1. Chem. 31 – 2/25 Lecture

2. Announcements I Exam 1 –On Monday (3/2) –Will Cover the parts we have covered in Ch. 1, 3 and 4 plus parts of Ch. 6 (through Le Châtelier’s Principle) –Some of HW1.3 postponed (see posted solutions) –Help session (11:00 to 12:00 on Friday – after office hours)

3. Announcements II Today’s Lecture –Chapter 6 Material – Le Châtelier’s Principle (last part on Exam1) –Review of Material on Exam 1 (including Equation) –Chapter 6 Material not on Exam 1 (Sparingly soluble salts)

4. Le Châtelier’s Principle Stess Number Two: Dilution Side with more moles is favored at lower concentrations Example: HNO 2 (aq) ↔ H + + NO 2 - If solution is diluted, reaction goes to products If diluted to 2X the volume: So Q<K, products favored

5. Le Châtelier’s Principle Stess Number Two: Dilution – Molecular Scale View H+H+ NO 2 - Concentrated Solution H+H+ NO 2 - H+H+ H+H+ H+H+ Diluted Solution – dissociation allows ions to fill more space H+H+ NO 2 - H+H+ H+H+ H+H+ H+H+

6. Le Châtelier’s Principle Stress Number 3: Temperature If ΔH>0, as T increases, products favored If ΔH<0, as T increases, reactants favored Easiest to remember by considering heat a reactant or product Example: OH - + H + ↔ H 2 O(l) + heat Increase in T

7. Some Le Chatelier’s Principle Examples Looking at the reaction below, that is initially at equilibrium, AgCl(s) ↔ Ag + (aq) + Cl - (aq) (ΔH°>0) determine the direction (toward products or reactants) each of the following changes will result in a)increasing the temperature b)addition of water c)addition of AgCl(s) d)addition of NaCl

8. Review for Exam Know the following (from Ch. 1) –Common base units (m, kg, s, mol, K) + common multipliers (nano to kilo) –How to convert between different units* –Definitions of main concentration units (M, weight fractions including % and ppm, and mass/volume) –How to convert between concentration units* –Steps to make solutions of known concentration + calculations for solution preparation* –How to do stoichiometry problems (involving solids or solutions)* Note: *means need quantitative knowledge

9. Review for Exam – cont. Know the following (from Ch. 1 – cont.) –Titration definitions (titrant, equivalence point, end point, standardization titrations, analyte titrations, back titrations) –Practical titration requirements –How to solve normal and back titration problems* Know the following (from Ch. 3) –Rules for significant figures (including for calculations with +, -, *, or / and when uncertainties are given)* –Definitions for: systematic and random error, accuracy and precision, uncertainty, relative error and relative uncertainty

10. Review for Exam Know the following (from Ch. 3 – cont.) -How to do propagation of uncertainty problems (+, -, *, /, exponent, and mixed operations) and to convert between absolute and relative uncertainty* Know the following (from Ch. 4) -What a Gaussian distribution represents -How to calculate mean values and standard deviations (can use calculators)* -The differences between populations and samples -How to calculate Z values*

11. Review for Exam – Ch. 4 (cont.) How to use Table 4-1 and Z values to calculate probabilities between limits* How to determine confidence intervals* + factors which influence confidence intervals Difference between Z and t based confidence intervals (lecture only) What confidence intervals tell you How to perform t-test (case 1)* How to recognize and select a proper test (3 t tests, F test and Grubbs test) How to deal with poor data points (including use of Grubbs test)*

12. Review for Exam Chapter 4 – cont. -How method of least squares works (qualitatively) -Steps to the calibration process -Assumptions required for least squares analysis -How to determine concentrations of unknowns* + limitations in this Chapter 6 -Be able to write equilibrium equations from given equilibrium reactions -Manipulate equilibrium reactions/equations* -Definitions of changes in Enthalpy, Entropy and Free Energy plus predictions given reaction

13. Review for Exam – Ch. 6 (cont.) Chapter 6 Be able to determine K from  G° or visa versa Be able to calculate  G from  H and  S Be able to predicts shift in equilibrium due to changes in conditions (Le Châtelier’s Principle)

14. Review for Exam Equations I will give -Basic propagation of uncertainty equations (e.g. for Y = a + b, Y = a·b, and Y = x n ) -Equation for calculation of standard deviation -Grubb’s test equation -Equation for converting K to  Gº

15. Ch. 6 – Solubility Problems Why Solubility is Important Use in gravimetric analysis (predict if precipitation is complete enough) Use in precipitation titrations (not covered) Use in separations (e.g. separation of Mg 2+ from Ca 2+ in tap water for separate analysis) Understand phase in which analytes will exist Problem Overview Dissolution of sparingly soluble salts in water Dissolution of sparingly soluble salts in common ion Precipitation problems (and selective precipitation problems)

16. Solubility Product Problems - Solubility in Water Example: solubility of Mg(OH) 2 in water Solubility defined as mol Mg(OH) 2 dissolved/L sol’n or g Mg(OH) 2 dissolved/L sol’n or other units Use ICE approach: Mg(OH) 2 (s) ↔ Mg 2+ + 2OH - Initial0 0 Change +x +2x Equilibriumx 2x Note: x = [Mg 2+ ] = solubility

17. Solubility Product Problems - Solubility of Mg(OH) 2 in water Equilibrium Equation: K sp = [Mg 2+ ][OH - ] 2 K sp = 7.1 x 10 -12 = x(2x) 2 = 4x 3 (see Appendix F for K sp ) x = (7.1 x 10 -12 /4) 1/3 = 1.2 x 10 -4 M Solubility = 1.2 x 10 -4 M = [Mg 2+ ] Conc. [OH - ] = 2x = 2.4 x 10 -4 M

18. Solubility Product Problems - Solubility of Mg(OH) 2 in Common Ion If we dissolve Mg(OH) 2 in a common ion (OH - or Mg 2+ ), from Le Châtelier’s principle, we know the solubility will be reduced Example 1) What is the solubility of Mg(OH) 2 in a pH = 11.0 buffer? No ICE table needed because, from pH, we know [OH - ] eq and buffer means dissolution of Mg(OH) 2 doesn’t affect pH.

19. Solubility Product Problems - Solubility of Mg(OH) 2 at pH 11 – cont. [H + ] = 10 -pH = 10 -11 M and [OH - ] = K w /[H + ] = 10 -3 M K sp = [Mg 2+ ][OH - ] 2 Moles Mg(OH) 2 dissolved = moles Mg 2+ [Mg 2+ ] = K sp /[OH - ] 2 = 7.1 x 10 -12 /(10 -3 ) 2 [Mg 2+ ] = 7 x 10 -6 M

20. Solubility Product Problems - Solubility of Mg(OH) 2 in Common Ion Example 2) Solubility of Mg(OH) 2 in 5.0 x 10 -3 M MgCl 2.

21. Solubility Product Problems Precipitation Problems What occurs if we mix 50 mL of 0.020 M BaCl 2 with 50 mL of 3.0 x 10 -4 M (NH 4 ) 2 SO 4 ? Does any solid form from the mixing of ions? What are the concentrations of ions remaining?