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# Chapter 16 Acid – Base Equilibria

Chapter 16 Acid – Base Equilibria. End-of-Chapter Problems: See Next Page; will answer questions on homework when through with Ch 16. Online HW 16a (due Friday, 4/19) &amp; 16b (due Friday, 4/26) Quiz #1 on Friday, 4/26/2013 &amp; Exam #1 on Monday, 4/29/2013.

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## Chapter 16 Acid – Base Equilibria

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1. Chapter 16Acid – Base Equilibria End-of-Chapter Problems: See Next Page; will answer questions on homework when through with Ch 16. Online HW 16a (due Friday, 4/19) & 16b (due Friday, 4/26) Quiz #1 on Friday, 4/26/2013 & Exam #1 on Monday, 4/29/2013

2. End-of-Chapter Problemspg 700 - 708, Chapter 16 These are good problems; a couple of exam questions might come from these 1, 2, 3, 5, 6; 8 through 20; 26, 29, 30, 31, 35, 37, 41, 43, 50, 59, 60, 63, 65, 71, 75, 79, 85, 87, 115, 117a (& Ka if pH=4.89 – 1/2 way to end pt)

3. I. Ka Values A. Introduction • * In ch 15 we calculated [H+] for strong acids & strong bases: [H+] is = Mof the strong acid; [OH-] = M of the strong base. • * Weak acids (HA) - ch 16 - only partially ionize. In order to determine [H+] of HA, we need to use the equilibrium reaction and equilibrium expression (review chapter 14 if necessary). • HA H+ + A-or HA (----) H+ + A-(eqrxn) • Ka = [H+] [A- ] (product on top; exponents = balancing coefficients) [HA] (this is the EQUILIBRIUM EXPRESSION) -Ka = weak acid equilibrium constant

4. I. Ka Values A. Introduction & B. Obtaining - Ka = Large for Strong Acids, > 1 Ka = Small for Weak Acids, < 1 - Examples: (only need to memorize Kw for water) Ka = 1.7 x 10-5for HC2H3O2Ka = 6.8 x 10-4for HF Ka > 50for HClO4 B. Obtaining Ka Values (& % Ionization): 1) Ka values from Table: (Table 16.1, Page 667 or A-13&14) 2) Spectrophotometric Exp. (in 122 lab) & Titration Curves Exp. (in 123lab) 3) pH Measurements Example: Calculate a) Ka & b) % ionization of 0.10 M phenol in water if pH = 5.43. Phenol = C6H5OH (continued on next page) HA A- + H+

5. Example: Calculate a) Ka & b) % ionization of 0.10 M phenol in water if pH = 5.43. Phenol = C6H5OH (continued) • pH = 5.43for 0.10 M HA.[H+] = 10-5.43 = 3.7 x 10-6 M • If x = M of phenol (HA) that ionizes, then: x = 3.7 x 10-6 M = [H+] = [A-] HA H+ + A- Initial M: 0.10 0 0 Equilibrium M: 0.10 – 3.7 x 10-6 ≈ 0.10 3.7 x 10-6 3.7 x 10-6 [H+] [A-] [3.7 x 10-6] [3.7 x 10-6] a) Ka = = = 1.4 x 10-10 [HA] [0.10] b) % ionization = Part x 100% = 3.7 x 10-6M x 100% = 3.7 x 10-3 % Total 0.10 M

6. II. Calculations Involving HA Introduction - two solution methods – 1) quadratic & 2) approximation - If Ka is known, then can calculate equilibrium concentrations of HA, A-, H+ - Will frequently let X= mole/L (M) of HA that dissociates - May end up with quadratic equation when solving for X & can solve two ways: 1) Quadratic: a x2 + b x + c = 0 2)Approximation: If have Y± x, can drop x if small compared to Y. For HA + H+ A-[HA] – x ≈ [HA] ifx is small (causes, < 5% error by dropping). - Can use text rule to see if x is small enough to drop: [HA] / Ka > 100 - If [HA] / Ka > 100 then can drop xwrt [HA]. - Dropping x may greatly simply the math.

