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Chem. 31 – 3/16 Lecture. Announcements I More on Additional Problem + Quiz –When stoichiometry is the same, K sp gives solubility (e.g. K sp (AgCl) =

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Presentation on theme: "Chem. 31 – 3/16 Lecture. Announcements I More on Additional Problem + Quiz –When stoichiometry is the same, K sp gives solubility (e.g. K sp (AgCl) ="— Presentation transcript:

1 Chem. 31 – 3/16 Lecture

2 Announcements I More on Additional Problem + Quiz –When stoichiometry is the same, K sp gives solubility (e.g. K sp (AgCl) = 1.8 x 10 -10 and K sp (AgI) = 8.3 x 10 -17 ) –When stoichiometry is different, one must look at reaction (e.g. Ag 2 CrO 4 – K sp = 1.2 x 10 -12 vs. BaCrO 4 K sp = 2.1 x 10 -10 ) –For Sparingly Soluble Salts, any further reactions of solubility products lead to greater solubility –Ca in quiz was tricky because the stoichiometry stayed 1:1 CaSO 4 (s) ↔ Ca 2+ + SO 4 2- ↔ CaSO 4 (aq) Water Hardness Lab Report –Turn in completed report form –Due Wednesday

3 Announcements II Today’s Lecture – Chapter 7 “Advanced Equilibrium Theory” –Replacement Equations: Activity and Activity Coefficients –Consideration of Activity in Solving Equilibrium Problems –The Real Equation for pH –The Systematic Method Examples of failures 6 steps to method More on Mass Balance

4 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

5 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 +

6 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

7 Ionic Strength Effects Real Equation for pH pH = -log A H+ = -log(  H+ [H + ]) Example Problem: Determine the pH of a solution containing 0.0050 M HCl and 0.020 M CaCl 2. Note: H 2 O  H + + OH - also affected by ionic strength

8 Second Part to Chapter 7 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 MgCO 3 –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

9 The Systematic Method Solubility of MgCO 3 – Why did it fail? MgCO 3  Mg 2+ + CO 3 2- x x Equil. (in ICE) So x = (K sp ) 1/2 = 1.87 x 10 -4 M (neglecting ionic strength effects) Problem is both ions can react further: CO 3 2- + H 2 O  HCO 3 - + OH - And HCO 3 - + H 2 O  H 2 CO 3 + OH - Also, Mg 2+ + OH -  MgOH + And Mg 2+ + CO 3 2-  MgCO 3 (aq) Finally, we also have H 2 O  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 “[CO 3 2- ]” from titration = [CO 3 2- ] + 0.5[OH - ] + 0.5[HCO 3 - ] + [MgCO 3 ] + 0.5[MgOH + ] The “further” reactions makes [Mg 2+ ] ≠ [CO 3 2- ], so ICE method fails (or needs modification by ICE tables for other reactions) Actual solubility is greater than ICE method finds [Mg 2+ ] total = solubility ~ 3.3 x 10 -4 M (from systematic approach) Predicted HCl needed = 3.3 mL (close to that measured) 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 enhancements: (% over rxn 1 only) 90% 0% 9% 16%

10 The Systematic Method The Six Steps 1.Write out all relevant reactions 2.Write a “Charge Balance Equation” 3.Write “Mass Balance Equations” 4.Write out all equilibrium equations 5.Check that the number of equations (in 2 to 4 above) = (or maybe >) the number of unknowns (undefined concentrations) 6.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

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

12 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 0.010 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 + ] AgCl(s) Ag + Cl - Ph 1,10-phenathroline Ag + Notes: with rxn 1) only, 2 Ag + s = 2 Cl - s; with rxn 2) also, 3 Cls = 2 Ags + 1 Ag(Ph) 2 2 nd Mass Balance Equation: [Ph] o = 0.010 M = [Ph] Total = [Ph] + 2[Ag(Ph) 2 + ] Initially 4 Phs, then 2 Phs + one complex containing 2 Phs (so total # of Phs remains constant)

13 The Systematic Method 2 nd Example An aqueous mixture of CdCl 2 and NaSCN is made –Initial concentrations are [CdCl 2 ] = 0.0080 M and [NaSCN] = 0.0040 M –Cd 2+ reacts with SCN - to form CdSCN + K = 95 –HSCN is a strong acid –Ignore any other reactions (e.g. formation of CdOH + ) –Ignore activity considerations –Determine the concentrations of all species


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