Chem 222 #9 Ch 7, 13 Feb 8, 2005

Transcription

1 Chem 222 #9 Ch 7, 13 Feb 8, 2005

2 Announcement The answer key for quiz 2 is at the web site

3 Quiz 2 2(b) 2(b) John put excess amount of Cu 4 (OH) 6 (SO 4 ) into H 2 O and adjusted ph of the solution at 9.0 with NaOH. Calculate [Cu 2+ ] of this saturated Cu 4 (OH) 6 (SO 4 ) solution. K sp for Cu 4 (OH) 6 (SO 4 ) is Cu 4 (OH) 6 (SO 4 ) (s) 4Cu OH - + SO 4 2- x x/4 This is from Home Work 6-17.

4 Spectrophotometric Titration (Lab. Exp 19) Absorbance: A = εbc (p411) Path length Concentration Fe 3+ Binding Site of Transferrin Apo-transferrin + 2Fe 3+ (Fe 3+ ) 2 transferrin Show absorbance at 465 nm! Titration of 2.00 ml of apotransferrin With M ferric nitrilotriacetate

5 203 µl Titration of 2.00 ml of apotransferrin With M ferric nitrilotriacetate Apo-transferrin + 2Fe 3+ (Fe 3+ ) 2 transferrin 3+ Moles of Fe Moles of protein = C C Fe protein V V Fe protein = 2 moles 1moles C protein = CFeVFe 1.79mM 203µ L = 2 V mL protein To construct the above graph the dilution effect must be corrected. (by a dilution factor) Corrected Abs = Observed V Abs V total initial

6 UV/Vis Absorbance in Normal Proteins Pheε 250 nm = 400 Tyrε 274nm = 1400 Trpε 280nm = 4500 If your protein contains 2 Tyr residues ε 274nm ~ =

7 Ex. The absorbance measured after adding 125 ul of ferric nitrilotriacetate to ml of apotransferrin was Calculate the corrected absorbance that should be plotted in the Figure 7-5. Corrected Abs = Observed V Abs V total ini = mL 2.000mL

8 7-4 The precipitation Titration Curve The titration curve is a graph showing how the concentration of one of the reactions varies as a titrant is added Because concentration of the reagent varies over many orders, it is most useful to plot pfunction: px = -log 10 [X]

9 Q. Consider the titration of ml of M I - with M Ag + I - + Ag + AgI(s) (7-9) titrant Suppose we are monitoring Ag + with an electrode. (7-9) is the reverse of the dissolution of AgI(s) I - +Ag + ; K sp =[I - ][Ag + ]= Almost 100 % forms AgI(s) Q. What volume of Ag + titrant is needed to reach the equivalence point? C I V I = C Ag V Age V Age = (Q1 M) ml/ M = ml

10 How does [Ag + ] change before the end point? Q. What is [Ag + ] when V Ag = ml? At the equivalence point at V Ag e = ml. A. Before the equivalence point, we have an excess of I -. [Ag + ] is obtained from [Ag + ] = K sp /[Q1] = /[Q1] [I - ] = (Moles of Original I-) - (Moles of Total volume CIVI CAgVAg = V + V I Ag Ag+ ) If V Ag = ml Moles of I - = (25.00 ml) ( M) - (49.00 ml)( M) = mmol mmol =0.050 mmol V total = (25.00 ml ml) = ml [I - ] = [Q2]/74.00 ml = M See P135 [Ag + ] = [Q3]/ M = pag + = -log[ag + ] = 12.91

11 At the equivalence point, what is [Ag + ]? At equivalence point, V Age C Ag = V I C I Little l - and Ag + are left, and [Ag + ] = [l - ]. We define [Ag + ] = x. Then, K sp = = [Ag + ][I - ] = [Q1] x = M pag + = -log( ) = Q2. What is [Ag + ] when V Ag = 51 ml

12 After the end point Q. What is [Ag + ] when V Ag = 51 ml Now, all the I - ions are consumed, we have an excess of Ag +. Moles of Ag + = (C Ag V Ag [Q1] )= (51.00 ml)( M) - (25.00 ml) ( M) =1.00 ml M = 0.05 mmol [Ag+] = 0.05 mmol/(total volume) = 0.05 mmol/(51 ml + 25 ml) = M pag+ = 4 log6. 6 = 3.1 8

13 Titration curve Excess I - Excess Ag + For 1:1 stoichiometry of reactants, the equivalence point can be indicated by

14 Ion dependence (pag + ) e = -Log(K sp 1/2 )

15 12-1 Strong acid-strong base Titration NaOH + HCl H 2 O + NaCl Excess OH - Excess H + ph V a (ml) Excess OH - Calculate [OH - ] [H + ] =Kw/[OH - ] Equivalence Point K w = x 2 ph = 7.0 Excess H + Calculate [H + ] = (C HCL V HCl C NaOH V NaOH )/(V HCl +V NaOH )

16 P141 Calculation of titration from spread sheet Known [M + ] [X - ] = Ksp/[M + ] Then [M + ] & [X - ] V M

17 Ch 13 EDTA Titration EDTA: EthylenDiamineTetraAcetic acid EDTA 4- + M 2+ EDTA-M 2+ EDTA forms a strong 1: 1 complex with most metals Widely used in industries Important Player in Environmental Science

18 ATP: fuel compounds for many biological compounds ATP-Metal Complex

19 13-1 Metal-Chelate Complex Metal Ions are Lewis acids, accepting electron-donating ligands that are Lewis bases. Cyanide (CN-) is called a monodendate because it binds to a metal ion through only one atom (the carbon atom). A ligand that attaches to a metal ion through more than one atoms Is called multidenate ( many toothed ). A simple chelating ligand: ethylenediaminie is bidenate

20 Structures of chelating agents The chelating effect is the ability of multidendate to form more stable metal complexes than those of similar mono dendate ligand.

21 Entropy Effects in Chelating Less random H = kj/mol S = - 2 kj/mol More random H = -58 kj/mol S = - 71 kj/mol G = H - T S K = exp(- G/RT)

22 EDTA EDTA is a hexa-potic system. A commonly used reagent is the disodium salt, Na 2 H 2 Y 2H 2 O H 6 Y 2+, H 5 Y +, H 4 Y (neutral acid), K 1 = [H + ][H 5 Y + ]/[H 6 Y 2+ ] K 3 = [H + ][H 3 Y - ]/[H 4 Y] etc. Q.When EDTA is dissolved into water, what is the fraction of Y 4-, α Y4-?