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Weak Acid Titration Calculations

Weak Acid Titration Calculations. Paul Gilletti Ph.D. Mesa Community College Mesa, AZ. Strong and Weak acids. Equilibrium and pH. Strong Acids: HClO 4 H 2 SO 4 HNO 3 HI HBr HCl HClO 3. Weak Acids: “The Rest”. 25.0 mL of 0.10M Acetic acid K a =1.8 x 10 -5. Starting pH.

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Weak Acid Titration Calculations

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  1. Weak Acid Titration Calculations Paul Gilletti Ph.D. Mesa Community College Mesa, AZ

  2. Strong and Weak acids Equilibrium and pH Strong Acids: HClO4 H2SO4 HNO3 HI HBr HCl HClO3 Weak Acids: “The Rest”

  3. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5 Starting pH Equivalence point 0.10 M NaOH

  4. Acetic acid is a weak acid. What would be the starting pH of the solution if it were a strong acid? (25.0 mL of 0.10M Acetic acid) Do a quick calculation as if acetic acid were a strong acid and compare this value to that on the graph. How do the two values compare? Weak Acid: Graph starting pH = 2.87 Calculated as Strong acid: pH = -log(0.10) = 1.0 How and Why are they different? “Let’s investigate!”

  5. Weak Strong

  6. How do the two values compare? Graph starting pH = 2.87 Strong acid calculation: pH = -log(0.10) = 1.0 What is the significance of this difference? Since the two values differ by ~2 pH units and pH is a log scale, the concentration of H3O+ in the strong acid calculation is 1 x 102 or 100 times greater than that observed on the graph. Let’s see WHY they are different?

  7. Strong Acids: 100% ionized (completely dissociated) in water. HCl + H2O  H3O+ + Cl- Note the “one way arrow”. Weak Acids: Only a small % (dissociated) in water. HC2H3O2 + H2O H3O+ + C2H3O2- Note the “2-way” arrow. Why are they different?

  8. Strong Acids: HCl HCl HCl HCl HCl (H2O) ADD WATER to MOLECULAR ACID

  9. Strong Acids: Cl- H3O+ (H2O) Cl- H3O+ H3O+ Cl- Cl- H3O+ H3O+ Cl- Note: No HCl molecules remain in solution, all have been ionized in water.

  10. Weak Acid Ionization: HC2H3O2 HC2H3O2 (H2O) HC2H3O2 HC2H3O2  HC2H3O2 Add water to MOLECULES of WEAK Acid

  11. Weak Acid Ionization: HC2H3O2 HC2H3O2 (H2O) HC2H3O2  H30+ C2H3O2- HC2H3O2 HC2H3O2 Note: At any given time only a small portion of the acid molecules are ionized and since reactions are running in BOTH directions the mixture composition stays the same. This gives rise to an Equilbrium expression, Ka

  12. pH calculations: To do pH calculations one must consider both the nature and condition (amount of ionization) of the species in the solution and then calculate the concentration of the hydronium ion (H+ or H3O+).

  13. Strong acid: [H3O+] = concentration of acid so: pH = -log [H3O+] = -log[acid] Weak Acid: one must calculate the [H3O+] from an equilibrium ionization expression. HA + H2O  H3O++ A- Ka = _ [H3O+][A-] [HA] These are equilibrium concentrations.

  14. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5 2 4 3 1. Starting pH, only acid and water. 0.10 M NaOH

  15. Approximations: [H+][A-] 4 2 pOH=-Log[XS OH-] 3 [OH-]=[HA] [H+]=[A-] 1

  16. Problem: Calculate the pH of 25.0 mL of 0.10M acetic acid (HOAc). The Ka of HOAc = 1.8 x 10-5 Which region of a titration curve would this be?

