Chem_329_3
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Chem_329_3

Course Number: CHEM 329, Fall 2008

College/University: Wisconsin

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Chemical Equilibrium Read Chapter 6-1 to 6-5 and Chapter 8-1 to 8-4. We will begin with a review of equilibria and then introduce activities (8-1 to 8-4). Consider the reaction aA + bB cC+dD The equilibrium constant is [ C ] [ D] K= a b [ A] [ B ] c d Last semester you used Ks for Strong acids Weak acids Buffers This semester we extend this to include solubilities, complex ion formation combined with...

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Equilibrium Read Chemical Chapter 6-1 to 6-5 and Chapter 8-1 to 8-4. We will begin with a review of equilibria and then introduce activities (8-1 to 8-4). Consider the reaction aA + bB cC+dD The equilibrium constant is [ C ] [ D] K= a b [ A] [ B ] c d Last semester you used Ks for Strong acids Weak acids Buffers This semester we extend this to include solubilities, complex ion formation combined with acid-base equilibria. Equilibrium Constants What determines the size of an equilibrium constant? [ C ] [ D] K= a b [ A] [ B ] c d Equilibrium constants come from thermodynamics. G = H TS K = exp G / RT [ ] What are the dimensions of the equilibrium constant? Each quantity is really a ratio of concentrations. Equilibrium Constants When we write the equilibrium constant [ C ] [ D] K= a b [ A] [ B ] c d what we are really writing (for a liquid) is [ C ] /(1M ) [ D] /(1M ) K= a b a b [ A] /(1M ) [ B ] /(1M ) c c d d See page 100 of text for more details. Lets consider a few examples. Hg2(IO3)2(s) init. conc. Solid final conc. Solid Hg22+ + 2IO30 x 0 2x Solve for x in Ksp = [Hg22+][IO3- ]2=x(2x)2 Ksp = 1.3x10-7M x=[Ksp/4]1/3 Solubilities If you have a salt AB it will dissolve to some extent in water. Ksp values tell you to what extent the salt will dissolve. The equation Ksp=[A][B] only applies if there is solid present and it is in equilibrium with the A and B ions in solution. As a chemist, you will find that solubility rules are useful. These rules are now given. Solubility Rules 1. Salts containing Group I elements are soluble (Li+, Na+, K+, Cs+, Rb+). Exceptions to this rule are rare. Salts containing the ammonium ion (NH4+) are also soluble. 2. Salts containing nitrate ion (NO3-) are generally soluble. 3. Salts containing Cl -, Br -, I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. Thus, AgCl, PbBr2, and Hg2Cl2 are all insoluble. 4. Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are common soluble salts of silver; virtually anything else is insoluble. 5. Most sulfate salts are soluble. Important exceptions to this rule include 6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble. 7. Most sulfides of transition metals are highly insoluble. Thus, CdS, FeS, ZnS, Ag2S are all insoluble. Arsenic, antimony, bismuth, and lead sulfides are also insoluble. 8. Carbonates are frequently insoluble. Group II carbonates (Ca, Sr, and Ba) are insoluble. Some other insoluble carbonates include FeCO3, PbCO3. Carbonates become soluble in acid solution. 