Chem 222 #29 Review Apr 28, 2005

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Angelica Fields
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1 Chem 222 #29 Review Apr 28, 2005

2 Announcement Please meet me after the class if you have any conflicts with the final exam schedule. You can expect similar questions with Quiz 6 in the final exam. If you have not received your notebook, reports, or quiz 5 back, please contact your section TA Your section TA is supposed to notify your current grade by this Saturday. If not please contact TA. I have a office hour this Friday and next Monday. I will be also available at T/R in class time.

3 What is next? Chem 343 Physical Chemistry Lab (Instrumental analysis by UV/Vis, FT-IR, GC-Mass, NMR, Gas effusion, Flash photolysis) Research in Lab - Choose the area of your interest - Longer term is better (1 year)

4 Final Exam 250 points ~ 40 points extra Curve? If top 14 students gain 90 % Linear scaling factor ~1.10 If top 14 students gain 97 % Scaling factor ~1.0

5 3. KHP is a salt of the intermediate form of phthalic acid. What is approximate PH of 0.15M of KHP? (pk 1 = 2.950, pk 2 = 5.408) a PH = 2.2 b PH = 4.2 c PH = 6.2 d PH = 8.2 e PH = The amino acid, Arginine, has the following forms. H 3 Arg H 2 Arg HArg Arg - Pk 1 =1.82 Pk 2 = 8.99 Pk 3 = What are the principal form and the second most abundant form of arginine at PH=10.0? Principal the second most abundant form a) H 2 Arg + H 3 Arg 2+ b) H 2 Arg + HArg c) H 2 Arg + Arg - d) HArg Arg - e) HArg H 2 Arg +

6 5. There are two different acids in each beaker. One beaker contains weak acid ( M, ml, pka = 6.00). The other contains strong acid ( M, ml). These acids are titrated with strong base, M NaOH. Choose the answer that gives closest PHs for weak acid and strong acid at the equivalence point? PH of weak acid PH of strong acid a) b) c) d) e) AH + OH - A - + H 2 O For week acids A - + H 2 O AH + OH- F x x V e =100 ml x = (F K b ) 1/2 = { M (100 ml /200 ml) } 1/2 = M poh = 5

7 When V =0, [Ag + ] = K sp /[X - ] (1) pag + = -Log[Ag + ] (2) [X - ] = 0.1 M Higher pag + Q1. (a. Higher b Lower) [Ag + ] Q2. (a. Higher b Lower) K sp When V = V e, [Ag + ] = [X - ] [Ag + ] = (Q3) Higher pag + Q4. (a. Higher b Lower) [Ag + ] Q5. (a. Higher b Lower) K sp

8 6. When you are titrating 0.1 M KI, KCl and KBr solutions with 0.05 M AgNO 3, you obtained the following titration curve. Then, K sp of AgI, AgCl and AgBr will be (p138) a. K sp (AgI)> K sp (AgCl)> K sp (AgBr) b. K sp (AgCl) > K sp (AgBr) > K sp (AgI) c. K sp (AgBr) > Ksp(AgI) > K sp (AgCl) d. AgI, AgCl and AgBr all have same K sp e. None of the above [Ag + ] = K sp /[X - ] How much is [X-] in the equivalence point?

9 8. From the following reduction potentials: I 2(s) + 2e - 2 I - Eº = V I 2 (aq) + 2 e - 2 I - Eº = V I e - 3 I - Eº = V Choose all the correct statement(s). No points unless you answer all correctly. (Hint: E 0 = ( /n)LogK ). a) The equilibrium constant of reaction: I 2 (aq) + I - I 3 - is 300. b) According to the equilibrium constant of I 2 (aq) + I - I 3 -, 0.1 g of molecular iodine in solid forms can be dissolved in 100 ml water completely. c) The reaction: I 2(aq) + I - I 3 - is spontaneous at standard condition. d) The reaction: I 2 (s) + I - I 3- is spontaneous at standard condition. e) The equilibrium constant of reaction I 2 (s) + I - I 3- is 1.0.

