Ekstraksi cair-cair Koefien distribusi Angka banding distribusi

Ekstraksi cair-cair
Koefien distribusi
Angka banding distribusi
Persen terekstraksi
Definisi ?
Prinsip pemisahan ?
Hukum Distribusi Nernst (1891)
Suatu zat terlarut X akan mendistribusikan dirinya
diantara dua pelarut yang saling tidak bercampur
sedemikian rupa, sehingga setelah kesetimbangan
distribusi tercapai, perbandingan konsentrasi2 X di
dalam kedua fasa, pada suhu yang konstan, akan
merupakan suatu tetapan, dengan syarat X
mempunyai berat molekul yang sama pada kedua fasa
Bila suatu zat X yang terlarut dalam pelarut 1 dikocok
dengan pelarut 2 yang saling tidak bercampur di
dalam corong pisah, maka sebagian zat X akan
terdistribusi pada pelarut 2. Proses ini merupakan
proses yang bolak-balik (setimbang). Setelah
kesetimbangan tercapai maka :
[X]1 / [X]2 = KD
KD = koefisien distribusi
Proses Ekstraksi
Prinsip Ekstraksi CairCair
5
Here is the universal rule:
At a certain temperature, the ratio of concentrations of a solute
in each solvent is always constant.ﾊAnd this ratio is called the
distribution coefficient, K.
(when solvent1 and solvent2 are immiscible liquids
For example,Suppose the
compound has a distribution
coefficient K = 2 between
solvent1 and solvent2
By convention the organic
solvent is (1) and waater is
(2)
(1) If there are 30 particles
of compound , these are
distributed between equal
volumes of solvent1 and solvent2..
(2) If there are 300
particles of compound , the
same distribution ratio is
observed in solvents 1 and 2
(3) When you double the
volume of solvent2 (i.e., 200
mL of solvent2 and 100 mL of
solvent1),
the 300 particles of
compound distribute as shown
If you use a larger amount of extraction solvent, more solute is
extracted
What happens if you extract twice with 100 mL of solvent2 ?
In this case, the amount of extraction solvent is the same volume as was
used in Figure 3, but the total volume is divided into two portions and
you extract with each.
As seen previously, with 200 mL
of solvent2 you extracted 240
particles of compound . One
extraction with 200 mL gave a
TOTAL of 240 particles
You still have 100 mL of solvent1,
containing 100 particles. Now you
add a second 100 mL volume of
fresh solvent2. According to the
distribution coefficient K=2, you
can extract 67 more particles
from the remaining solution
An additional 67 particles are
extracted with the second portion
of extraction solvent
(solvent2).The total number of
particles extracted from the first
(200 particles) and second (67
particles) volumes of extraction
solvent is 267.This is a greater
number of particles than the
single extraction (240 particles)
using one 200 mL portion of
solvent2!
It is more efficient to carry out
two extractions with 1/2 volume
of extraction solvent than one
large volume!
If you extract twice with 1/2 the volume, the extraction is more
efficient than if you extract once with a full volume. Likewise,
extraction three times with 1/3 the volume is even more efficient….
four times with 1/4 the volume is more efficient….five times with 1/5
the volume is more efficient…ad infinitum
The greater the number of small extractions, the greater the
quantity of solute removed. However for maximum efficiency the
rule of thumb is to extract three times with 1/3 volume
Separatory Funnel Extraction Procedure
Separatory funnels are designed to facilitate the mixing of immiscible liquids
Konsentrasi dari S yang ada fasa air setelah dilakukan ekstraksi
adalah :
Konsentrasi S di dalam fasa organik adalah :
1. Zat terlarut A memiliki KD antara air dan
kloroform sebesar 5,00. 50 ml sampel larutan fasa
air dengan konsentrasi 0,050 M diekstraksi dengan
menggunakan 15 ml kloroform.
(a) Hitunglah efisiensi ekstraksi untuk
pemisahan ini ?
(b) Berapa konsentrasi zat terlarut pada keadaan
akhir pada tiap fasa ?
(c) Berapa volume kloroform yang dibutuhkan
untuk mengekstrak 99,9 % zat terlarut?
penyelesaian
a. Fraksi zat terlarut di dalam fasa air yang tersisa
setelah dilakukan ekstraksi
 Didapat harga qaq1, Didapat harga qorg1 ,
 Pebandingan antara harga qaq dan qorg merupakan harga
efisiensi ektraksi
b. Mol zat terlarut dalam fasa air sebelum ekstraksi
c. Untuk mengekstrak 99,9 %, maka zat terlarut yang
ada dalam haruslah 0,001
2. Dari soal no 1, Tentukan :
a. Efisiensi ekstraksi jika dilakukan 3 kali ekstraksi
b. Jumlah ekstraksi yang dilakukan agar didapat
persen terekstraksi sebesar 99,9 %
What type of organic compounds can be made water-soluble?
Compounds belonging to the following solubility classes can be
converted to their water-soluble salt form
(1) Organic acids include carboxylic acids (strong organic acids)
and phenols (weak organic acids).
(2) Organic bases includes amines
Latihan
1.
2.
3.
4.
Jelaskan perbedaan angka banding distribusi dan koefisien distribusi
A solute, S, has a distribution ratio between water and ether of 7.5.
Calculate the extraction efficiency if a 50.0-mL aqueous sample of S
is extracted using 50.0 mL of ether as (a) a single portion of 50.0 mL;
(b) two portions, each of 25.0 mL; (c) four portions, each of 12.5 mL;
and (d) five portions, each of 10.0 mL. Assume that the solute is not
involved in any secondary equilibria.
What volume of ether is needed to extract 99.9% of the solute
in problem 24 when using (a) one extraction; (b) two
extractions; (c) four extractions; and (d) five extractions.
What must a solute’s distribution ratio be if 99% of the solute
in a 50.0-mL sample is to be extracted with a single 50.0-mL
portion of an organic solvent? Repeat for the case where two
25.0-mL portions of the organic solvent are used.