1 / 34

Chapter 8

Chapter 8. Activity. Homework Chapter 8 - Activity. 8.2, 8.3, 8.6, 8.9, 8.10, 8.12. Question 8.2. Q: Which statements are true? In the ionic strength, m , range of 0-0.1 M, activity coefficients decrease with: a) increasing ionic strength b) increasing ionic charge

louise
Download Presentation

Chapter 8

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 8 Activity

  2. HomeworkChapter 8 - Activity • 8.2, 8.3, 8.6, 8.9, 8.10, 8.12

  3. Question 8.2 Q: Which statements are true? In the ionic strength, m, range of 0-0.1 M, activity coefficients decrease with: a) increasing ionic strength b) increasing ionic charge c) decreasing hydrated radius All are true!!

  4. Question 8.3 • Calculate the ionic strength of • 0.0087 M KOH • 0.0002 M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0087M(+1)2+ 0.0087M(-1)2] = 0.0087 M Remember for +1/-1 systems: Ionic strength, m = Molarity, M

  5. Question 8.3 (cont’d) • Calculate the ionic strength of • 0.0087 M KOH • 0.0002 M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0002M(+3)2+ 0.006M(-1)2] = 0.0012 M

  6. Question 8.6 Calculate the activity coefficient of Zn2+ when m = 0.083 M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. a) =-0.375 g=0.422

  7. Question 8.6 (cont’d) Calculate the activity coefficient of Zn2+ when m = 0.083 M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. = 0.432

  8. Question 8-9 Calculate the concentration of Hg22+ in saturated solutions of Hg2Br2 in 0.00100 M KNO3. Hg2Br2(s)D Hg22+ + 2Br- Ksp=5.6x10-23 some - - -x +x +2x some-x +x +2x I C E

  9. 8-10. • Find the concentration of Ba2+ in a 0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 Ksp = 7.11 x 10-11 some - 0.100 -x +x +2x some-x +x +2x I C E

  10. 8-10. • Find the concentration of Ba2+ in a 0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 Ksp = 1.5 x 10-9 some - 0.1 -x +x +2x some-x +x 0.1+2x I C E

  11. 8-10. • Find the concentration of Ba2+ in a 0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 Ksp = 1.5 x 10-9 some - 0.1 -x +x +2x some-x +x 0.1+2x I C E X = 6.57 x 10-7

  12. Question 8-12 • Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities? (m) = ½ (c1z12+ c2z22 + …) = ½ [0.010M(+1)2+ 0.010M(-1)2+0.0120M(+1)2+ 0.0120M(-1)2] = 0.0220 M gOH = 0.873 pH = AH = [H+]gH

  13. Question 8-12 (cont’d) • Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities? pH = 11.94

  14. Question 8-12 (cont’d) • Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities? pH ~ -log[H+] =

  15. Finally • Calculate the pH of a solution that contains 0.1 M Acid and 0.01 M conjugate base • Calculate the pH of a solution containing 0.1 M Acid and 0.05 M conjugate base. • Calculate the pH of a solution containing 0.1 M Acid and 0.1 M conjugate base.

  16. Acid/Base Titrations

  17. Titrations • Titration Curve – always calculate equivalent point first • Strong Acid/Strong Base • Regions that require “different” calculations • B/F any base is added • Half-way point region • At the equivalence point • After the equivalence point

  18. Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • First -find Volume at equivalence • M1V1 = M2V2 • (0.050 L)(0.02000M) = 0.1000 V • V = 10.0 mL

  19. Strong Acid/Strong Base • 50.00 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Second – find initial pH pH = - logAH ~ -log [H+] pOH = -logAOH ~ -log [OH-] pH = 12.30

  20. (~6 ml) Limiting Reactant Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Third– find pH at mid-way volume • KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.0006000 mol Before After 0.000400 mol 0 mol 0.0006000 mol 0.0006000 mol pH = 11.8

  21. Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Fourth – find pH at equivalence point • KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.0010000 mol Before After 0 mol 0 mol 0.0010000 mol 0.0010000 mol pH = 7.0

  22. Limiting Reactant Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Finally – find pH after equivalence point 12 ml • KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.001200 mol Before After 0 mol 0.0002000 mol 0.0010000 mol pH = 2.5

  23. Titration of WEAK acid with a strong base

  24. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • First, calculate the volume at the equivalence-point • M1V1 = M2V2 • (0.0250 L) 0.1000 M = 0.1000 M (V2) • V2 = 0.0250 L or 25.0 mL

  25. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Second, Calculate the initial pH of the acetic acid solution

  26. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Third, Calculate the pH at some intermediate volume

  27. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Fourth, Calculate the pH at equivalence

  28. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Finally calculate the pH after the addition 26.0 mL of NaOH

More Related