7. 1. Write equilibrium rxn& equilibriumexpression with given chemical as reactant. 2. This step is where a reaction (rxn) occurs before equilibrium – like in a titration. Assume rxn goes to completion & calculate new initialM’s. 3. Let X = moles/L of chemical that reacts. 4. Write down initial Mof each & then equilibriaMof each. 5. Make approximation & plug equilibriaM into equilibrium expression. 6. Solve for X. General Problem Solving Method for all Ch 16 Problems

8. II. Calculations Involving HA – Outline of rest of chapterNote: All problems worked by method on previous page • A. Weak Acid & Salt of Weak Base • B. Weak Base & Salt of Weak Acid • C. Summary • D. Common Ion (two starting chemicals) • E. Buffer(similar to common ion problem) • F. Titration(1. allow rxn to occur & calc. new initial M ) (2. determine equilibriavalues & solve for X .)

9. A. Overview: Weak Acid, Ka = small 1. Write equil. rxn & expression: HA H+ + A-Ka = [H+][A-] / [HA] 2. Let x = M of HA that ionizes; obtain initial & equil. molarities: [HA] [H+] [A-] Initial M: Y 0 0 EquilM: Y-x ≈ Y xxif [Y] / Ka > 100 3. Plug into EquilExp &Solve for x: Ka = [H+][A-] = x2 ≈ x2x = √ [Y] Ka [HA] [Y-x] [Y]

10. A. Weak Acids – Example 1 • Calculate the pH of 0.10 M Acetic Acid (HA), Ka = 1.7 x 10-5 HA H+ + A-1.7 x 10-5 = [H+ ][A- ] / [HA] Let x = M of acetic acid that ionizes, then At equilibrium: Initial M0.01 0 0 Equil M0.10 – x x x HA H+ + A- Check to see if can drop x wrt 0.10: [HA]/Ka = 0.10 / 1.7 x 10-5 = 6000 Yes, >100 1.7 x 10-5 = [H+ ][A- ] = [x][x] = [x][x] = x2 [HA] [0.10 - x] [0.10] [0.10] x = √ [0.10] (1.7 x 10-5) = 1.3 x 10-3M = [H+] pH = - log (1.3 x 10-3) = 2.89

11. A. Weak Acids - Notes Note: 1) For 0.10 M acetic acid with a Ka = 1.7x10-5, the approximation was OK. - When will it NOT be OK? When the answer causes > 5% error or when [HA]/Ka < 100 - Generally will NOT be OK when: 1)[HA] is small &/or 2) Ka is large Example:1.0x10-4Macetic acid. [HA]/Ka = [1.0x10-4] / 1.7x10-5 = 5.9 Here we can’t drop x & have to include [1.0x10-4 – x] & use quadratic eqn. Note:2) May be able to make a time saving assumption with polyprotic acids Polyprotic Acid = one which releases more than one H+ per molecule Example:H3PO4 H+ + H2PO41- H+ + HPO42- H+ + PO43- Ka1 = 7x10-3 Ka2 = 6x10-8 Ka3 = 4x10-13 Most of acid comes from 1st step; so can ignore contributions from other steps if causes less than a 5% error (true when Ka1 value is >1000 x Ka2).

12. A. Weak Acids - Polyprotic Acid (H2A) - Example 2 • Calculate [H+] of 2.0 M H2A. Given: Ka1 = 1.0x10-5 Ka2 = 2.0x10-10 H2A (-------) H+ + HA- & HA- (-------) H+ + A-2 - Ka1 > 1000xKa2so can ignore 2nd ionization - [H2A] / Ka1 = 2.0/1.0x10-5 = 2x105 so x is small wrt 2.0 • Let x = M of H2A that reacts; x = M of H+ & Mof HA- • Calculate initial & equilibrium M’s H2A H+ + HA- Initial: 2.0 0 0 Equil: 2.0-x ≈2.0 x x 1.0x10-5 = [H+][HA-] = x2x = √2.0x10-5x= [HA-]=[H+] = 4.5 x 10-3M [H2A] 2.0

13. A. Weak Acids (Salts of Weak Bases) - Example 3, NH4Cl • NH3 = weak base, Kb = 1.8x10-5; the salt or conjugate is NH4+= weak acid, but its Ka value is usually not in a table. [ NH3 + H2O NH4+ + OH- ] • Note: Kw = Ka x KbKa = Kw/Kb = 1.00x10-14/1.8x10-5 = 5.6x10-10. • Now Treat NH4+ like a weak acid • Calculate pH of 2.0 MNH4Cl = 2.0 M NH4+ + 2.0 MCl-(can ignore Cl-. Why?) [ HA H+ A- ] NH4+H+ + NH3Ka = [H+][NH3] 2.0-X ≈2.0 X X [NH4+] 5.6x10-10= [H+][NH3] = X2 X = √ 1.12x10-9 [NH4+]2.0 X = [H+] = 3.35x10-5pH = -log 3.35x10-5 = 4.48

