1. Chem. 31 – 3/11 Lecture

2. Announcements I Exam 1 –Grading error on p. 3 (problem 4); was graded as though 10 pts for entire problem – not just part a) –That was the reason for getting 96 pts total and giving everyone 4 additional points –Average = 77 Cl lab report –Due Today Homework Set 2 – Turn in Problem 2.1

3. Announcements II Quiz 3 Today Today’s Lecture – Chapter 7 “Advanced Equilibrium Theory” –Why Equilibirum Theory Can Fail –Ionic Strength: What is it and how do we calculate it –Replacement Equations: Activity and Activity Coefficients

4. Chapter 7 “Adjustments” to Equilibrium Theory There are two areas where the general chemistry equilibrium theory can give wrong results: –When the solution has high concentrations of ions –When multiple, interacting equilibria occur –I had planned a demonstration, but due to distance to class plus finding equipment, I’m skipping the demonstration this year

5. Demonstration – Slide 1 Summary of Observation: –Two saturated solutions of MgCO 3 are prepared. –One is prepared in water and the other is prepared in ~0.1 M NaCl. –5.0 mL of each solution was transferred (and filtered) into a beaker. –3.5 mL of 0.002 M HCl needed for saturated MgCO 3 and 6.0 mL needed in 0.1 M NaCl Saturated MgCO 3 Saturated MgCO 3 in NaCl(aq)

6. Demonstration – Slide 2 Did the moles of HCl used match expectations? and Why did the solution containing NaCl need more HCl? –First Question: How many mL of HCl were expected? MgCO 3 (s)  Mg 2+ + CO 3 2- K sp = 3.5 x 10 -8 T = 25°C K sp = 3.5 x 10 -8 = [Mg 2+ ][CO 3 2- ] since [Mg 2+ ] = [CO 3 2- ] (assuming no other reactions), [CO 3 2- ] = (3.5 x 10 -8 ) 0.5 = 1.87 x 10 -4 M n(HCl) = (2 mol HCl/mol CO 3 2- )(1.87 x 10 -4 mmol/mL)(5.0 mL) = 0.001875 mmol HCl Calculate V(HCl) = 0.001875 mmol HCl/[HCl] = 0.001875 mmol HCl/0.002 mmol/mL = 0.935 mL Actual V(HCl) > 1 mL Conclusions It takes more HCl than expected, so more CO 3 2- dissolved than expected. Also, the NaCl increased the solubility of MgCO 3

7. Demonstration – Slide 3 What was the affect of the NaCl? –More CO 3 2- (and Mg 2+ ) was found to dissolve in the 0.10 M NaCl Why? –The Na + and Cl - ions stabilize CO 3 2- and Mg 2+ ions

8. Ionic Strength Effects Spheres Surrounding Ions Mg 2+ Low Ionic Strength CO 3 2- HO H  HO H HO H Ion – dipole interaction HO H HO H HO H Mg 2+ CO 3 2- HO H HO H HO H  High Ionic Strength Na + Stronger ion – ion interaction replaces ion - dipole Cl - HO H

9. Ionic Strength Definition :  = 0.5*  C i Z i 2 where i is an ion of charge Z and molar concentration C. But What is Ionic Strength –A measure which allows us to correct for ion – ion effects Examples: –0.10 M NaCl –0.010 M MgCl 2 –0.010 M Ce(SO 4 ) 2

10. Effects of Ionic Strength on Equilibria Equilibrium Equation Learned Previously: –for reaction A ↔ B, K = [B]/[A] Replacement Equation: –K = A B / A A –So what is A X ? –A X is the activity of X –A X =  X [X], where  X = activity coefficient –The activity coefficient depends on the ionic strength

11. Determination of Activity Coefficients Use of Debye-Hückel Equation: - where Z x = ion charge,  x = hydrated ion radius (pm) - useful for 0.0001 M <  < 0.1 M Can also use Table 7-1 for specific  value Calculate  (Mg 2+ ) at  = 0.050 M  (Mg 2+ ) = 800 pm

12. Factors Influencing  Ionic Strength: as  increase,  decreases Charge of Ion: a larger decrease in  occurs for more highly charged ions Size of Ion: Note: very small ions like Li + actually have large hydrated spheres Li + Rb + ion Hydrated sphere

13. Ionic Strength Effects on Equilibria Qualitative Effects An increase in ionic strength shifts equilibria to the side with more ions or more highly charged ions Example Problems: (predict the shift as  increases) –NH 3 (aq) + H 2 O(l) ↔ NH 4 + + OH - –Cu 2+ + 4OH - ↔ Cu(OH) 4 2- –2HSO 3 - ↔ S 2 O 3 2- + H 2 O(l) – HSO 4 - ↔ SO 4 2- + H +

14. Ionic Strength Effects Effects on Equilibrium - Quantitative Calculate expected [Mg 2+ ] in equilibrium with solid MgCO 3 for cases both with and without NaCl. –Go to Board