Page 1 of 4 Chem201 Winter 2006 INTR O DU C TION TO ANA LY TI C A L SEPA R A TIO NS Extraction is a separation technique, based on different solubility of solute(s) in two immiscible solvents. This technique is used for the separation and isolation of different solutes, concentrating of solute. Since solvents used for the extraction are immiscible, they form a two-phase system (phase 1 and phase 2). Solute is partitioned between these phases: S ( phase 1) ! S ( phase 2) and this partition is governed by the corresponding equilibrium constant: K= [S]2 [S]1 If total amount of the solute in the system is m , then q is the fraction of it in phase 1 at equilibrium and (1-q) is the fraction of the same solute at equilibrium in phase 2. Therefore, concentration of the solute in the phase 1 is: S1 = qm V1 and concentration of the solute in phase 2 is: S2 = (1! q )m V2 Thus, the expression for the equilibrium is: (1! q )m [S] (1! q )V1 V2 K= 2= = qm [S]1 qV2 V1 If we solve this equation for q: q= V1 V1 + KV2 When n extractions are performed: ! V1 $ n q =# & "V1 + KV2 % Sample 1. Solute A is extracted by toluene (V2 ) from water (V1 = 100 mL). In this system K value for the solute is equal 3. We can add 500 mL of toluene in one time, or 100 mL of toluene 5 times to perform the extraction. What is the best scenario? 1. q= V1 100 = = 0.062 V1 + KV2 100 + 3 ! 500 The fraction remaining in the aqueous phase is 6.2%. Page 2 of 4 Chem201 Winter 2006 ! V1 $ n ! $5 100 q =# = # & = 0.00098 & "V1 + KV2 % "100 + 3 '100 % 2. The fraction remaining in the aqueous phase is 0.098%. The conc lusion is, it is b ette r to do f e w sma ll extr a c tions than one big extr a c tion. E f f e c t of the pH If solute is acid or base, its charge changes as pH is changed. Usually charged species are more soluble in aqueous phase and non-charged – in organic phase. Base in aqueous solution undergoes the following equilibria: + BH ! B + H + [B]aq [H + ]aq Ka = [BH + ]aq The fraction of the neutral form present in aqueous phase is: !B = [B]aq = [B]aq + [BH + ]aq [B]aq Ka = + [B]aq [H ]aq K a + [H + ] [B]aq + Ka Neutral form is soluble in organic phase. And its distribution is governed by distribution coefficient D : D= [B]org concentration in organic phase = concentration in aqueous phase [B]aq + [BH + ]aq It is similar to partition coefficient, but used instead when dealing with species present in more than one chemic a l form . When [BH + ] is expressed as function of the K a and placed into the expression for D, we can end up with the following equation: ! $ Ka D = K# = K' B + & " K a + [H ] % If acid has to be extracted then equation for the D is: ! [H + ] $ D = K# = K' HA + & " K a + [H ] % Sample 2 . Partition coefficient (K) for the base, B, is 3. This base dissolved in water (pKa = 9), its concentration is 0.01M. 50 mL of this base is extracted with 100 mL of solvent. What would be the forma l concentration of the base in aqueous phase at: (a) pH 10 and (b) pH 8? For the p H 10: Page 3 of 4 Chem201 Winter 2006 ! $ ! 10'9 $ Ka D = K# = 3 = 2.73 & # + '9 '10 & " K a + [H ] % "10 + 10 % Since charged species are involved, we have to use D in place of K to find q: q= V1 50 = = 0.15 V1 + KV2 50 + 2.73 !100 [B]aq = 0.15 ! 0.01 = 0.0015M For the p H 8: ! $ ! 10'9 $ Ka D = K# = 3# '9 = 0.273 + & '8 & " K a + [H ] % "10 + 10 % Since charged species are involved, we have to use D in place of K to find q: q= V1 50 = = 0.65 V1 + KV2 50 + 0.273 !100 [B]aq = 0.65 ! 0.01 = 0.0065M The conc lusion is, to extr a c t b ase w ith high e f f ic iency one has to inc r e ase the pH o f the aqueous phase . E f f e c t of the ch e lating agent Selective chelating is a common practice to separate metal ions. However, charged chelated ions are better dissolved in aqueous phase, which makes extraction impossible. Therefore, special chelating agents are used for this purpose. This time we should consider more equilibrium processes: a. in aqueous phase + HL aq ! H aq + L aq " nL! + M n + " ML n [H + ]aq [L" ]aq Ka = [HL ]aq #= [ML n ]aq [nL! ]n aq [M n + ] Each metal may complex with the ligand, but pH alters the selectivity. b. between aqueous and organic phase HL aq ! HL org KL = [HL ]org [HL ]aq Page 4 of 4 Chem201 ML n aq ! ML n org KM = Winter 2006 [ML n ]org [ML n ]aq Assuming that Mn+ is in aqueous phase and MLn is in organic phase, the distribution coefficient is: D= total metal concentration in organic phase [ML n ]org = total metal concentration in aqueous phase [M n + ]aq [ML n ]org = K M [ML n ]aq = K M ![L" ]n aq [M n + ]aq Also, [L! ]aq = K a [HL ]aq [H + ]aq By substituting last equation into the previous: " n [ML n ]org = K M ! [L ] aq [M n+ ]aq = K M ![M n+ # K [HL ] & n ]aq %% a + aq (( $ [H ]aq ' If we put last equation into the expression for D: " K [HL ] % n K M ! [M ]aq $$ a + aq '' " K [HL ] % n # [H ]aq & = = K M !$$ a + aq '' [M n + ]aq # [H ]aq & n+ D= [ML n ]org [M n + ]aq Most of the HL is in organic phase, and therefore: [HL ]aq = [HL ]org KL By substitution we have final equation for D: " K [HL ] % n K !K n [HL ]n org D = K M !$$ a + aq '' = M n a ( [H + ]n aq KL # [H ]aq & The conc lusion is, in o rde r to extr a c t m e tal io n w ith high e f f ic iency one has to va ry the p H and concentr a tion of the ch e lating agent.
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