We will briefly go over Answer Key

Transcription

1 Chem 310 Test % of Grade (I ll decide later) Tests will be returned no later than 1 week after taken Reminder on regrades We will briefly go over Answer Key

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3 Oops forgot to do this!! So answer is < 1%!!

4 Test 1 Grade Breakdown High 98 Low 26 Mean s 3 80s 4 70s 9 60s 8 50s 5 40s 6 <40 6 If I had to give letter grades On this test A B B C C D I will not look at the above Again!!!!!!!!!!!!!!

5 Recommended Reading & Problems; Section IIa Chp 6 Sections All but 6.6(pages , ) Problems 6.1-4, 6.6-9, , 6.19, 6.21, , , 6.41, , Chp 7 Sections (pages ) Problems , 7.24, Chp 8 Sections (pages ) Problems - none Chp 9 Sections (pages ) Problems 9.1-3, 9.7-9,

6 Chapter 6 - Chemical Equilibrium Recall that the chemical and physical properties of analytes are invoked for Chemical Analysis The Chemical indicates the importance of equilibria Classes of Equilibria (K eq ) Acid Base (K a & K b ); Complexation (K f ); Solubility Product (K sp ); Phase Distribution (K D ) Etc. aa + bb cc + The Equilibrium Constant dd Use Molar concentrations If gases use bar Solids are omitted K = [ ] C C [ D] [ A] A [ B] B D

7 Aside Acid Rain Result of nitrogen and sulfur oxides from fuel consumption e.g., NO 2 (g) + H 2 O(l) + O 2 (g) HNO 3 (aq) nitric acid Limestone helps lakes with This problem CaCO 3 (s) is a base Your book give similar example on page 99

8 Manipulation of Equilibrium Constants Consider a reaction with an Equilibrium Constant, K. If the reaction is written in the opposite direction (right to left), the Equilibrium Constant is 1/K. If Reaction 1 (K 1 ) and Reaction 2 (K 2 ) are added, the sum of the two Reactions (K 3 ) is: K 3 =K 1 K 2 Problems will work these at board

9 What does it mean if I write Q instead of K? (for a given reaction, the equations look the same) When I write Q, I am saying that the system is NOT at Equilibrium. To calculate Q, I write in the appropriate concentrations at a particular point in time. If I wait for a while, the reaction will continue until I get a value of K. Then I am at equilibrium! G = H T S Thermodynamics and Equilibria K = e G 0 / RT

10 G = H T S K = e G For spontaneous reactions, G is negative. 0 / RT Negative H favors a reaction exothermic Positive H endothermic More entropy (disorder) favors a reaction Temperature effects depend on the entropy term If G 0 is negative, K is greater than 1. G 0 implies that all reaction components are in Standard State. What is G at equilibrium??

11 Le Châtelier s Principle When a stress is applied to a system at equilibrium the system shifts to reduce the stress (e.g., why does one ice skate with so little friction?) Of course, we are interested in how the Principle is applied to traditional equilibria Add something to one side of reaction shift way Take something from one side of reaction shift toward Heat used as reactant or product to determine temperature effects Common ion effect is really a matter of Le Châtelier s Principle How will adding a common ion effect the concentration of the other ions in an equilibrium? E.g., what effect does ph have on the solubility of AgCN K sp ~ K a ~ 10-9 We will work this at the board

12 Separation of Ions in Solution by Precipitation Example: Ions will precipitate in order of the K sp values. Ag + added to 0.1 M Cl -, Br -, I - the first to precipitate is I - K spagi = 8x10-17 ; K spagbr = 5x10-13 ; K spagcl = 2x10-10 If K sp values differ enough the I - concentration will be negligible at the point that the Br - begins to precipitate Problem 6-24: What concentration of carbonate must be added 0.10 M Zn2+ precipitate 99.9% of it? ZnCO 3 (s) Zn 2+ + CO 3 2- K sp = [Zn 2+ ] [CO 3 2- ] we need zinc conc (0.10) M 1.0 x = (1.0 x 10-4 ) [CO 3 2- ] [CO 3 2- ] = 1.0 x 10-6 M

13 Acid-Base Basics Bronsted-Lowery Definition: Acids are proton donors and bases are proton acceptors in acid base reactions The reactions are neutralizations that often form a salt HCl + NH 3 NH 4+ Cl - The products of this reaction are also conjugates ; i.e. NH4+ is the conjugate acid of ammonia and Cl- is the conjugate base of hydrochloric acid. What can be said about the strengths of these conjugates? Here is another B-L rxn of hydrochloric acid HCl + H 2 0 Cl - + Hydronium ion ph is really -log[h 3 O + ]

14 Acid-Base Basics H 2 0 H + + OH - K w = [H + ] [OH - ] (ok I m being lazy about K w ~ 1.0 x the hydronium ion thing) ph + poh = 14 What is the Kb of NH 3? (your book only lists Kas)

