1 / 43

Now that that is out of the way, we are ready to begin…. Start praying now!

Now that that is out of the way, we are ready to begin…. Start praying now!. 17.1 Common-Ion Effect- sample 17.2 17.2 Buffered solutions resist change in pH Need: #1 Weak acid or weak base #2 Salt #3 Common ion #1 & #2. Buffers. Ex. NH 4 Cl + NH 4 OH 

uriel-bradshaw
Download Presentation

Now that that is out of the way, we are ready to begin…. Start praying now!

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. Now that that is out of the way, we are ready to begin…. Start praying now! • 17.1 Common-Ion Effect- sample 17.2 • 17.2 Buffered solutions resist change in pH • Need: • #1 Weak acid or weak base • #2 Salt • #3 Common ion #1 & #2

  2. Buffers • Ex. NH4Cl + NH4OH  NaCl + HCl  HF + KF  NaC2H3O2 + HC2H3O2 

  3. How it works • HX  H+1 + X-1 • Add H+ shift to the left: H++X-  HX • Add OH-  shift to right OH- + HX  H2O + X-

  4. Buffer problem • What is the pH of a mixture of 0.040M HF and 0.70M KF solution? (Ka = 3.53 x 10-4) • HF  H+1 + F- KF  K+ + F- • 0.040 x x 0.70 0.70 0.70 • Ka = [H+1][F-] = (x)(0.70 + x) =3.53 x 10-4 • [HF] (0.040 – x ) • 0.70x = 1.41 x 10-5 • x = 2.02 X 10-5 pH = -log(2.02 X 10-5) • pH = 4.69 (4.70 w/ quad) +x -x

  5. Buffer Capacity • Depends on concentration HC3H5O3 H+ + C3H5O3- .12 M x x NaC3H5O3  Na+ + C3H5O3- .10 M .10 .10 Ka = [H+][C3H5O3-1] = (x)(.10+x) = 1.4x10-4 [HC3H5O3] (.12 – x ) 1 x= 1.7x 10-4 pH= 3.77

  6. OR Use the Henderson- Hasselbalch Hasselhoff equation log [CB] [Acid] • pH= pKa + log [Base] [Acid] =-log(1.4x 10-4) + log (.10/.12) = 3.85 + (-.08) = 3.77 log [A-] [HA]

  7. Sample 17.4 Kb = [NH4+][OH-] pH= 9.00 [NH3] pOH= 5.00 (OH-)= 1x10-5M 1.8x 10-5= (x)(1.0x 10-5) .10M X= .18 M/L * 2.0 L = .36 mol NH4Cl

  8. Addition of SA. Or SB. To Buffers HC2H3O2 H+ + C2H3O2-1 HAc  H+ + Ac-1 .3 x x NaAc  Na+ + Ac- .3 .3 .3 HC2H3O2 + OH-1  H2O + C2H3O2-1 Int. .300 .02 .300 Rxn. -.02 -.02 +.02 End .28M 0 .32 M Ka= (x)(.28) = 1.8x 10-5 x= 2.1x10-5 (.32) pH= 4.68

  9. 17.3 Titration Calculations • Titration

  10. Strong acid vs strong base

  11. Weak acid vs strong base

  12. pH = pKa @ half equil point

  13. Strong acid vs weak base

  14. Phosphoric acid vs NaOH

  15. Graph of the first devirative

  16. Titration calc. SA vs SB • Given: 50.0ml of 0.100M HCl. • 1.) What is the pH of this solution before any base is added? • HCl  H+ + Cl- • 0.100 0.100 0.100 • pH = -log (0.100) • = 1.00

  17. 2.) What is the pH after 25.0ml of 0.100MNaOH has been added? • 50.0ml x 1 liter x 0.100 mol = 0.005 mol H+ • 1000ml 1 liter • 25.0ml x 1 liter x 0.100 mol = 0.0025 mol OH- • 1000ml 1 liter • H+ +OH-  H2O start .005mol .0025mol 0

  18. ∆ -.0025mol -.0025mol +.0025mol • End .0025 0 .0025 • pH = -log [ H+] • = -log(.0025/.075liters) • = 1.48 • 3.) ….pH after 49.9ml of 0.100M NaOH has been added? • H+ +OH-  H2O • I .005mol .00499 0

  19. ∆ -.00499 -.00499 +.00499 • End .00001 mol 0 +.00499 • pH = -log (.00001mol/.0999liter) • = 4.00 • 4.) …after 50.0ml of 0.100M of NaOH? • pH = 7.00 • Mol of acid = mol of base

  20. 5.) …after 50.1ml of 0.100MNaOH have been added? • H+ +OH-  H2O • I .00500mol .00501 0 • ∆ -.00500 -.00500 +.00500mol • E 0 .00001mol +.00500 XS is BASE pOH = -log (.00001mol/.1001liter) = 4.00 pH = 10.00

  21. 6.)…after 75.0ml of NaOH have been added? • pH = 12.30

  22. WEAK ACID vs STRONG BASE • 25.0ml of 0.100M HC2H3O2 (Ka = 1.8 x 10-5) is being titrated with 0.100M NaOH. • 1.) What is the pH before any NaOH has been added? (pH of a weak acid) • HC2H3O2  H+ + C2H3O2- • 0.100 x x • x2= Ka = 1.8 x 10-5 • 0.100

  23. pH = 2.87 • 2.) What is the pH after 10.0ml of NaOH have been added? • 25.0ml x 1 liter x 0.100 mol = 0.0025mol HC2H3O2 • 1000ml 1 liter • 10.0ml x 1 liter x 0.100 mol = 0.00100mol OH- • 1000ml 1 liter

