Chem 222

#20 Ch 23, Ch26, Ch11Mar 29, 2005

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.

Exp DU 10-5

• Ion exchange separation and spectrophotometric determination of Ni2+ and Co2+

Separation by strong-base anion exchanger

In 9M HCl• Co2+ + 4Cl- CoCl42- Absorbed• Ni2+ Ni2+ Not Absorbed

In 4M HCl

CoCl42- Co2+ + 4Cl- Eluted

Polystyrene resin

Attracts cations-SO3

-H+

26-1 Ion-Exchange Chromatography

Attracts anions-CH2N+(CH3)3Cl-

Ion Exchangers

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 extractionIsolate or concentrate the analyte

Partition Coefficient:

K = 1

2

1

2

][][SS

AA

S

S =

q: Fraction of S in the phase 1M: moles of S V1, V2: volume of solvent 1 & 2

♦ [S]2= (1 – q)m/V ♦ [S]1 = qm/V

The fraction in Phase 1 after 1 extractionq = V1/(V1 + KV2) (K ~ 0 q ~ 1)

(K >> V1/V2 q ~ 0)The fraction in Phase 1 after 2 extraction

q2 = {V1/(V1 + KV2)}2

The fraction in Phase 1 after N extractionqN = {V1/(V1 + KV2)}N

Partition Coefficient:

K = 1

2

1

2

][][SS

AA

S

S =

Ex. (p550)

• Solute A has a partition coefficient of 3 between toluene and water. Suppose that 100 mL of a 0.010 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)} = 0.062.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 = 0.00098 = 0.1 %

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 coefficientD = [B]2/{[B]1 + [BH+]1}

= KKa/{Ka + [H+]}

11-1 Diprotic Acids and BasesH3N+

-O2C

CH-R Substituent\

/

where R is a different group for each amino acid.

pH = pKa + Log[A-]/[AH] [A-]/[AH] = 10(pH-pKa)

Diprotic systems

R pKa1 R pKa2 R| 2.33 | 9.74 |

H3N+CHCO2H ↔ H3N+CHCO2- ↔H2NCHCO2

-

H2L+ HL L-

Equilibrium in Diprotic SystemsH2L+ ↔ HL + H+ Ka1≡K1

HL ↔ L- + H+ Ka2 ≡K2

L- +H2O ↔ HL + OH- Kb1=KW/[Q1]HL +H2O ↔ H2L- + OH- Kb2 =[Q2]

How much is the pH of 0.05 M of L2H solution for L of K1 = ?

P206 pH & Compositions of 0.05 M H2L+, HL, L- Solution

The acidic form H2L+

Leucine hydrochloride contains protonated species. Because K1 =4.69×10-3, H2L+ is a weak acid. HL is an even weaker acid: k2 = 1.79 × 10-10. It appears that the H2L+ will dissociate only partly, and the resulting HL will hardly dissociate.

H2L+ HL + H+

F – x x x

Ka2 = [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- +H2O HL + OH- Kb1= Kw/Ka2 = 5.59 x 10-5

HL +H2O H2L+ + OH- Kb2= Kw/Ka1 = 2.13 x 10-12

Kb1 shows that L- will not hydrolyze very much to give HL. Kb2 shows H2L+ is generated even less.

L- +H2O HL + OH- Kb1 = [Q1]F-x x x Kb2 = [H2L+][OH-]/[HL]

K1 = [Q1]/(F –x) x=1.31×10-2 M

• K1 = x2/(F –x)K1 =4.69×10-3 & F = 0.050 Mx2 + K1x –FK1 =0

x = {-K1 + (K12 + 4FK1)1/2}/2

= 1.31×10-2

pH = 1.89

x ~ (K1F)1/2 = 1.53 × 10-2

Fractional Composition Diagram (p216)

Intermediate Form, HLA solution prepared from leucine, HL, is more

complicated.HL H+ +L- ; Ka2 = 1.79×10-10 (11-8)HL + H2O H2L+ +OH- ; Kb2=2.13×10-12

(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+] +[H2L+] = [L-] +[OH-]

[H2L+] = [HL][H+]/Ka1[L-] = [HL]Ka2/[H+][OH-] = Kw/[H+]

[H+] + [HL][H+]/Ka1 = [HL]Ka2/[H+] + Kw/[H+]{1+([HL]/Ka1)}[H+]2 = ([HL]Ka2 +Kw)

♦Major species is [HL] [HL] ~ F [H+] = {(FKa2 +Kw)/{1+(F/Ka1)}}1/2

= (11-11)

2/1

1

112

++FKKKKFK

a

waaa

Simplified Calculation for the Intermediate Form

• [H+] = 2/1

1

112

++FKKKKFK

a

waaa

Two additional assumptions

Ka2F >> Kw

[H+] =

Ka1 << F

[H+] =

2/1

1

12

+FKKFK

a

aa

( ) 2/121

2/112

aaaa KK

FKFK

=

pH = -Log[H+] = -Log(Ka1Ka2)1/2

= {-LogKa1 –LogKa2}/2 = [Q1] (11-12)

(Remember This )

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 0.010 M KPH.

F >> K1 = 10[Q1] & FK2 >> KW = 10-14

pH = [Q2] = 4.18

2/1

1

112

++FKKKKFK

a

waaapH =

Check pH with the equation (11-11)

= 4.18 for F =0.10 M4.20 for F =0.010M

Intermediate Form, HA-

SummaryA solution prepared from the intermediate

form HA-, is more complicated.HA- H+ +A2- Ka2 (11-8)HA- + H+ H2A ; 1/Ka1

(11-9)

pH = (pKa1 + pKa2)/2 [H+] = 10-pH

♦ [HA-] = F♦ Ka2 = [H+][A2-]/[HA-] [A2-] = Ka2[HA-]/[H+] ♦ 1/Ka1 = [H2A]/[HA-][H+]

[H2A] = [H+][HA-]/Ka1

Read the summary in p211

(pK1+pK2)/2

Exp KH 5-1(due next M/R)

From Nernst’s equation• E =K + m Log[I-] (5-1)

m ~ -0.05916

For [I-] of 1.00 ×10-2, 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

Concentration of Unknown

• E = K + mLog[I-]

E2 = K + mLog[I-]addedE1 = K + mLog[I-]unknown

∆E ≡ E2 – E1= 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)