14. B. Overview: Weak Bases including Salts (KCN) of Weak Acids - Group I Salts of weak acids (NaC2H3O2, KCN, KF, LiNO2) are soluble & ionize 100 %, but the anions (C2H3O2-,CN-, F-, NO2-) are weak bases & pull some H+ from water leaving OH-. - If place KCN in H2O, CN- is a base & pulls H+ off water to yield OH- CN- + H2O HCN + OH- Kb = [HCN][OH-] Note: these Kb’s usually not in table & [CN-] H2O drops out of equil. expression. - Get Kb from Ka value of HCN (4.9x10-10) using: Ka x Kb = Kw (know this) Kb = Kw/Ka = 1.00x10-14 / 4.9x10-10 = 2.0x10-5 for KCN/NaCN/CN-(in water)

15. B. Weak Bases Example 1: Calc [OH-] & [H+] of 1.0 M KCN (in water); Given: Ka=4.9x10-10 for HCN Note: 1.0 M KCN = 1.0 M CN- Kb = Kw/Ka= 1.00x10-14/ 4.9x10-10= 2.0x10-5 Let x = M of CN- that reacts; assumption OK ie[CN-]/Kb = 1.0/2.0x10-5 = 50000 CN-+ H2O HCN + OH- Init.M:1.0 - 0 0 EquilM: 1.0-x ≈ 1.0 - x x 2.0x10-5 = [HCN] [OH-] = x2 [CN-] 1.0 x = √ 2.0x10-5 x 1.0 = 4.5x10-3 = [OH-] Now calculate [H+] 1.00x10-14 = [H+] [OH-] = [H+] [4.5x10-3] [H+] = 2.2x10-12

16. C. Weak Bases - Example 2 • Calculate a) the [OH-] & b) the pH of 0.50 M NH3in water. Given: Kb = 1.8x10-5 a) Let x = M of NH3 that reacts Note: [NH3]/Kb = 0.5 / 1.8x10-5 > 100 (drop x) NH3 + H2O (-----) NH4+ + OH- EquilM: 0.50-x ~ 0.50 x x 1.8x10-5 = [NH4+][OH-] = x2x = [OH-] = √ (1.8x10-5) x 0.50 = 3.0x10-3M [NH3]0.50 b) [OH-] = 3.0x10-3M now calculate the pH Kw =[H+] [OH-] [H+] = Kw / [OH-] = 1.00x10-14 / 3.0x10-3 = 3.3x10-12M pH = - log 3.3x10-12 = 11.5

17. C. Summary - Weak acid, HA: calculate [H+] using equilibrium rxn & exprn. -Weak acids can be salts of weak bases, like NH4Cl; want to give up H+. - Ka x Kb = 1.00x10-14 can be used to obtain needed K value. - Weak base, NH3 or CN-, react with H2O increasing [OH-]; calculate [OH-] then [H+] if needed (from 1.00x10-14 = [H+][OH-]) - Group Ia ions & Salts of STRONG acidslikeNaCl, LiI, KNO3DO NOT INFLUENCE pH & are Neutral. • What are acidities of NaClO4, KCl, NH4Cl, KF in water? NaClO4 & KCl = neutralNH4Cl = acidicKF = basic

18. D. Overview: Common Ion Problem (Two Initial Chemicals) - Previous problems had one initial component; now figure acidity if have two initial components such as HA & A- - Method: a) set up equilibria b) let x = M of HA that reacts c) make simplifications & solve for x - Example 1: 1 M HA & 2 M NaA (2 M A-); Calculate [H+] if Ka=8x10-5 (Let x = M of HA that ionizes.) HA H+ + A- Init M: 1 0 2 Equil M:1-x ≈ 1 x 2+x ≈ 2 8x10-5 = [H+] [A-] = [x] [2] x = [H+] = 8x10-5 / 2 = 4x10-5 [HA] [1]

