EXPERIMENT 2 MUTUAL SOLUBILITY OF BINARY SYSTEM PHENOL-WATER A. PURPOSE 1. Getting a composition curve of phenol-water system against temperature at constant pressure. 2. Determining Determining the critical temperature of of mutual solubility solubility of phenol-water phenol-water system. system. B. BASIC THEORY The binary system phenol - water is a system that shows the nature of the mutual solubility between phenol and water at a certain temperature and pressure remain. Solubility (solubility) is the ability of a certain chemical substances, dissolved substances (solutes), to dissolve in a solvent (solvent). Solubility is expressed in the maximum amount of solute dissolved in a solvent at equilibrium. Resulting solution is called a saturated solution. Certain substances can be dissolved by any comparison against a solvent. An example is ethanol in water. This trait is more in English more accurately called miscible. The solvent is usually a liquid that can be either pure substances or mixtures. The mixture consists of some kind. In view of the phase, the phenol-water binary system, there are two types of mixtures can be paid in certain circumstances. A phase is defined as part of a uniform or homogeneous system between its macroscopic state of the sub, but really - completely separate from the other parts of the system by a clearly defined and well. A mixture of two solids or liquids that do not mix with each other to form a separate phase. While the mixture of gases is a phase because the system is homogeneous. Common symbol for the number of phases is P, (SK Dogra & Dogra S, 2008). Solubility of a substance in another substance is affected by: 1.Type of solvent and solute. Substances with a similar chemical structure, generally intermingled well being which is not usually difficult to mix. Water and alcohol mix perfectly (completely misible), misible), water and ether mixed partial (partially miscible), while water and oil do not mix at all (completely immiscible). (Sukardjo, Chemical Physics, p. 142) 2. Temperature Most solids become more soluble in a liquid, when the temperature is raised, for example kaliumnitrat (KNO3) in water, but there are some solid solubility decreases as the temperature is raised such as the formation of an aqueous solution of seriumsulfat (ce2 (SO4) 3). -Gas in liquid solubility of a gas in a liquid usually decreases with increasing temperature. (Keenan, Charles W., Chemistry For the University, University, Keena edition of Volume I, p. 383) 3. Pressure Pressure is not so affect the solubility solubility of a substance in the liquid, but the effect on gas solubility. solubility. (Sukardjo, 1986). These types of solutions are important there are 4, namely: 1. Solution gas in a gas gas gas mingled always perfect form a solution. The properties of the solution are additive, as long as the total pressure is not too great. 2. Solution of gas in liquid. Depending on the type of gas, the type of solvent, pressure and temperature. Solubility of N 2, H2, O2, and He in water, very small. While HCl and NH3 are very large. This is because the gas does not react with water first, while the second gas react to form hydrochloric acid and ammonium hydroxide. Type of solvent also affected, such as N2, O2, and CO2 more soluble in alcohol than in water, while the NH3 and H2S more soluble in water than alcohol.
1
3. Liquid in liquid solution. When the two liquids are mixed, these substances can be perfectly mixed, mixed partially, or not at all mixed. Liquid solubility in the liquid depends on the type of fluid and temperature. Example: Substances similar dissolution power besar.Benzena-Toluene, Water-Alcohol, Water-Methyl. Different substances can not mix water-Nitro Benzene, Air-chloro benzene. 4. Solids in a liquid solution. The solubility of solids in a liquid depends on the type of solute, type of solvent, temperature, and a little pressure. Limit of solubility is the concentration of a saturated solution. The concentration of saturated solution for a variety of very different substances in the water, depending on the type of substance. Generally, the solubility of organic substances in water is greater than in organic solvents. Generally, the solubility increases with increasing temperature because most agents have a positive dissolution heat. (Soekardjo, 1986). Mutual solubility is the solubility of a partially mixed solution when the temperature is below the critical temperature. If the critical temperature is reached, then the solution can be perfectly mixed (homogeneous) and if the temperature has passed the critical temperature, the system will return the solution is mixed with some conditions. One example of the reciprocal temperature is the solubility of phenol in water forming a parabolic curve that is based on the increase in% phenol in any changes in either temperature below the critical temperature. If the temperature of the solubility of phenol distilled water is raised above 50 ° C, the composition of the solution is a solution of the system will change. The content of phenol in the water to the top layer will increase (more than 11.8%) and phenol content of the bottom layer will be reduced (less than 62.6%). By the time the temperature reaches 66 ° C t he solubility of the composition of the solution is a balanced system, and both can be mixed with sempurna.Temperatur certain critical temperature is rising which will be obtained by the composition of the solution is in equilibrium. (Olaf A. Hougen, Chemical Process Principles, p: 167-168). There are two kinds of solutions, namely: 1. Homogeneous solution, ie when the two kinds of substances can form a uniform solution so that its structure can not be observed the different parts, even though the optical microscope. Or the solution can be said to be mixed uniformly (miscible). 2. Heterogeneous solution, ie when the two kinds of substances are mixed there are certain surfaces that can be detected between the parts or separate phases.