10 K = 10 (E0*n/ ) where E 0 = E +0 E - 0 I 2(aq) + 2 e - 2 I - E + º = V - I e - 3 I - E - º = V I 2 (aq) + I - I 3 E 0 = V K = E 0 > 0: Spontaneous I 2(s) + 2 e - 2 I - E + º = V - I e - 3 I - E - º = V I 2(s) + I - I 3 E 0 = 0.00 V K = 1 Not spontaneous

11 10. The following is the unbalanced reaction: Cr 2 O Fe 2+ Cr 3+ + Fe 3+ Which statement(s) is correct for the reaction? No points unless you answer all correctly. a) The balanced reaction is: 1Cr 2 O Fe H + 2Cr Fe H 2 O b) The balanced reaction is: 1Cr 2 O Fe H + 2Cr Fe H 2 O c) Cr 2 O 7 2- acts as a strong oxidant in this reaction. Fe 2+ Fe 3+ + e - d) Overall reaction, Cr 2 O 7 2- was oxidized by Fe 2+ Cr(VI) 2 O 7 2-2Cr 3+ e) Fe 2+ was reduced by Cr 2 O 7 2-.

12 Q 11 Switch P Light bulb Switch R - + Cd electrode Salt bridge Ag electrode Cd(NO 3 ) 2 (aq) Ag(NO 3 ) (aq) :Battery 1.5V Cd e - Cd (s) E o = V Ag + + e - Ag (s) E o = V Consider the following statements about the above setup with 1M of electrolyte (AgNO 3 and Cd(NO 3 ) 2 ) A. When switch P is closed or turned on (switch R is open or turned off) the setup works as a galvanic cell and the light turns on. B. When switch R is closed or turned on (switch P is open or turned off) the silver ion is reduced. C. When the switch P is closed or turned on (switch R is open) Cd electrode functions as an anode. From the above statements the incorrect statements are a) A, B and C b) A and Cc) A and B d) B and C e) Other than a)-d)

13 12 Assume that the light bulb in the above circuit is replaced by an analog voltmeter and switch P is closed or turned on (switch R is off). Both the electrolyte solutions are M. The reading in the voltmeter could be a) V b) V c) V d) V 13 The potential of a galvanic cell could be affected by A. Temperature of the electrolytes in half cells B. Concentration of the electrolytes (ions) in half cells C. What kind of electrodes are used a) A, B and C b) A and B c) only B d) B and C

14 Final Exam 5 Sections Each section contains 6 multiple choices and 1 long question Multiple choice 3 each 6 = 18 Long question 40 each 58 5 = 290 You can gain up to sections Ch 0-6 Ch 7, 13 Ch Ch 16, 18, 23, 25, 26 Ch 14, 15

15 Ch 0-6 Significant figures Sig fig in arithmetic (Log, etc) Log( ) (see Ch 3) Q-test Quiz 6 Definition and calculation of ph K sp K sp for CuCl 2 is K sp = [Q1 ]

16 Long q from Ch 0-6 Ch 6-3, 6-4, 6-5 Solubility, Common Ion, Separation by precipitation When pk sp = 3.00, K sp = [Q1] How many sig fig do you need? How much Hg 2 I 2 can you dissolve in 1L of solution when K sp = ? K sp = [Hg 2+ 2 ][Q1] = x (Q2) x =??

17 Ch 7(titration), Ch 13 (EDTA) Word definitions in 7-1 (end point, equivalence point, etc) Titration curves in Ch 7 Spectroscopic titration 7-3 Dilution effect Precipitation titration

18 Ch 7, 13 Long q EDTA titration (13-3) - Conditional formation constant - pmetal in a titration curve One topic from the previous list

19 Ch 10, 11, 12 Acid base & titration How to calculate ph for - strong acid/base [H + ] [OH - ] - weak acid/base K a /K b - buffer ph = pk a + Log(Q1) a. [A - ]/[HA] b. [HA]/[A - ] Polyprotic acid

20 Mixing a weak acid and its conjugated base HA H + +A - pk a = 4.00 K a = 10 [Q1] F- x x x F = 0.01 M x 2 /(F-x) = K a x = Fraction of dissociation α =[A - ]/{[A - ] +[HA]} = x/f = =3.1 % What happens if you add A - to H 2 O HA H + +A - How much fraction of A - reacts with water in a solution containing 0.10 M of A -? A - +H 2 O HA + OH - pk b =[Q2] F y y y y 2 /(F -y) = K b y = & α =