  17. Problem: Calculate the pH of 25.0 mL of 0.10M acetic acid (HOAc). The Ka of HOAc = 1.8 x 10-5 Ka = [H+][OAc-] [HOAc] HOAc  H+ + OAC- I C E 0.10 0 0 -x +x +x Ka = [x][x] 0.10-x 0.10-x x x Small,drop, WHY? Ka = x2 0.10 x = 0.00134 = [H+] pH = -log 0.00134 =2.87

  18. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5 2. Some base has been added This is also the “buffer region” 4 3 pKa = pH at this point (halfway point) 1. 0.10 M NaOH

  19. Region 2 Theory and Calculations: Base has been added and some of the acid has been neutralized. What has changed and WHY? 1. The moles of acid is decreased. 2. The moles of acid anion (A-) is increased. HA + OH-  H2O + A- 3. The total volume has increased. HA + OH- H2O + A- +XSHA note: the 1:1 ratio

  20. Region 2 Theory and Calculations: Base has been added and some of the acid has been neutralized. HA + OH- H2O + A- intial RXN Equilibrium XS HA + H2O  H3O+ + A- 2 sources always (calculate from equilibrium) Leads to: Ka= [H+][A-] [HA] Why? and [H+][A-]

  21. Region 2 Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5). Solution: HOAc + OH- H2O + OAc- 1:1 ratio 1. Find initial moles of Acid: 0.10mol HOAc = mol HOAc 0.02500L 0.0025 1L 2. find moles of OH-: 0.01400LNaOH 0.080molNaOH 1molOH- _____________________________= mol OH- 0.00112 1L 1molNaOH

  22. Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5). Solution: 1. We found initial mol of Acid: HOAc + OH- H2O + OAc- 1:1 ratio 0.10mol HOAc _____ = mol HOAc 0.02500L 0.0025 1L 2. We found mol of OH-: which =‘s initial moles of OAC- anion. 0.080molNaOH 1molOH- 0.01400LNaOH _____________________________________= mol OH- 0.00112 1molNaOH 1L 3. Subtraction gives what? No, not the numerical answer! moles of XS acid (initially present for OUR Eq. calculation).

  23. Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5). Solution: 1. Find initial moles of Acid: HOAc + OH- H2O + OAc- 1:1 ratio 0.10mol HOAc _____ = mol HOAc 0.02500L 0.0025 1L 2. find moles of OH-: which =‘s initialmoles of OAC- anion. 0.080molNaOH 1molOH- 0.01400LNaOH _____________________________________= mol OH- 0.00112 1L 1molNaOH 3. Subtraction gives moles of XS acid (initial for I.C.E.). 0.0025 - 0.00112 = 0.00138 mole acid INITIAL 4. and..we have 0.00112 mole OAc- INITIAL for I.C.E.

  24. Problem: Calculate the pH of a solution that is prepared by combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5). Solution: FROM PREVIOUS SLIDE. 1. initial M of Acid: HOAc + OH- H2O + OAc- 1:1 ratio 0.00138 mole/0.039L = 0.0354 M acid after RXN =INITIAL 2. M of OH-: which =‘s initial M of OAC- anion. 0.00112 mole/0.039L = 0.0287 M OAc-after RXN =INITIAL** [HOAc]  [H+] + [OAc-] I** C E 0.0354 0 0.0287 -x +x +x 0.0354-x x 0.0287+x Can we drop these? x = 0.0000222 =[H+] pH = - Log [0.0000222] = 4.65

  25. Region 2 summary: 1. write balanced equation for initial weak/strong RXN. 2. find moles of acid 3. find moles of base 4. subtract to determine XS (and limiting reactant which equals initial moles of salt formed) 5. Change XS to M and limiting Reactant to M 6. Write balanced Chemical Equilibrium Equation: HA  H+ + A- 7. Use values from #5 a initial values in I.C.E. Chart 8. Work as Equilibrium problem.

