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Today . Reminder: (Ka)(Kb) = Kw Common Ion Effect – Lecture and Problems BUFFERS. Fake quiz question. Calculate the hydronium ion concentration and pH when 50.0 mL of 0.10 M acetic acid is mixed with 50.0 mL of 0.10 M NaOH. Ka for acetic acid = 1.8 x 10 -5 .

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Today

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  1. Today • Reminder: (Ka)(Kb) = Kw • Common Ion Effect – Lecture and Problems • BUFFERS

  2. Fake quiz question • Calculate the hydronium ion concentration and pH when 50.0 mL of 0.10 M acetic acid is mixed with 50.0 mL of 0.10 M NaOH. • Ka for acetic acid = 1.8 x 10-5. • Kb for the acetate ion: (Ka)(Kb) = Kw • Hydronium ion concentration = 2 x 10-9 M • pH = 8.7 • 1 pt for Kb = 5.6 x 10-10 • 2 pts for right answers!

  3. More About Chemical EquilibriaAcid-Base ReactionsChapter 15

  4. Stomach Acidity &Acid-Base Reactions

  5. Acid-Base Reactions What is relative pH before, during, & after reaction? a) Common ion effect and buffers b) Titrations • Strong acid + strong base HCl + NaOH ----> • Strong acid + weak base HCl + NH3 ---> • Weak acid + strong base HAc + NaOH ---> • Weak acid + weak base HAc + NH3 --->

  6. 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 ____________. The pH will go _____________. After all, NH4+ is an acid!

  7. pH of Aqueous NH3 QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O ↔NH4+(aq) + OH-(aq) 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 xx

  8. pH of Aqueous NH3 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 = [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

  9. pH of NH3/NH4+ Mixture 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) We expect that the pH will decline on adding NH4Cl. Let’s test that! [NH3] [NH4+] [OH-] initial change equilib

  10. pH of NH3/NH4+ Mixture 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) We expect that the pH will decline on adding NH4Cl. Let’s test that! [NH3] [NH4+] [OH-] initial 0.25 0.10 0 change -x +x +x equilib 0.25 - x 0.10 + x x

  11. pH of NH3/NH4+ Mixture 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 = (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. approximation

  12. Additional Problems We have a solution of ammonia. What is the effect of adding ammonium chloride to this solution? • 1. The pH increases. • 2. The concentration of NH3 increases. • 3. The concentration of H3O+ increases a) 1 only b) 2 only c) 3 only d) 1 and 2 e) 2 and 3 . . . . . . . . .

  13. Practice Problem 2 What is the pH of a mixture containing 0.30 M HNO2 and 0.15 M NaNO2? (Ka for HNO2 = 4.5  10-4) • 3.05 b) 4.05 c) 4.35 d) 4.65 e) 5.01 . . . . . . . . . a

  14. Practice Problem 3 What is the pH of the solution that results from adding 15 mL of 0.50 M NaOH to 25 mL of 0.50 M HF? (Ka for HF = 7.2  10-4) • 3.32 b) 3.49 c) 4.61 d) 7.53 e) 10.86 . . . . . . . . .a

  15. Practice Problem 4 What is the pH of the solution that results from adding 25 mL of 0.33 M HCl to 25 mL of 0.58 M NH3? (Kb for NH3 = 1.8  10-5) • 3.87 b) 4.62 c) 8.99 d) 9.13 e) 9.38 . . . . . . . . .d

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

  17. Buffer Solutions A buffer solution is a special case of the common ion effect. The function of a buffer is to resist changes in the pH of a solution. Buffer Composition Weak Acid + Conj. Base HOAc + OAc- H2PO4- + HPO42- Weak Base + Conj. Acid NH3 + NH4+

  18. 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 gobbles up OH- !!!!

  19. Buffer Solutions Consider HAc/Ac- to see how buffers work CONJ. BASE USES UP ADDED H+ HAc + H2O ↔ Ac- + 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 Ac- completely snarfs up H+ !

  20. Buffer Solutions Problem: What is the pH of a buffer that has [HAc] = 0.700 M and [Ac-] = 0.600 M? HAc + H2O ↔ Ac- + H3O+ Ka = 1.8 x 10-5 [HAc] ↔ [Ac-] [H3O+] initial change equilib 0.700 0.600 0 -x +x +x x 0.700 - x 0.600 + x

  21. Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HAc + H2O ↔ Ac- + H3O+ Ka = 1.8 x 10-5 [HAc] ↔ [Ac-] [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

  22. Buffer Solutions Notice that the expression for calculating the H+ conc. of the buffer is Notice that the H+ or OH- concs. depend on K and the ratio of acid and base concs.

  23. Henderson-Hasselbalch Equation Take the negative log of both sides of this equation The pH is determined largely by the pKa of the acid and then adjusted by the ratio of acid and conjugate base.

  24. 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 [HAc] = 0.700 M and [Ac-] = 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 = 3.00

  25. 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 [HAc] = 0.700 M and [Ac-] = 0.600 M (pH before = 4.68) Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + Ac- (from buffer) ---> HAc (from buffer) The reaction occurs completely because K is very large (think back to previous problem where K = 1/Ka)

  26. 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 (pH = 3.00) b) 1.00 L of buffer that has [HAc] = 0.700 M and [Ac-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 1—Stoichiometry [H3O+] + [Ac-] ---> [HAc] Before rxn Change After rxn 0.00100 0.600 0.700 -0.00100 -0.00100 +0.00100 0 0.599 0.701

  27. 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 (pH = 3.00) b) 1.00 L of buffer that has [HAc] = 0.700 M and [Ac-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HAc + H2O ↔ H3O+ + Ac- [HAc] ↔ [H3O+] + [Ac-] Before rxn Change After rxn 0.701 0 0.599 -x +x +x 0.701-x x 0.599 + x

  28. 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 (pH = 3.00) b) 1.00 L of buffer that has [HAc] = 0.700 M and [Ac-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HAc + H2O ↔ H3O+ + Ac- [HAc] ↔ [H3O+] + [Ac-] After rxn 0.701-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.

  29. 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 (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- [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed on adding HCl to the buffer!

  30. 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.

  31. 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 HAc / Ac- 1.8 x 10-5 HCN / CN- 4.0 x 10-10 Best choice is acetic acid / acetate.

  32. Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Solve for [HAc]/[Ac-] ratio Therefore, if you use 0.100 mol of NaAc and 0.278 mol of HAc, you will have pH = 4.30.

  33. 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. Result: diluting a buffer solution does not change its pH What is the difference between a ratio that is 0.1 : 0.1 and one that is 0.5 : 0.5? Buffering capacity: the amount of resistance to change in pH

  34. Preparing a Buffer Buffer prepared from 8.4 g NaHCO3- weak acid 16.0 g Na2CO32-conjugate base HCO3- + H2O  H3O+ + CO32- What is the pH?

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