15 Know strong vs weak acids (what it means) and what is a polyprotic acid

16 A titration is a volumetric method of wet chemical analysis using a solution of known concentration as a titrant. The titrant must react with the analyte with known stoichiometry and reasonable kinetics. why? The process is depicted here Titration Basics

17 Titration Basics Other Terms Standard solution - the titrant (solution concentration known with high accuracy). Equivalence point - point in a titration when there are equivalent amounts of titrant and analyte End point - point in a titration when some noticeable signal indications equivalence point. Note - There can be errors - that is end point and equiv point may not be the same Primary standard - highly purified compound that is used as a reference materials to standardize solutions. Other characteristics are stabile, readily available, good solubility, and large molar mass Why Normality & Equivalents - we will use molarity and moles only

18 Titration Basics Shapes (types) of titration curves e.g., ph, poh, pmetal e.g., colorimetric titration A + B C

19 More terms & concepts Titration Basics An indicator (e.g. phenolphthalein) changes in an observable way near the equivalence point. (Acid-base indicators are themselves acids or bases!) The difference between the end point (what you detect) and the equivalence point (what you WANT to detect) is called the titration error. (Causes of systematic titration error?) A blank titration allows us to determine and eliminate the titration error. Simply repeat the experiment in the absence of analyte. Detecting the end point usually depends on a rapid change in concentration of the analyte or titrant near the equivalence point. Back titration is going past the equivalence point with excess titrant then titrating the excess (next page K-method problem)

20 Kjeldahl Nitrogen Analysis - An old technique that is still used to measusre Nitrogen in samples & illustrates some titration concepts - Convert all the nitrogen containing compounds in the sample to NH 4+. Digest the organic compounds in the sample (decompose, dissolve) in boiling H 2 SO 4. Nitrogen in the organic molecules is converted to NH Titrate to determine the amount of NH 4+ in the sample. You know the amount of sample you used. You determine the amount (moles, mass) of nitrogen in that sample based on the titration to determine NH Make the solution basic to convert NH 4+ to NH 3. Then distill the NH 3 (volatile) into a flask containing a known amount of HCl (more HCl than NH 3!): NH 3 +HCl NH 4+ +Cl - - There will be HCl left over. Titrate the HCl with NaOH. This is a back titration. - If you know how much HCl remained (by titration) after distillation of NH 3, you can calculate how much NH 3 must have been distilled into the receiver.

21 Titration Basics Problems

22 All Concentrations are not Created Equal! Chemical Activity As it turns out spectator ions can actually influence equilibria Activity not Molar concentration should be used in equilibrium expressions α = γ [ ] A B + C K = γ c [C] γ B [B] / γ A [A] The activity coefficient depends on ionic strength (µ = 0.5 Σ c i z i2 ) Fortunately, in relatively dilute solutions γ one and α ~ [ ] That s All For Chapter 8!!

23 Systematic Treatment of Equilibrium (Chapter 9) Write out what you want to know. Write out all the other information you have that is relevant. Use this information to answer your questions. Write charge balance equations Write mass balance equations Write any relevant equilibrium expressions (Any balanced reaction involving real species in solution is valid although perhaps not significant) Use this information to answer your questions. (you will need as many independent equations as you have unknowns fortunately approximations can be made)

24 Flow Chart Mass Balance Write an equation to show that the sum of the concentrations of the different forms of a chemical you have added to solution add up to concentration of the parent compound you added to solution. Charge Balance The total charge in a solution should be 0. Write an equation to show that all the charged chemical species in solution add up to a total charge of 0. The sum of the positive charged species equals the sum of the negative charged species.

25 Some Chp 9 Problems

26 Calculate the Concentration of Ca 2+ in a solution that is saturated in CaF(s) and buffered to ph 3.0. Write relevant solubility and acid dissociation reactions (how many different ions?) CaF 2 (s) Ca2+ + 2F- HF H+ + F- Write the Ksp (4 x ) and Ka (7 x 10-4 ) expressions K sp = [Ca2+] [F-] 2 K a = ([H+] [F-]) / [HF] Write the mass balance equation for Calcium [Ca2+] = ½ ([F-] + [HF])

27 Calculate the Concentration of Ca 2+ in a solution that is saturated in CaF(s) and buffered to ph 3.0. [Ca2+] = ½ ([F-] + [HF]) Use the Ka expression to substitute for HF and then the Ksp expression to substitute for F - [Ca2+] = ½([F-] + ([H+][F-])/ K a ) = ½[F-](1 + [H+]/K a ) [Ca2+] = ½ (K sp /[Ca2+]) 1/2 (1 + [H+]/K a ) [Ca2+] 1.5 = (K sp /4) 1/2 (1 + [H+]/K a ) Solve for [Ca 2+ ]; how does this differ from value if HF were strong acid? [Ca2+] 1.5 = (4x10-11 / 4) 1/2 ( /7x10-4 ) = 7.7x10-6 [Ca2+] = 4x10-4 M this is the solubility s If no Ka 4x10-11 = s (2s) 2 s = 2x10-4

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