Chem 222 #23 Ch12, 13, 14 Apr 7, 2005

Size: px

Start display at page:

Download "Chem 222 #23 Ch12, 13, 14 Apr 7, 2005"

Milton Boone
6 years ago
Views:

1 Chem 222 #23 Ch12, 13, 14 Apr 7, 2005

2 Announcement Quiz 5 will be held on Apr. 14, next Thursday Quiz 6 will be held on Apr. 26. If you need any corrections for the midterm, please finish it by the next Tuesday. You need to submit your notebook on Apr 18/19 with report for KH 4-4 Submit your DNA report on Apr. 21 at this class Finish DNA early

3 Warning A student who exits during the class may have a penalty of no attendance (5 % reduction).

4 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 V e = 10.0 ml Excess OH-: [OH - ] = [Q1]C OH [Q2] [Q1]: Fraction of remaining OH =? [Q2] = V KOH /(V KOH + V) Excess H + : [H+] = C HBr (V V e )/(V+V KOH )

5 Titration of 50.0 ml of M KOH with M HBr Region 1: Before the equivalence point When V = 3 ml, what is the ph? [OH - ] ( Ve V ) C V = OH Ini e V V OH OH + V 10.0 [ Q2] 50.0 = (0.0200M ) Fraction of Remaining OH- = M Region 2: After the equivalence Beyond the equivalence point, we add excess HBr. When V = ml [H + ] = C H Ini ( V V e ) V OH 1 + V 1 = (0.1000M )[ Q3] ml ( ) ml Moles of Excess H + = M

6 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 Pure [Q1]

7 Ex. Titration of Pyridine with HCl (p231) Q1. Find the ph when V a = 4.63 ml ph = pk a + Log[B]/[BH + ] (a) pka + Log{(V e -V a )/(V a )} (b) pka + Log{V a /(V e -V a )} (c) pka + Log(V a /V e ) Fraction of BH + = 4.63 ml/19.6 ml Fraction of B = ( )/19.6

8 Fraction B (100 %) Add V a of H + [B] : [BH + ] = (V e V a ) : V a When V e = 100 ml [B]: [BH + ] =? :? V = 20 ml

9 Q2 Find the ph at V =V e. F is the original pyridine concentration. [H + ] = a. (K a F) 1/2 b. {K a FV pyridine /(V pyridine +V e )} 1/2 Q3 What is the concentration of pyridine in the above case? pyridine Q4 Find the ph at V = ml (V e =19.6mL) [H + ] = a. C HCl V/(V pyridine + V) b. C HCl c. C HCl (V V e )/(V pyridine + V)

10 12-3 Titration of weak bases with strong acids V = 0 [Weak Bases] V < V e [Buffer] V = V e [Weak Acids] V > V e [Strong Acids]

11 13-3 EDTA Titration Curves (p265) If K f is large, we can consider the reaction to be complete at each point in the titration. Essentially, the same as strong acid strong base titration ph pm (Except for Region 2)

12 When pm 2+ is, is [M 2+ ] reduced or increased? How do you calculate [M 2+ ] when M 2+ is in excess?

13 Titration of 50.0 ml of M Ca 2+ (buffered to ph 10.0) with M EDTA Ca 2+ + EDTA CaY 2- V e = 50.0 ml 0.040M /0.080 M = 25.0 ml K f = α Y4- K f = (0.36)( ) = Region 1: Before the equivalence point (Excess Ca 2+ ) When V =5.0 ml [Ca 2+ ] = pca 2+ = -Log[Ca 2+ ] =1.54 = M

14 Region 3: After the equivalence point In this region, virtually all the metal is in CaY 2- with excess unreacted EDTA. When V = 26.0 ml (V e = 25.0 ml)

15 Region 2: At the Equivalence Point (Mostly CaY 2 Ca 2+ + Y 4- ) When V = 25.0 ml Calculate F for CaY 2+ F = [CaY 2+ ] INI = Ca 2+ + Y 4- CaY 2- Initial F Final x x F-x (F-x)/x 2 = K f x = (F/K f ) 1/2 pca 2+ = -Log(x)

16 K dependence of the titration curve [Zn 2+ ] = α M C Zn K f = [ZnY 2- ]/{[Zn 2+ ][EDTA] At Equivalence, ZnY 2- (Zn2+ mixture) + Y 4- ; K f F-x x x

17 KH 4-4 Glucose in Blood Serum Little Absorption Absorb light at 630 nm A = εbc glucose =kc glucose 1)Determine k from standard Samples 2) A for the blood sample C glucose

18 A = kc + A 0 One value

19 DNA Finger Printing Group experiment You are supposed to analyze the data (Follow lesson 4) and answer questions in Lesson 3 and questions in Lesson 4 in your notebook.

20 Basic Idea 1. Digest DNA with restriction enzymes 2. Separate DNA according to size by electrophoresis

21 3 Stain DNA and compare electrophoresis to find identical DNA (DNA left at Crime Scene) 4. Make a standard curve with standard DNAs with known size 5. Quantify size of DNA with the curve

22 Gel electrophoresis (p28 of handout) DNA are negatively charged + DNA Shorter DNA move more quickly + AAAGGTTT AAGG

23 Ch 14. Fundamentals of Electrochemistry 14-1 Basic Concepts A redox reactions involves transfer of electrons from one species to another. Fe 3+ + V 2+ Fe 2+ + V 3+ Oxidizing Reducing agent agent (oxidant) (reductant) Electric Charge (Faraday constant) q = n F [Coulombs] [moles] [Coulombs/moles] A mole of electrons has a charge of F = ( C) ( /mol e - ) = C/moles

24 Ex. (p284) Ex. If you have one moles of Fe 3+ and V 2+, how many coulombs of charge have been transferred from V 2+? Fe 3+ + V 2+ Fe 2+ + V 3+ q = 1 moles F [C/moles] = C

25 Electric Current The quantity of charge flowing each second through a circuit is called the current, I. I = q 1 /t [A], where q 1 is the charge [C] that flow through a circuit during t sec. [A] = [C/sec] I When I = 1 A, How many moles of Sn 4+ is reduced to Sn 2+ per second? N = 1 A/ ([Q1] F) Try Ex in p285

26 Voltage, Work, and Free Energy Electric potential between two points, E: Work needed when moving one coulomb of electric charge from one point to another. Work [J] = E [V] q [C] (14-3) J: joule V: volts 1.5 V 0.5 A How much work does the battery do for 1 min? Work =1.5 V 0.5 A 60 s In chemical reaction, free energy changes. G [J] = Eq = E(-nF) = -nfe (14-5)

27 Ohm s law: I = E/ R [Ω] Power: P = Work/s =Eq/s = EI I : How much is the current (I)? P: How much is the power delivered by the battery?

28 14-2 Galvanic Cells Anode Cathode Cd/Ni(OH) 4 NiCd Battery Half Reaction G = -150 kj/mole of Cd Ex. Calculate the voltage measured above E = - G/nF = 150 kj/ {[Q1] F }

29 On next Tuesday, the subjects to be covered will be up to Ch 14-5.

Chem 222 #29 Review Apr 28, 2005

Chem 222 #29 Review Apr 28, 2005 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