9. Chromates are frequently insoluble. Examples: PbCrO4, BaCrO4 10. Phosphates are frequently insoluble. Examples: Ca3(PO4)2, Ag2PO4 11. Fluorides are frequently insoluble. Examples: BaF2, MgF2, PbF2. Thanks to Professor Kenneth W. Busch from whose Web page these data were extracted. Memorize rules 1-4. Lets consider a 2nd example. Consider an aqueous solution made by adding sufficient quantities of Hg2(IO3)2 and Hg2I2 salts that equilibrium is reached. Will the mercurous ion concentration be greater than or smaller than the previous problem? What is the I- concentration at equilibrium? For Hg2(IO3)2 Ksp = 1.3x10-7 Hg2I2 For Ksp = 1.1x10-28 Common Ion Problem Write down eq. expressions for Hg2(IO3)2 and Hg2I2. To find all the concentrations, first treat each problem independently assuming the other salt is not present. For Hg2(IO3)2 Ksp = 1.3x10-7 [Hg2+2]=1.3x10-4 For Hg2I2 Ksp = 1.1x10-28 Ksp = [Hg22+][I- ]2=x(2x)2 [Hg2 +2]=3.0x10-10 If we had a mixture of both, all the mercurous ions would be due to the Hg2(IO3)2 . To solve for [I- ] assume [Hg2 +2]=1.3x10-4 and solve the following Common Ion Problem Continued To solve for [I- ] assume [Hg2+2]=1.3x10-4 and set up the following table Hg2I2(s) init. conc. Solid final conc. Solid Hg22+ + 2I1.3x10-4 0 1.3x10-4+x 2x Solve for x in Ksp = [Hg22+][I- ]2= (1.3x10-4+x )(2x)2 Ksp = 1.1x10-28 = 1.3x10-4 (2x)2 x = 4.6x10-13 Always test for self consistency. Activities This is where we learn that life is more complex than we have previously thought. Lets consider a 3rd example. Consider an aqueous solution made by adding sufficient Hg2(IO3)2 that equilibrium is reached. Then add enough KNO3 so that the solution has a .1M potassium concentration. What is the concentration of the mercurous ion after equilibrium is reached? Based on our expression for K, adding potassium nitrate should have no effect on the [Hg22+] concentration. In fact we find that the concentration goes up. Why? Here is another example of this effect. Fe3+ + SCNFe(SCN)2+ An ionic atmosphere surrounds ions in solution. The size of determines the interaction between the two spheres, and is a function of the ionic strength . As the number of ions (of any flavor) in solution increase (i.e. increases) what happens to the interactions between the two shaded regions. How does this affect the concentration of the ions in solution? Equilibrium Constants This equilibrium constant expression must be wrong. [ C ] [ D] K= a b [ A] [ B ] c d Equilibrium constants come from thermodynamics. G = H TS C D A B a c d b K = exp G / RT [ ] The correct expression for K is K= [C ] [ D] = b [ A] [ B ] c c C aa A d d D b B (8-5) The key equations K= C D A B a c d b [C ] [ D] = b [ A] [ B ] c c C aa A d 2 d D b B The extended Debeye-Huckel equation is log = 0.51z 1 + ( / 305) How do we solve for concentrations? 1 1 2 2 2 Where c1 z1 + c2 z 2 + ...) = ci zi =( 2 2i Sample problem Calculate the Ca2+ conc. obtained by dissolving CaF2 in a 0.05M soln of NaClO4 CaF2(s) init. conc. Solid final conc. Solid 0 x Ca2+ + 2F0 2x Solve for x in Ksp = [Ca2+]Ca[F- ]2 2F Ksp = 3.9x10-11M Ignore the contribution of Ca2+ and F- to ionic strength. Calculate the ionic strength for a 0.05M soln of NaClO4 1 2 2 2 = (c1 z1 + c 2 z 2 + ...) = ci z i 2 i 1 2 2 = (.05 *1 + .05 *1 ) = .05 2 we find Ca=.485 and F=.81 Calculate the Ca2+ conc. obtained by dissolving CaF2 in a 0.05M soln of NaClO4 CaF2(s) init. conc. Solid final conc. Solid Ca2+ + 2F0 x 0 2x Solve for x in Ksp = [Ca2+]Ca[F- ]2 2F = x Ca (2x)2 2F 3.9x10-11= x(.485)(2x)2(.81)2 3.9x10-11/(.485)(.81)2=4x3 x=3.1x10-4 check for consistency by recalculating . Recalculate the ionic strength for the solution 1 = (.05 *12 + .05 *12 + .00062 *12 + .00031* 2 2 ) = .05 2 Since the ionic strength has not changed we are confident our answer is correct. Practice calculating ionic strengths (prob. 8-5) and activity coefficients (prob. 8-6), and concenctrations (prob. 8-12). See also problem 5 of homework.

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