1 / 13

Unless otherwise stated, all images in this file have been reproduced from:

Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry , 2007 (John Wiley)      ISBN: 9 78047081 0866. CHEM1002 [Part 2]. Dr Michela Simone Weeks 8 – 13 Office Hours: Monday 3-5, Friday 4-5

talia
Download Presentation

Unless otherwise stated, all images in this file have been reproduced from:

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. Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille,Chemistry, 2007 (John Wiley)     ISBN: 9 78047081 0866

  2. CHEM1002 [Part 2] Dr Michela Simone Weeks 8 – 13 Office Hours: Monday 3-5, Friday 4-5 Room: 412A (or 416) Phone: 93512830 e-mail:michela.simone@sydney.edu.au

  3. Summary of Last Lecture • Solubility Equilibria I • Saturated solutions contains solid salts in equilibrium with the maximum amount of the ions in solution that is possible • The solubility product (Ksp) is the equilibrium constant for this equilibrium situation • The solubility product (Ksp) gives the maximum concentrations of the ions in the solution and the maximum amount of solid that will dissolve • The ionic product (Qsp) has the same form has the solubility product and is used to test whether more solid will dissolve or precipitation will occur • If Qsp < Ksp, more ions can enter solution and more solid can dissolve • If Qsp > Ksp, precipitation must occur

  4. Solubility Equilibria • Lecture 10: • Solubility • Blackman Chapter 10, Sections 10.4 • Lecture 11: • Common ion effect • Blackman Chapter 10, Sections 10.4

  5. Reminder: Solubility Equilibria PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+(aq)][Cl-(aq)]2 solubility product constant • If Pb2+(aq) and Cl-(aq) are present in the same solution at the same time, their concentrations can never be so large that [Pb2+(aq)][Cl-(aq)]2 is bigger than Ksp

  6. Solubility PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+(aq)][Cl-(aq)]2 solubility product constant • If x mol of PbCl2(s) dissolves in 1 L of solution, then • [Pb2+(aq)] = x mol L-1 and [Cl-] = 2x mol L-1 • Ksp = [Pb2+(aq)][Cl-(aq)]2 = (x)(2x)2 = 4x3 • x = (Ksp/4)1/3 = solubility • As Ksp = 1.6 x 10-5, • solubility = (1.6 x 10-5/4)1/3 = 0.016 mol L-1

  7. The Common Ion Effect and Solubility PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+(aq)][Cl-(aq)]2 solubility product constant • If NaCl is added to the solution, [Cl-(aq)] increases so • equilibrium shifts to the left: less PbCl2 dissolves • [Pb2+(aq)][Cl-(aq)]2+ must remain constant so [Pb2+(aq)] decreases

  8. The Common Ion Effect and Solubility PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+(aq)][Cl-(aq)]2 solubility product constant • If NaCl is added to the solution, to give [Cl-(aq)] = 0.2 M • If x mol of PbCl2(s) dissolve in 1 L of solution, then • [Pb2+(aq)] = x mol L-1 and [Cl-] = 0.2 mol L-1 • Ksp = [Pb2+(aq)][Cl-(aq)]2 = (x)(0.2)2 = 0.04x • As Ksp = 1.6 x 10-5, • solubility = 1.6 x 10-5/0.04 = 0.0004 mol L-1

  9. Solubility and pH • Because of the common ion effect, solubility will be pH dependent if dissolution involves H+ and OH- • e.g. Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) • In this case solubility will be higher at low pH values.

  10. Separation of Cations • Fe3+ and Zn2+ can be separated using the pH dependence of the solubilities of their salts. • For example, Fe(OH)3 and Zn(OH)2 can be separated at a pH of 4.76 (achieved by with a buffer). • Ksp {Fe(OH)3} = 1.0 x 10-38 = [Fe3+][OH-]3 • pOH = 14.00 - 4.76 so [OH-] = 10-9.24 M • [Fe3+] = 1.0 x 10-38/(10-9.24)3 M = 5.2 x 10-11 M Fe(OH)3 is highly insoluble at this pH.

  11. Separation of Cations • Ksp {Zn(OH)2} = 1.0 x 10-15= [Zn2+][OH-]2 • [OH-] = 10-9.24 M • [Zn2+] = 1.0 x 10-15/(10-9.24)2 M = 3.0 x 103 M Fe(OH)3 is highly insoluble at this pH. Zn(OH)2 is highly soluble at this pH

  12. Summary: Solubility EquilibriaII • Learning Outcomes - you should now be able to: • Complete the worksheet • Apply solubility equilibria(qualitative and quantitative) • Use ionic product to determine solubility • Apply the common ion effect • Answer review problems 10.49 - 10.75 in Blackman • Next lecture: • Complexes

  13. Practice Examples • 2. The Ksp for scandium(III) hydroxide is 2 x 10. What is the solubility of Sc(OH)3 (in mol L-1) of a solution buffered at 6.7?

More Related