2
T
daerah 1 fasa
L1
B2
A2
Suhu
L0
A1
T2 B1
T1
To XA = 1
XC
XF = 1
Figure 1. Graph of mutual solubility phenol and water Description: L1 = phase phenol in water L2 = water phase in the phenol xA = mole fraction of water mole XF = mole fraction of phenol xc = mole fraction of the components on the critical point (TC) C. USED EQUIPMENT 1. Reaction tube with diameter 4 cm 1 pc 2. Thermometer 1pc 3. Stirrer 1 pc 4. Beaker glass 400 mL 1 pc 5. Waterbath 1 set 6. Buret 50 mL 1 pc
D. CHEMICAL SUBSTANCE USED 1. Phenol 2. Aquades
3
E. PROCEDURE
weigh fenol 5 grams
make biuret preparation and filled with water
titrating phenol with aquadest, until turbid next is weigh volume is added
Temperature record (t1) when solution became clear, wait until t2 (t1 + 4)
doing experiment with variation of added water
tube is outside, record the temperature of tube when solution became turbid
calculate t average
calculate Xfenol and Xair , and makes graph biner of the diagram
F. RESULT AND DISCUSSION Room temperature: 26 oC Phenol used: 99 % gram/mol Mass phenol = 5 gram Mr phenol = 94 % phenol = 99,9% Mass of phenol =99,9%x 5 gram
= 4.995 gram
4
Mol phenol
=
= 0.053
Graph x water versus temperature
Graph Xwater Vs T Average 70 T 65 Tc
a v e
60 55 50
r 45
T average
a 40 g 35 e 30 0.6
0.7
0.8
Xc
0.9
1
x water
In a mixing experiment was conducted with a certain composition in which the mixture - a mixture of these experienced heating and cooling at a temperature solubility of each - respectively. In mixing water - phenol solution was obtained which is not intermingled forming two layers, the top layer of the water and the bottom layer is a phenol, it is caused because the water has a lower density than the fenol.Setelah mixing occurs between water and phenol in different tubes with different kompsisi comparison, in the warm up and then cooling, in which the current reaches a certain temperature this solution will be mixed and will each split and form two phases again, where the solution becomes cloudy again. This experiment will prove the binary system phenol-water solubility. Phenol and water kelarutanya will change when it is added to the mix with one of the constituent components of the phenol and water. Change the color of the solution becomes clear and cloudy from clear to cloudy substance indicates that solubility changes are affected by temperature changes. In this experiment the water component being added and the amount of phenols fixed so that the solution changes from clear to cloudy or otherwise occurred in a changing climate. Changes in temperature depends on the composition or mole fractions of the two substances. From the data between the temperature (T) and mole fractions obtained from the experiment can be graphed binary system phenol - water, between the mole fraction vs temperature (T). Graphics should be formed where the parabolic peak is reached at the critical temperature when having a mole fraction of a particular component. In the experiment the critical temperature is 65º C with a composition of the mixture is the mole fraction of phenol 0.155 and 0.845 mole fraction water. This shows that the temperature of 65º C, the components inside the curve is the two-phase systems and components outside of the curve or beyond a critical component of a single-phase system.
5
The components are in a single phase when mixed with soluble homogeneous (clear), while the two components are in phase when the addition of water to produce two layers (cloudy). Graphics are formed in this experiment is less than perfect because of its curves are not symmetrical and more dominant on the left. At least it tends to form a parabolic curve. This curve is the curve of solubility of phenol in water and showed no mutual solubility of phenol to water. Shape of the curve obtained less according to the theory, this might be due to the following. 1. less conscientious practitioner at experiments, for example, when reading the thermometer. 2. The validity of the tools used. 3. Error analysis of the data. After this experiment, it was when phenol is added to the water with the ratio of the volume of phenol and water volume remained different, higher temperatures resulting in a total volume of water solutions at most. Changes of this solution is shown, the solution color change from cloudy to clear after a heated and from clear to cloudy after settling. Discoloration is caused because the compounds solubility changes are affected by temperature changes. The analysis that we used in this experiment include qualitative analysis and quantitative analysis. Qualitative analysis can be defined as the analysis based on observations by our senses to prove the presence or absence of analyte. While quantitative analysis is an analysis based on a mathematical calculation, such as temperature measurement, calculation of moles of water and phenol, as well as the calculation of the mole fraction.
