O H3O+ + Bz- Ka = 6.3 x 10-5 [H3O+] = { [HBz] / [Bz-] } Ka At the half-way point, [HBz] = [Bz-], so [H3O+] = Ka = 6.3 x 10-5 pH = 4.20

  18. 18 The Common Ion Effect 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 ____________. LEFT Closer to 0 The pH will go _______. After all, NH4+ is an acid!

  19. QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) The Common Ion Effect 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

  20. QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) - 2 + [NH ][OH ] x -5 4 K 1.8 x 10 = = = b [NH ] 0.25 - x 3 The Common Ion Effect Assuming x is << 0.25, we have [OH-] = x = [Kb(0.25)]1/2 = 0.0021 M This gives pOH = 2.67 and so - - - pH = 14.00 - 2.67 = 11.33 for 0.25 M NH3

  21. QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) The Common Ion Effect We expect that the pH will decline on adding NH4Cl. Let’s test that at 0.10M ! [NH3] [NH4+] [OH-] initial change equilib 0.25 0.10 0 -x +x +x 0.25 - x 0.10 + x x

  22. QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) - + [NH ][OH ] x(0.10 + x) -5 4 K 1.8 x 10 = = = b [NH ] 0.25 - x 3 The Common Ion Effect Because equilibrium shifts left, x is MUCH less than 0.0021 M, the value without NH4Cl. [OH-] = x = (0.25 / 0.10)Kb = 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.

  23. Buffer Solutions HCl is added to pure water. HCl is added to a solution of a weak acid H2PO4- and its conjugate base HPO42-.

  24. 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 HC2H3O2 + C2H3O2- H2PO4- + HPO42- Weak Base + Conj. Acid NH3 + NH4+

  25. Buffer Solutions Consider HOAc/OAc- to see how buffers work. ACID USES UP ADDED OH-. We know that OAc- + H2O HOAc + OH-has 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 uses up the OH- !!!!

  26. Buffer Solutions Consider HOAc/OAc- to see how buffers work. CONJUGATE BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ has 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 uses up the H+ !

  27. Buffer 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 0.700 0.600 0 -x +x +x 0.700 - x 0.600 + x x

  28. + [H O ](0.600) -5 3 K 1.8 x 10 = = a 0.700 Buffer 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 [H3O+] = 2.1 x 10-5 and pH = 4.68 [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have

  29. Orig. conc. of HOAc + [ H O ] • K = 3 a - Orig. conc. of OAc [Acid] [Base] + - [ H O ] • K [ OH ] • K = = 3 a b [Conj. base] [Conj. acid] Buffer Solutions Notice that the expression for calculating the H+ concentration of the buffer is This leads to a general equation for finding the H+ or OH- concentration of a buffer. Notice that the H+ or OH- concentrations depend on K and the ratio of acid and base concentrations.

  30. [Acid] + [ H O ] • K = 3 a [Conj. base] [Acid] pH pK - log = a [Conj. base] [Conj. base] pH pK + log = a [Acid] Henderson-Hasselbalch Equation This is called the Henderson-Hasselbalch equation. Take the negative log of both sides of this equation OR

  31. [Conj. base] pH pK + log = a [Acid] 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.

  32. Adding an Acid to a Buffer Problem: What is the new 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) Calculate [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 = 3.00

  33. Adding an Acid to a Buffer 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) 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.

  34. Adding an Acid to a Buffer 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) 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

  35. Adding an Acid to a Buffer 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) Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [OAc-] [H3O+] Before rxn 0.701 0.599 0 Change -x +x +x After rxn 0.710-x 0.599+x x

  36. Adding an Acid to a Buffer 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) [HOAc] [OAc-] [H3O+] 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. Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc-

  37. [HOAc] 0.701 + -5 [ H O ] • K • ( 1 . 8 x 10 ) = = 3 a - 0.599 [OAc ] Adding an Acid to a Buffer 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) [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed significantly upon adding HCl to the buffer! Solution to Part (b): Step 2—Equilibrium HOAc + H2O OAc- +H3O+

  38. 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). You get the exact [H3O+] by adjusting the ratio of acid to conjugate base.

  39. Preparing a Buffer Solution Buffer prepared from HCO3- weak acid CO32- conjugate base HCO3- + H2O H3O+ + CO32-

  40. 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.

  41. [HOAc] + -5 -5 [ H O ] 5.0 x 10 = (1.8 x 10 ) = 3 - [OAc ] Preparing 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.

  42. Preparing 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. This simplifying approximation will be correct for all buffers with 3<pH<11, since the [H]+ will be small compared to the acid and conjugate base.

  43. REVIEW PROBLEMS • Calculate the pH of a 0.10 M HNO2 solution before and after making the solution 0.25 M in NaNO2. • Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

  44. REVIEW PROBLEMS • Calculate the pH of a solution that is 0.18 M in Na2HPO4and 0.12 M in NaH2PO4. 6.2 x 10-8 • Suggest an appropriate buffer system for pH 5.0. SOLUTIONS

  45. Titrations 45 pH Titrant volume, mL

  46. Acid-Base Titrations Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly.

  47. Acid-Base Titrations Additional NaOH is added. pH rises as equivalence point is approached.

  48. Acid-Base Titrations Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point.

  49. 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? Equivalence point pH of solution of benzoic acid, a weak acid

  50. 2 x -10 K 1.6 x 10 = = b 0.020 - x Acid-Base Titrations 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? 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