19. D. Common Ion Effect. Example 2 • In the past, our calculations have involved only one component of an equilibria. Now predict the result when two components are added – a weak acid and its salt (common ion or conjugate) or a weak base and its salt (common ion or conjugate). • Calculate the pH of a solution which is 0.40 M in acetic acid (HOAc) and 0.20 M in sodium acetate (NaOAc). Let x = M of acetic acid that dissociates. Ka=1.7x10-5 Initial M: 0.40 0 0.20 HOAc H+ + OAc- EquilM:0.40-x x 0.20+x [HOAc]/Ka > 100 for 0.20; drop x 1.7x10-5 = [H+][OAc-] = [x][0.20] x = [H+] = 3.4x10-5M [HOAc] [0.40] pH = -log 3.4x10-5 = 4.47 Notes: 1) pH in 0.4 MHOAc= 2.6; pH in 0.4 MHOAc& 0.2 MNaOAc = 4.5 2) Le Chatelier’s Principle shows why adding 0.20 MNaOAc is more basic.

20. Review Problems. Ka of HF = 7x10-4 & X is small wrt # for each 1. Calculate [H+] of 1.0 M HF. 1.0-X X X HF H+ + F-7x10-4=[H+][F-]/[HF] = X2/1.0 X=[H+]= 0.03 M 2. Calculate [OH-] of 1.0 M KF. Kb = 1.0x10-14/7x10-4 = 1.4x10-11 1.0-X X X F-+ H2O HF + OH-1.4x10-11=[HF][OH-]/[F-] = X2/1.0 X=[OH-]= 4x10-6M 3. Calculate [H+] of mixture of 1.0 M HF & 2.0 M KF. 1.0-X X 2.0+X HF H+ + F- 7x10-4=[H+][F-]/[HF] = [X][2.0]/[1.0] X=[H+]= 4x10-4M

21. E. Overview: Buffer Problems - Buffer Definition: A solution which resists a change in pH. - Mixtures of weak acids & their salts or weak bases & their salts are buffers (are really just common ion problems). - Why does a weak acid (HA) & its salt (NaA or A-) resist a change in pH when either additional H+ or OH- are added? Because a buffer consumes both H+ & OH-. HA reacts with the OH-HA + OH- -----) A- + H2O A- reacts with the H+A- + H+-----) HA - If HA and A- are available, added H+ or OH- are consumed .

22. E. Buffers – General Features Example: Water – H2O If add 1 drop 10 M HCl to 10 mL water the pH goes from 7 to 1 If add 1 drop 10 M NaOH to 10 mL water the pH goes from 7 to 13 Water is not a buffer; the ΔpH was 6 in both cases; or the Δ[H+] = 106M change! Example: 1 M Acetic Acid and 1 M Sodium Acetate - HA & NaA If add 1 drop 10 M HCl to 10 mL of the above, the pH goes from 4.8 to 4.7 If add 1 drop 10 M NaOH to 10 mL of the above, the pH goes from 4.8 to 4.9 The Δ pH = 0.1 in both cases, or the Δ[H+] = 10-1M change! Notes: a) the buffer capacity is a measure of the amount of strong acid or base consumed by the buffer; this depends upon the initial M of the HA and A-. Best to have both large M of HA & A- as well as a 1:1 ratio of HA to A-. b) the effectivepH range of the buffer = pKa ± 1 pH unit. pKa of acetic acid = 4.8 so, the effective buffer range = 3.8 to 5.8

23. E. Buffers Henderson-Hasselbalch Equation - Used for Buffers & Common Ion For a weak acid: HA H+ + A- ~1907 A.D. Ka = [H+ ][A- ] [H+] = Ka [HA] Now take negative [HA] [A-] log of both sides. pH = pKa –log [HA] pH = pKa + log [A- ] [A-] [HA] pH = pKa + log [A- ] ~1917 A.D. [HA] Henderson-Hasselbalch Equation (H-H eqn) Base Acid

