1 / 41

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) & 16b (due Friday, 4/26) Quiz #1 on Friday, 4/26/2013 & Exam #1 on Monday, 4/29/2013.

inocencia
Download Presentation

Chapter 16 Acid – Base Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related