  24. HC2H3O2 + NaOH  NaC2H3O2 + HOH • HC2H3O2 + OH-C2H3O2-+HOH • .0025 .00100 0 • -.00100 -.00100 +.00100 • 0.0015 0 +.00100 • USE THE HENDERSON-HASSELBALCH EQUATION

  25. Use the Henderson- HasselbalchHasselhoff equation

  26. pH = pKa + log([CB]/[Acid]) • = -log(1.8 x 10-5) + log(.001/.0015) • = 4.74 + (-0.176) • = 4.56 • 3.) …after 12.5ml of 0.100M NaOH added? • HC2H3O2 + OH- C2H3O2-+HOH

  27. 12.5ml x 1 liter x 0.100 mol = 0.00125mol OH- 1000ml 1 liter • HC2H3O2 + OH- C2H3O2-+HOH • .0025 .00125 0 • -.00125 -.00125 +.00125 • .00125 0 +.00125 • pH = pKa + log(.00125/.00125) • = 4.74

  28. pH = pKa @ ½ eq point • Law of COSINE • c2 = a2 + b2 – 2ab cos C • cos 90o = • Zero • c2 = a2 + b2 • Leads to the Pythagorean Theorem

  29. 4.) ….after 25.0ml added? • @ EQ point. • pH of a salt…. • 1.0x 10-14 = x2 • 1.8 x 10-5 .10/2 • pH = 8.72

  30. 5.) …after 26.0ml added? • (strong base in XS) • Do not use the HENDERSON-HASSELBALCH EQUATION • Use only when the XS is WEAK. • 25.0ml x 1 liter x 0.100 mol = 0.0025mol HC2H3O2 • 1000ml 1 liter • 26.0ml x 1 liter x 0.100 mol =0.00260 mol OH- • 1000ml 1 liter

  31. HC2H3O2+ OH-  C2H3O2-+H2O • 0.0025mol .0026 0 0 • -.0025 -.0025 +.0025 • 0 .0001 • pOH = -log (.0001mol/.051 liters) • = 2.71 • pH = 11.29

  32. 6.) …after 15.0ml added? • 25.0ml x 1 liter x 0.100 mol = 0.0025mol HC2H3O2 • 1000ml 1 liter • 15.0ml x 1 liter x 0.100 mol = 0.00150 mol OH- • 1000ml 1 liter • HC2H3O2+ OH-  C2H3O2-+H2O • .0025mol .0015 0 0 • -.0015 -.0015 +.0015 • .0010 0 +.0015

  33. pH = pKa + log([CB]/[Acid]) • = - log(1.8 x 10-5 ) + log(.0015mol/.001mol) • = 4.74 + .176 • = 4.92

  34. 17.4 Solubility Equilibrium • Ksp = solubility product • BaSO4 (s) Ba2+(aq) + SO42-(aq) • Ksp = [Ba2+][SO42-] called expression [BaSO4] Ca3(PO4)2  3Ca2+ + 2PO43- Ksp = [Ca2+ ]3[PO43-]2

  35. Calculate the Ksp of Ag2SO4 if the silver conc is 0.00013 M. • Ag2SO4  2Ag+ + SO42- • ______ .00013M .000065M • Ksp = [Ag+]2[SO42-] • = (1.3 x 10-4)2(.65 x 10-4) • = 1.1 x 10-12

  36. What is the solubility of CaF2 in g/l if the Ksp is 3.9 x 10-11 • CaF2 Ca2+ + 2F- • x x 2x • Ksp = [Ca2+ ][F- ]2 • 3.9 x 10-11 = (x)(2x)2 = 4x3 • x = 2.1 x 10-4 M x 78.1 g CaF2 = 1.6 x 10-2 g/l • 1 mol

  37. Will a ppt form? • Sample 17.15 • 0.10 L of 0.0080 M Pb(NO3)2 is mixed w/ 0.40 L of 0.0050 M of Na2SO4. Will a ppt form? • Pb(NO3)2 +Na2SO4  PbSO4+ NaNO3 • 0.10 L x 0.0080 mol/L = 8.0 x 10-4 mol Pb2+ • [Pb2+] = 8.0 x 10-4 mol = 1.6 x 10-3M • 0.50 L

  38. 0.40 L x 0.0050 mol/L = 2.0 x 10-3 mol SO42- • [SO42-] = 2.0 x 10-3 mol = 4.0 x 10-3M • 0.50 L • Ksp = [Pb2+] [SO42-] = 6.3 x 10-7 • = (1.6 x 10-3M)(4.0 x 10-3M) • Q = 6.4 x 10-6 • ppt will form!

  39. Q • If Q > Ksp …then ppt. will form. • If Q < Ksp …then no ppt. • If Q = Ksp…then @ equilibrium.

  40.  • 17.5 Common ion effect • The presence of either Ca2+ (aq) or F- (aq) in a solution reduces the solubility equilibrium of CaF2 to the left. • The solute conc. decreases by the presence of a 2nd solute w/ a common ion.

  41. Sample 17.12 • Calculate the molar solubility (M) of CaF2 at 25oC in a solution that is • a.) 0.010M in Ca(NO3)2 • Ca(NO3)2 Ca2+ + 2NO3- • 0.010 0.010 0.020 • Ksp = 3.9 x 10-11 Ksp = [Ca2+][F-]2 • Ksp = 3.9 x 10-11 = (0.010)(2x)2 • x = 3.1 x 10-5M

  42. b.) 0.010M in NaF • NaF  Na+ + F- • Ksp = 3.9 x 10-11 Ksp = [Ca2+][F-]2 • Ksp = 3.9 x 10-11 = (x)(0.01)2 • x = 3.9 x 10-7M

More Related