24. E. Buffers – Notes (important) 1) A buffer works best when [HA]/[A-] = 1:1 = 1.00 When a) [HA] = [A-] = small, or when b) ratio of HA to A- is far from 1:1, then the buffer is poor. When generating a buffer, choose a buffer with a pKawithin ±1.0 of the desired acidity. Which acid will buffer better at pH=4? HF (Ka = 7x10-4) or HCN (Ka = 5x10-10) 2) A quick lab way to get pKa is to measure pH of a 1:1 mix of M HA & A-(true at half way to the end point in a titration). pH = pKa + log [A- ]/[HA] pH = pKa + log 1/1 pH = pKa + 0 pH = pKa 3) Note: Can use either moles or M in HH equation; OK since volume cancels out. 4) A weak base and its salt such as NH3 & NH4Cl (NH4+) is also a buffer. 5) Buffer Problem Solving Method when adding strong H+ or OH- : a) allow the added strong acid or base to react completely with buffer b) calculate new moles (or M) of HA & A- left over after the reaction c) plug into the HH Equation & solve for pH

25. E. Buffers - Example Problem 1 • Calculate the pH of a buffer made from adding 0.50 mole of acetic acid and 0.30 mole of sodium acetate to enough water to make 2.0 L. pH = pKa + log [A-]/[HA] = -log 1.7x10-5 + log [0.30]/[0.50] pH = -log 1.7x10-5 + log (0.60) = 4.77 + (-0.22) = 4.55 Notes: 1) we made OK assumption that [HA] – x ≈ [HA] and [A-] + x ≈ [A-] 2) volume drops out of H-H eqn; so, can use either moles or M • Biological buffers: Proteins like egg albumin or milk casein H2PO4-1 & HPO4-2 H2CO3 and HCO3-

26. E. Buffers - Example Problem 2, part 1 • 0.100 moles of HA (Ka=1.00x10-8) is mixed with 0.200 moles NaA. Calculate the pH. Calculate the pH. Note: can use moles in H-H eqn & HA/Ka = 107 Initial moles: 0.100 ? 0.200 HA H+ + A- Equil. moles: 0.100-x = 0.100 x 0.200+x = 0.200 pH = pKa + log [A- ] / [HA] = -log (1.00x10-8) + log [0.200/0.100] pH = 8.00 + 0.301= 8.30

27. E. Buffers - Example Problem 2, part 2 (same prob. but add NaOH ) • 0.100 moles of HA (Ka=1.00x10-8, pKa = 8.00)is mixed with 0.200 moles NaA. Now Calculate the pH after adding 0.0100 moles of NaOH. (HA/Ka = 107) Note: a) First allow OH- to react with HA, producing more A-. Then calculate new initial MOLES of HA & A-. b) Solve using Henderson Hasselbalch equation. Initial moles: 0.100 0.0100 0.200 -- a) Reaction (before equil):1HA + 1OH- -----) 1A- + 1H2O New Initial moles: 0.090 0.0000 0.210 New initial moles HA = 0.100 mole – 0.0100 mole = 0.090mole HA New initial moles A- = 0.200 mole + 0.0100 mole = 0.210 mole A- New initial moles OH- = 0.0100 mole - 0.0100 mole = 0.0000 mole OH- b) Solve for pH (after equil): HA (----) H+ + A-HA=0.090, A-=0.210, H+=X pH = pKa + log [A- ] / [HA] = 8.00 + log [0.210/0.090] = 8.00 + 0.37 = 8.37

28. F. Titrations – general information Definition: Volumetric determination of the amount of an acid or base by addition of a standard acid or base until neutralization has occurred. Notes: 1) Neutralization Point = End Point or Equivalence Point (equal m acid & base) 2) A burette (buret) is usually used to add the titrant & can be read to ±0.01 mL 3) a plot of pH vs volume of titrant is called a pH curve or a titration curve. 4) End Point can be determined by either an indicator or by use of a pH meter. Pick an indicator that changes color near to the pH at the end point, or if using a pH meter, the steepest portion of the curve is the end point (can also plot the 1st or 2nd derivative vs volume added). 5) Can easily determine Ka since pH = pKa half way to the end point for a weak acid. 6) A quick examination of the pH curve will tell if titrating an acid or base as well as if they are strong or weak.

