Ekstraksi cair-cair Koefien distribusi Angka banding distribusi
Transcription
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.