  26. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5 2. 4 3. Equivalence point 1. 0.10 M NaOH

  27. Region 3. Theory and Calculations. Mole of acid = moles of OH- added. Equivalence pt. HA + OH- H2O + A-no XS of either The titration curve indicates the pH to be above 7. What causes this? Hydrolysis of the A- anion: A- + H2O  HA +OH- Since this equilibrium involves OH- being formed, it is a Kb problem. º

  28. Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. 1. Find moles of Acid (which =‘s moles of base at Eq pt.) 0.10mol HOAc 25.00mL = mol HOac 0.0025 1000mL 2. Find mol of OH- 31.25mL __ = mol OH- 0.0025 0.080mol NaOH 1molOH- 1000mL 1mol NaOH 3. Since the # of moles are the same, this is at the Eq. Pt. and therefore the initial moles of OAc- ion = 0.0025 mol

  29. Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. moles of Acid = moles of base = 0.0025 = Eq. Pt. New Volume = 0.05625L..and [OAc-] = 0.0025/0.05625=0.0444M  Question: What equilibrium(ia) gives rise to the pH? 1. hydrolysis of the acetate anion: OAc- + H2O  HOAc + OH- [OAc-] [HOAc] [OH-] 0.0444 0 0 ICE -x +x +x 0.0444-x x x

  30. Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. moles of Acid = moles of base = 0.0025 = Eq. Pt. New Volume = 0.05625L...... and [OAc-] =0.0444M Question: What equilibrium(ia) gives rise to the pH? 1. hydrolysis of the acetate anion: OAc- + H2O  HOAc + OH- OAc- HOAc OH- 0.0444 0 0 ICE -x +x +x 0.0444-x x x Can we drop this x? How do we find Kb? Also: KaKb=Kw=1.0 x 10-14 @25oC so Kb= 5.56 x 10-10

  31. Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5. moles of Acid = moles of base = 0.0025 = Eq. Pt. New Volume = 0.05625L...... and [OAc-] =0.0444M Question: What equilibrium(ia) gives rise to the pH? 1. hydrolysis of the acetate anion: OAc- + H2O  HOAc + OH- OAc- HOAc OH- 0.0444 0 0 ICE -x +x +x 0.0444-x x x pOH = 5.30 pH =14 - 5.30 = 8.70 x = 4.97 x 10-6 = [OH-]

  32. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5 2. 4 XS base 3. 1. 0.10 M NaOH

  33. Region 4 theory and Calculations: All the acid has been neutralized and XS base has been added HA + OH- H2O + A- +XS OH- What is the dominating factor that controls the pH? The XS strong base(OH-) Calculations: Find moles of XS OH- and then use the new volume to find [OH-] and then the pOH and pH.

  34. Region 4 problem: Calculate the pH of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 40.00mL of 0.080M NaOH. Ka(HOAc)=1.8 x 10-5. 1. Find moles of acid: 0.10mol HOAc 25.00mL = mol HOAc 0.0025 1000mL 2. Find moles of OH-: 0.080molOH- 40.00mL = mol OH- 0.0032 1000mL 3. Subtract to find XS: 0.0032 - 0.0025 = 0.0007 mole XS OH- 4. Find OH- concentration, pOH, and pH: pOH = -Log[0.0007/0.06500] = 1.97 and pH = 12.03

  35. Approximations: [H+][A-] 4 2 STUDY FOR QUIZ pOH=-Log[XS OH-] 3 [OH-]=[HA] [H+]=[A-] 1

  36. Titration Curves for Three Different Weak Acids Ka = 1.4 x 10-7 Ka = 1.4 x 10-6 pH Ka = 1.4 x 10-5 mL of Strong Base Added

  37. CO32- + H+  HCO3- HCO3- + H+  HCO3

  38. Region 4: Region 2: acid + baseXS acid + salt HA+NaOH HA(XS)+NaA pOH=-Log[OH-] Region 3: mol acid=mol base hydrolysis of the salt anion A- + H2OHA + OH- Region 1: only acid (HA) 0.10 M NaOH

  39. Region 4: XS base pOH=-Log[OH-] Region 2: acid + baseXSacid + salt HA+NaOH HA(XS)+NaA Region 3: mol acid=mol base hydrolysis of the salt anion A- + H2OHA + OH- Region 1: only acid (HA)

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