29. F. Overview: Titrations

30. F. Overview: Titration Curve (plot pH vs Vol Std) Calculations • pH Determination while Titrating a Strong Acid like HCl (or Base) • Before Titration Starts: Initial pH from M of HCl • During Titration & Before End Point: Allow reaction to occur; Calculate excess HCl left over; pH from excess HCl. • At End Point: Allow rxn to occur; Only NaCl left in water; pH = 7 • After End Point: Allow reaction to occur; Calculate Excess NaOH; pH from excess NaOH. • pH Determination while Titrating a Weak Acid like HA (or Base) • Before Titration Starts: Initial pH from M of HA • During Titration & Before End Point: Allow reaction to occur; Calculate excess HA & A- left over; pH from excess HA & A-; can use HH eqn. • At End Point: Allow reaction to occur; Only have A- ; pH from reaction of A- with H2O (get pH from MA- & Kb.) • After End Point: Allow reaction to occur; Calculate Excess NaOH; pH from excess OH- .

31. F. Titration #1 - 25.0 mL of 0.100 M HCl with 0.100 M NaOH (Strong Acid with a Strong Base). Note: Large change in pH near end point and pH at equivalence point = 7.00

32. F. Titration #2 - 25.0 mL of 0.100 M Nicotinic Acid (Ka=1.4x10-5) with 0.100 M NaOH (Weak Acid with a Strong Base). Note: Smaller change in pH near end point & pH at equival. point = 8.78

33. F. Titration of a Strong Acid with NaOH: Calculations • Calculate pH for titration of 10. mL of 0.80 M HCl with 1.0 M NaOH at 0.0 mL, 3.0 mL, 8.0 mL & 12.0 mL 1.0 MNaOH added. • Notes: allow reaction to occur & then calculate pH. The pH is due to excess strong acid or base except at end point. • Titration Rxn: 1HCl + 1NaOH 1NaCl + 1H2O • 0.0mL NaOH added: Only have HCl. pH = - log [0.80] = 0.097 • 3.0 mL NaOH added: 0.0080 mol HCl + 0.0030 mol NaOH Yields 0.0050 moles of HCl + 0.0030 moles of NaCl left over in a total volume of 13.0 mL. 0.0050 m / 0.0130 L = 0.384 M HCl pH = - log [0.384] = 0.41

34. F. Titration of a Strong Acid with NaOH: Calculations - Continued 1HCl + 1NaOH 1NaCl + 1H2O • 8.0 mL NaOH added: have 0.0080 mol HCl & 0.0080 mol NaOH. All HCl & NaOH reacted (end pt)& have 0.0080 molNaCl(salt of strong acid & base) in 16.0 mL H2O. pH = pH of neutral water = 7.0 • 12.0 mL NaOH added: 0.0080 mol HCl + 0.012 mol NaOH. Yields 0.000 mol HCl, 0.004 mol NaOH, & 0.0080 mol NaCl in a total volume of 22.0 mL (after end pt.) 0.004 m / 0.022 L = 0.18 M NaOH. pOH = - log [0.18] = 0.74 pH = Kw - pOH = 14.00 – 0.74 = 13.3

35. F. Titration of a weak Acid with NaOH: Calculations • Calculate pH for titration of 10. mL of 0.80 M HA with 1.0 M NaOH at 0.0mL, 3.0mL, 8.0 mL & 12.0 mL 1.0 M NaOH added. Ka=1.0x10-6 • Notes: allow reaction to occur & then calculate pH. Titration Rxn: 1HA + 1NaOH 1NaA + 1H2O Equil. Rxn: HA H+ + A- 1.0x10-6 = [H+][A-] / [HA] • 0.0 mL added: Only have weak acid. Let x = M of HA that reacts. At equilibrium: [H+] = [A-] = x; [HA] = [0.80-x] ≈ 0.80 1.0x10-6 = [H+][A-] = x2 x = 8.9x10-4 = [H+] [HA] 0.80 pH= - log [8.9x10-4] = 3.05 (start less acidic – higher pH than HCl titration)

36. F. Titration of a weak Acid with NaOH: Calculations - continued Notes: allow reaction to occur & then calculate pH. 1HA + 1OH- 1A- + 1H2O Rxn 1 HA H+ + A- 1.0x10-6 = [H+][A-] / [HA] Rxn 2 • 3.0 mL added: Calculate mols; allow OH- & HA to react - Rxn #1 0.0080 mol HA + 0.0030 mol NaOH yields: 0.0000 mol OH-+ 0.0050 mol of HA +0.0030 mol of A-left over in 13.0 mL. Note: 1) this is now a buffer - part way to end pt. 2) use Rxn #2, mols & HH eqn pH = pKa + log [A-] / [HA] = -log 1.0x10-6 + log [0.0030] / [0.0050] pH = 6.00 – 0.222 = 5.78