21 10-4 Weak base equilibria B + H 2 O BH + + OH - [Q1] x x [ BH ][ OH [ B] ] = x F + 2 x = K b Ex. (p188) Find the ph of 0.10 M Ammonia with pk b = NH 3 + H 2 O NH 4+ + OH - (F-x) x x X 2 /(F-x) = K b = x = ([ Q2 ] ) 1/2 = M [OH - ] = (Q1 ) [H + ] = (Q2)/[OH - ]

22 Ex. Effect of adding acid to a buffer We prepare 1L of solution containing M of tris and M of trishydrochloride with pka of ph = pk a + Log[B]/[BH + ] ph = (Q1) +Log(Q2 /0.0296) = Let s add 12 ml of 1M HCl to the solution

23 Ex. How to prepare a buffer solution (p194) How many milliliters of M NaOH should be added to 10.0 g ( moles) of tris hydrochloride to give a ph of 7.60 in a final volume of 250 ml? pk b of tris is BH+ + OH- B Initial x Final x x pka= (Q2) pkb [Q1] = pka + Log[B]/[BH + ] [B]/[BH + ] = 10 {(Q1) pka} x / ( x) = 10 {(Q1) pka}

24 Diprotic systems R pk a1 R pk a2 R H 3 N + CHCO 2 H H 3 N + CHCO 2- H 2 NCHCO 2 - H 2 L + HL L - Equilibrium in Diprotic Systems H 2 L + HL + H + K a1 K 1 HL L - + H + K a2 K 2 L - +H 2 O HL + OH - HL +H 2 O H 2 L - + OH - K b1 =K W /[Q1] K b2 =[Q2] How much is the ph of 0.05 M of L 2 H solution for L of pk 1 = 2.33?

25 Simplified Calculation for the Intermediate Form [H + ] = FK a2 K K a1 a1 + K + F a1 K w 1/ 2 Two additional assumptions K a2 F >> K w [H + ] = FK K a2 a1 K +F a1 1/ 2 K a1 << F 1/ 2 [H + ] = FKa2Ka 1 = ( K ) 1/ 2 a1ka2 F ph = -Log[H + ] (Remember This ) = -Log(K a1 K a2 ) 1/2 = {-LogK a1 LogK a2 }/2 = [Q1] (11-12)

26 Ch Long q Acid/Base Titrations Polyprotic acid/base Study also previous quiz

27 Ch 12 Acid-Base Titration (p225) 12-1 Titration of Strong Acid with Strong Base Titration of 50.0 ml of M KOH with M HBr At Equivalence Point V e [Q1] = V e = 10.0 ml ph =?

28 12-2 Titration of weak acid with strong base Titration of 50.0 ml of M MES with M NaOH V e M = 50.0 ml M V e = [Q1 ] ml

29 Ch 16, 18, 21, 23, 25, 26 Ch 16 Ch 18 Remember Equations Relationship of A, P, T Beer s law Ch 23, 25, 26 - Lecture notes - HPLC, plate #, resolution, particle size - Quiz 6 Previous quiz

30 Ch 16, 18, 21, 23, 25, 26 Long q Chapter 16 (Redox titration) Other Balance the following half reactions MnO - 4 Mn 2+ Mo 3+ MoO 2+ 2 Combine the net reactions Expect the potential in a titration

31 Standardization of KMnO 4 Suppose that g of Na 2 C 2 O 4 is dissolved in a 250.0mL volumetric flask. If ml of this solution require ml of KMnO 4 solution for titration, what is the molarity of KMnO 4 solution? Oxalate concentration C OX is Weight of Na C2O4 / MW Volume of solution 2 = g /(134.00g L C OX = mm / mol) Moles of Moles of Oxalate KMnO4 = C C OX MnO4 V V OX MnO4 = [Q1] moles 2 moles C = KMnO4 = C V 7 OX V OX KMnO4 mmol 2 5 = M 10mL 48.36mL

32 Ch 14, 15 Structure of ph electrode - What are involved as components ph combination electrode Quiz 6 Reference electrode Reaction spontaneous or not Long Question Nernst equation Line notation Cell reactions

Chem 222 Quiz 6 Apr 26, Chem 222 Quiz 6

Chem 222 Quiz 6 Apr 26, Chem 222 Quiz 6 Chem 222 Quiz 6 1. Do not open this page until you are asked to start. 2. When you submit the answer, return only page 2-6. 3. After your time is up, please bring the answers to the instructor within 1