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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

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Chapter 8

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  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

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