37. F. Titration of a weak Acid with NaOH: Calculations - continued • Notes: allow reaction to occur & then calculate pH. 1HA + 1NaOH 1NaA + 1H2O • 8.0 mL added: 0.0080 mol HA + 0.0080 mol NaOH (end pt.) Yields 0.0000 mol HA&OH-+ 0.0080 mol NaAin 18.0 mL. [A-] = 0.0080 m / 0.0180 L = 0.444M A-Only have A- in water: A- + H2O (------) HA + OH-Kb = Kw/Ka = 1.0x10-14 / 1.0x10-6 = 1.0x10-8 Let x = M of A- that reacts; [A-] = 0.44-x = 0.44; [HA] = [OH-] = x 1.0x10-8 = [HA][OH-]/[A-] = x2 / 0.44 x = [OH-] = 6.7x10-5 [H+] = Kw/[OH-]= 1.0x10-14 / 6.7x10-5= 1.5x10-10pH = 9.82(end pt.)

38. F. Titration of a weak Acid with NaOH: Calculations - continued Notes: allow reaction to occur & then calculate pH. 1HA + 1NaOH 1NaA + 1H2O • 12.0 mL added: 0.0080 mol HA + 0.0120 mol NaOH Yields 0.0000 mol HA + 0.0080 mol NaA + 0.0040 mol of NaOH left over in 22.0 mL.(after end pt.) [A-] = 0.0080 m / 0.0220 L = 0.36 M A-(weak base) [OH-] = 0.0040 m / 0.0220 L = 0.18 M OH-(strong base) Note: In mixture of weak & strong bases, pH mainly due to strong base: [H+] = Kw / [OH-] = 1.0x10-14 / 0.18 = 5.5x10-14 pH = 13.26

39. III. Summary Let X = M of reactant that reacts to reach equilibria • Weak Acids (HA): HA (---) H++A- Ka=[H+][A-]/[HA] • Bases & WA Salts (A-): A-+H2O (---) HA+OH- Kb=[HA][OH-]/[A-] • Common Ion/Buffer (HA&A-): HA (---) H++A- Ka=[H+][A-]/[HA] Notes: 1. Can also use H-H Eqn for common ion pH = pKa + Log[base]/[acid] 2. If add strong H+ or OH-a) allow to react, b) calculate new mol of HA & A-,c) plug into H-H eqn. • Titrations: Strong H+ with OH-. a) initially: pH due to strong H+. b) before end pt.: allow to react & calculate excess strong H+. c) at end pt.: pH = 7.0. d) after end pt.: allow to react & calculate excess OH-, pH due to strong OH-. • Titrations: Weak HA with OH-. a) initially: pH due to HA. b) before end pt.: allow to react & calculate excess HA & A-. c) at end pt.: pH due to A-. d) after end pt.: allow to react & calculate excess OH-, pH due to strong OH-.

40. 1. Write equil. rxn & equil. expression with given chemical as reactant. May have to get Kb , then : Ka x Kb = 1.00 x10-14 2. If a rxn takes place (titration or possibly buffer with strong acid or base added), then allow rxn to occur & calculate the new initial M of reactants and products. 3. Let X = M of chemical that reacts. 4. Write down initialM of each & equilibriaM of each (ICE table). 5. Make approx. & plug equilibriaM into equil. expression (or moles into HH eqn). 6. Solve for X. (May have to convert X to pH or pOH). III. Summary - One method for all weak acid-base problems

41. Given M of HA HA (---) H+ + A-Ka = [H+][A-] = x2 Solve for x [HA] [HA] Given M of A- A- + H2O (---) HA + OH-Kb = 10-14/Ka = [HA][OH-] = x2 [A-] [A-] Solve for x = [OH-] & convert to [H+] with 1.00x10-14 = [H+][OH-] Given M of both HA & A- HA (---) H+ + A- Ka=[H+][A-] or use pH = pKa + log [A- ] / [HA] [HA] Reaction : Acid & add OH- or Base & add H+ Allow to react; calculate new starting m or M & solve by one of methods above. III. Summary - Calculating [H+]Let x = M of species reaching equilibrium

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