1. Chem. 31 – 11/20 Lecture

2. Announcements Quiz 5 (last quiz) - Today
Labs Due on Wednesday (can turn in early)
Formal Part A 11/22
IC Part 1 11/22
Today’s Lecture
Chapter 8 – Advanced Equilibrium
Equilibrium problems – modified to account for activity – cont. from last time
Real equation for pH and use in calculations
Failure of the ICE method with two equilibria
The systematic method and its six steps

3. Ionic Strength Effects Effects on Equilibrium - Quantitative
Calculate expected [Mg2+] in equilibrium with solid MgCO3 for cases both with and without NaCl.
2nd case summary: [CO32-] = 4.66 x 10-4 M vs. [CO32-] = 1.87 x 10-4 M if assumed m = 0
What about 1st case? Isn’t m = 0 ?
No. Initially, yes, but as MgCO3 dissolves the ionic strength increases. This problem must be solved through iterations. Go over iterations

4. Ionic Strength Effects Real Equation for pH
pH = -logAH+ = -log(gH+[H+])
Example Problem: Determine the pH of a solution containing M Ba(OH)2

5. Second Part to Chapter 8 The Systematic Method
Question: Why can’t we apply the ICE (initial, change, equilibrium) method to any type of equilibrium problem?
Answer: That method is best designed for cases where there is only one relevant equilibrium reaction.
Examples of failures:
Solubility of MgCO3
pH of 5.0 x 10-8 M HCl solution (Show failure of Chem. 1B method)
Note: both problems can be solved using ICE method, but problem set up is more complicated

6. The Systematic Method Solubility of MgCO3 – Why did it fail?
MgCO3  Mg2+ + CO32-
x x Equil. (in ICE)
So x = (Ksp)1/2 = 1.87 x 10-4 M (neglecting ionic strength effects)
Problem is both ions can react further:
CO32- + H2O  HCO3- + OH-
And HCO3- + H2O  H2CO3 + OH-
Also, Mg2+ + OH-  MgOH+
And Mg2+ + CO32-  MgCO3 (aq)
Finally, we also have H2O  H+ + OH- re-establishing equilibrium
Each additional reaction results in greater dissolution
To properly solve problem we must consider 6 reactions not just 1
Measured “[CO32-]” from titration = [CO32-] + 0.5[OH-] + 0.5[HCO3-] + [MgCO3] + 0.5[MgOH+]
The “further” reactions makes [Mg2+] ≠ [CO32-], so ICE method fails (or needs modification by ICE tables for other reactions)
Actual solubility is greater than ICE method finds
[Mg2+]total = solubility ~ 3.3 x 10-4 M (from systematic approach)
Predicted HCl needed = 3.3 mL (vs. ~3.5 mL)
These calculations didn’t include activity which would lead to a ~10% increase in solubility (~3.6 mL HCl needed). In 0.1 M NaCl, I get 6.1 mL HCl needed (close to that observed)
enhancements: (% over rxn 1 only)
90% 0% 9%
16%

7. The Systematic Method The Six Steps
Write out all relevant reactions
Write a “Charge Balance Equation”
Write “Mass Balance Equations”
Write out all equilibrium equations
Check that the number of equations (in 2 to 4 above) = (or maybe >) the number of unknowns (undefined concentrations)
Solve for the desired unknown(s) by reducing the equations to one equation with one unknown. Then solve for remaining unknowns
Note: the emphasis of teaching the systematic method is steps 1 to 5. Step 6 will be reserved for “easy” problems with 2 to max 3 unknowns

8. The Systematic Method pH of 5.0 x 10-8 M HCl
Demonstrate Method on Board

9. The Systematic Method Conceptual Approach to Mass Balance Equations
With every source of related species, there should be one mass balance equation (or one set for ionic compounds)
Example:
Solubility of AgCl in water with M 1,10-phenathroline (Ph)
Reactions:
1) AgCl(s)  Ag+ + Cl-
2) Ag+ + 2Ph  Ag(Ph)2+
Mass Balance equations:
if only rxn 1) [Cl-] = [Ag+]
w/ rxn 2) [Cl-] = [Ag+] + [Ag(Ph)2+]
1,10-phenathroline
Ag+ Ph Ph Ph Ph
Ag+
Ag+
Cl-
Cl-
Ag+
Ag+
Cl-
Cl-
2nd Mass Balance Equation:
[Ph]o = M = [Ph]Total = [Ph] + 2[Ag(Ph)2+]
AgCl(s)
Notes: with rxn 1) only, 2 Ag+s = 2 Cl-s; with rxn 2) also, 3 Cls = 2 Ags + 1 Ag(Ph)2
Initially 4 Phs, then 2 Phs + one complex containing 2 Phs (so total # of Phs remains constant)

10. The Systematic Method 2nd Example
A student prepares a solution that contains mol of AgNO3 and mol NH3 in water with a total volume of 1.00 L. The AgNO3 is totally soluble, NH3 is a weak base, and Ag+ reacts with NH3 to form Ag(NH3)2+. Assume the Ag+ does not react with water or OH-. Go through the first 5 steps of the systematic method.