Chem 222 #20 Ch 23, Ch26, Ch11 Mar 29, 2005

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1 Chem 222 #20 Ch 23, Ch26, Ch11 Mar 29, 2005

2 Announcement Your midterm will be returned probably this week. Reports for Exp 13, 15 are due this W/R. Reports for KH 5-1 is due on next M/T. Today we will cover Ch 23(1-3) Analytical Separation and Ch 26 (1) Ion-exchange Chromatography and Ch 11.

3 Exp DU 10-5 Ion exchange separation and spectrophotometric determination of Ni 2+ and Co 2+ Separation by strong-base anion exchanger In 9M HCl Co Cl - CoCl 4 2- Absorbed Ni 2+ Ni 2+ Not Absorbed In 4M HCl CoCl 4 2- Co Cl - Eluted

4 26-1 Ion-Exchange Chromatography Ion Exchangers Polystyrene resin Attracts cations -SO 3- H + Attracts anions -CH 2 N + (CH 3 ) 3 Cl -

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6 23-1 Solvent Extraction Extraction is the transfer of a solute from one phase to another. Ex. Extraction of an aqueous solution with organic solvent The objective of extraction Isolate or concentrate the analyte Partition Coefficient: AS 2 [ S] 2 K = = A [ S] S1 1

7 Partition Coefficient: AS 2 [ S] 2 K = = A [ S] S1 1 q: Fraction of S in the phase 1 M: moles of S V 1, V 2 : volume of solvent 1 & 2 [S] 2 = (1 q)m/v [S] 1 = qm/v The fraction in Phase 1 after 1 extraction q = V 1 /(V 1 + KV 2 ) (K ~ 0 q ~ 1) (K >> V 1 /V 2 q ~ 0) The fraction in Phase 1 after 2 extraction q 2 = {V 1 /(V 1 + KV 2 )} 2 The fraction in Phase 1 after N extraction q N = {V 1 /(V 1 + KV 2 )} N

8 Ex. (p550) Solute A has a partition coefficient of 3 between toluene and water. Suppose that 100 ml of a M aqueous solution of A are extracted with toluene. What fraction of A remains in the aqueous phase (a) if one extraction with 500 ml is performed? K = [A] toluene /[A] H2O q = 100 ml /{100mL + (3)(500 ml)} = Only 6.2 % (b) With five 100-mL extractions, how much fraction of A is remaining? q = [100 ml /{100mL + (3)(100 ml)}] 5 = (0.25) 5 = = 0.1 %

9 ph Effects If a solute is an acid or base, its charge changes as the ph is changed. A neutral species is more soluble in an organic solvent Ex. B is soluble in aqueous (1) and organic (2) phase K = [B] 2 /[B] 1 BH + is soluble in aqueous phase. Distribution coefficient D = [B] 2 /{[B] 1 + [BH + ] 1 } = KKa/{Ka + [H + ]}

10 11-1 Diprotic Acids and Bases H 3 N + - O 2 C \ CH-R Substituent / where R is a different group for each amino acid. ph = pka + Log[A - ]/[AH] [A - ]/[AH] = 10 (ph-pka)

11 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 K 1 =?

12 P206 ph & Compositions of 0.05 M H 2 L +, HL, L - Solution The acidic form H 2 L + Leucine hydrochloride contains protonated species. Because K 1 = , H 2 L + is a weak acid. HL is an even weaker acid: k 2 = It appears that the H 2 L + will dissociate only partly, and the resulting HL will hardly dissociate. H 2 L + HL + H + F x x x K 1 = [Q1]/(F x) x= M K a2 = [H + ][L - ]/[HL] Basic Form L - The species, L -, is found in a salt such as sodium leucinate. In solution, the salt gives L -. L - +H 2 O HL + OH - K b1 = K w /K a2 = 5.59 x 10-5 HL +H 2 O H 2 L + + OH - K b2 = K w /K a1 = 2.13 x K b1 shows that L - will not hydrolyze very much to give HL. K b2 shows H 2 L + is generated even less. L - +H 2 O HL + OH - K b1 = [Q1] F-x x x K b2 = [H 2 L + ][OH - ]/[HL]

13 K 1 = x 2 /(F x) K 1 = & F = M x 2 + K 1 x FK 1 =0 x = {-K 1 + (K FK 1 ) 1/2 }/2 = ph = 1.89 x ~ (K 1 F) 1/2 =

14 Fractional Composition Diagram (p216)

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16 Intermediate Form, HL A solution prepared from leucine, HL, is more complicated. HL H + +L - ; K a2 = (11-8) HL + H 2 O H 2 L + +OH - ; K b2 = (11-9) A molecule that can both donate and accept a proton is said to be amphiprotic. To solve (11-8) & (11-9), we use charge [Q1] [H + ] +[H 2 L + ] = [L - ] +[OH - ] [H 2 L + ] = [HL][H + ]/K a1 [L - ] = [HL]K a2 /[H + ] [OH - ] = K w /[H + ] [H + ] + [HL][H + ]/K a1 = [HL]K a2 /[H + ] + K w /[H + ] {1+([HL]/K a1 )}[H + ] 2 = ([HL]K a2 +K w ) Major species is [HL] [HL] ~ F [H + ] = {(FK a2 +K w )/{1+(F/K a1 )}} 1/2 = 1/ 2 FK a2ka1 + Ka 1K w (11-11) Ka 1 + F

17 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)

18 Ex. ph of the Intermediate Form of a Diprotic Acid (p210) KPH is a salt of the intermediate form of phathalic acid. Calculate the ph of both 0.10 M and M KPH. F >> K 1 = 10 [Q1] & FK 2 >> K W = ph = [Q2] = 4.18 Check ph with the equation (11-11) ph = 1/ 2 FK a2ka1 + Ka 1K w Ka 1 + F = 4.18 for F =0.10 M 4.20 for F =0.010M

19 Intermediate Form, HA - Summary A solution prepared from the intermediate A a1 form HA -, is more complicated. HA - H + +A 2- K a2 (11-8) HA - + H + H 2 ; 1/K (11-9) ph = (pk a1 + pk a2 )/2 [H + ] = 10 -ph [HA - ] = F K a2 = [H + ][A 2- ]/[HA - ] [A 2- ] = K a2 [HA - ]/[H + ] 1/K a1 = [H 2 A]/[HA - ][H + ] [H 2 A] = [H + ][HA - ]/K a1

20 Read the summary in p211

21 (pk 1 +pk 2 )/2

22 Exp KH 5-1(due next M/R) From Nernst s equation E =K + m Log[I - ] (5-1) m ~ For [I - ] of , 10-3, 10-4 M, measure E and plot E vs Log[I - ] (or Log[KI]) K & m should be obtained by fitting a plot with y = K + mx

23 Concentration of Unknown E = K + mlog[i - ] E 2 = K + mlog[i - ] added E 1 = K + mlog[i - ] unknown E E 2 E 1 = mlog{[i - ] added /[I - ] unknown } A 10 E/m = [I - ] added /[I - ] unknown A = [I - ] added /X (X [I-] unknown ) [I-] added = (50.0/55.0) X + S (5.0/50.0)