G. CONCLUSION AND SUGGESTION 1. Conclusion a. Circumstances where the color change from cloudy to clear and back again from clear to cloudy was one example of mutual solubility. b. The temperature will be higher if the more volume of water used. c. That affect the state of being clear and cloudy instead of clear to cloudy the temperature change. d. Factors - factors solubility in this experiment include concentration, temperature, ionic namesake, agitation, and surface area. e. Mutual solubility of binary systems phenol - water has a critical temperature of 65OC. f. At the critical temperature phenol mole fraction values 0.155 and 0.845 mole fraction water. 2. Suggestion a. Before conducting the experiment, the experimentalist should should prepare carefully. b. When carrying out the experiment, the experimentalist should be more careful in making observations. c. Experimentalist should be more careful during the experiment, because the substances used are phenol when exposed skin can cause injury. 6
H. REFERENCE Hougen, A. Olaf, “Chemical Process Principles” , Part I: Material And Energy Balance, 3th Edition, John Wiley & Sons, New York, 1960. Dogra,S& Dogra SK .2008. Kimia Fisik dan Soal – Soal . UI –Press : Jakarta Sukardjo, 1987. Kimia Fisika. Rineka Cipta: Jakarta Keenan, Charles W. 1986. Kimia-Perguruan Tinggi. Erlangga: Jakarta Wahyuni, Sri, and Team Lecturer of Physical Chemistry. 2011. Practicum Guide of
Physical Chemistry . Semarang: UNNES.
I.
APPENDIXES A. CALCULATION AND TABLE 1. the addition of distilled water until the turbidity occurs first No. Aquades (mL) Observation T1 (oC) T2 (oC) Taverage 1 2 cloudy 2. The addition of distilled water after the turbidity Mass (grams) Temperature (oC) Aquades No. (mL) Phenol Water T1 T2 Taverage 1 0,1 4,989 2,1 49 33 41 2 0,2 4,989 2,3 44 43 43,5 3 0,2 4,989 2,5 51 50 50,5 4 0,1 4,989 2,6 53 50 51,5 5 0,3 4,989 2,9 58 53 55,5 6 0,3 4,989 3,2 58 57 57,5 7 0,2 4,989 3,4 59 58 58,5 8 0,4 4,989 3,8 59 58 58,5 9 0,4 4,989 4,2 63 62 62,5 10 1 4,989 5,2 66 64 65 11 1 4,989 6,2 65 63 64 12 1 4,989 7,2 64 60 62 13 1 4,989 8,2 63 59 61 14 1,5 4,989 9,7 61 59 60 15 1,8 4,989 11,5 59 58 58,5 No 1 2 3 4 5 6 7 8 9
X water 0,688 0,707 0,724 0,731 0,753 0,771 0,781 0,799 0,815
T average 41 43,5 50,5 51,5 55,5 57,5 58,5 58,5 62,5 7
10 11 12 13 14 15
0,845 0,866 0,883 0,896 0,91 0,923
65 64 62 61 60 58,5
3. Mole phenol (n p) Known: mass of phenol = 4,989 gram Phenol used = 99 % Mr phenol = 94 Mass of phenol = % x m = 99 % x 4,989 = 4.939 g mole phenol =
=
= 0.0525 mol
4. Mole water (nw) Mole water =
Mr water = 18 a. Mwater = 2.1 g nw1 = 0.117 mol b. Mwater = 2.3 g nw2 = 0.128 mol c. Mwater = 2.5 g nw3 = 0.139 mol d. Mwater = 2.6 g nw4= 0,144 mol e. Mwater = 2.9 g nw5 = 0.161 mol f. Mwater = 3.2 g nw6 = 0.178 mol g. Mwater = 3.4 g nw7 = 0.189 mol h. Mwater = 3.8 g nw8= 0.211 mol i. Mwater = 4.2 g nw9 = 0.233 mol j. Mwater = 5.2 g nw10 = 0.288 mol k. Mwater = 6.2 g nw11 = 0.344 mol l. Mwater = 7.2 g nw12 = 0,177 mol m. Mwater = 8.2 g nw13 = 0,456 mol n. Mwater = 9.7 g nw14 = 0,539 mol o. Mwater = 11.5 g nw15 = 0.639 mol 8
5. Mole fraction of phenol (XP) XP = a. b. c. d. e. f. g. h. i. j. k. l. m. n. o.
XP1 = 0.132 XP2 = 0.293 XP3 = 0.276 XP4 = 0.269 XP5 = 0.247 XP6 = 0.229 XP7 = 0.219 XP8 = 0.201 XP9 = 0.185 XP10 = 0.155 XP11 = 0.134 XP12 = 0.117 XP13 = 0.104 XP14 = 0.09 XP15 = 0.077
6. Mole fraction of water (XW) a. XW1 = 1 – XP1 = 1 – 0.312 = 0.688 b. XW2 = 1 – XP2 = 1 – 0.293 = 0.707 c. XW3 = 1 – XP3 = 1 – 0.276 = 0.707 d. XW4 = 1 – XP4 = 1 – 0.269 = 0.731 e. XW5 = 1 – XP5 = 1 – 0.247 = 0.753 f. XW6 = 1 – XP6 = 1 – 0.229 = 0.771 g. XW7 = 1 – XP7 = 1 – 0.219 = 0.781 h. XW8 = 1 – XP8 = 1 – 0.201 = 0.799 i. XW9 = 1 – XP9 = 1 – 0.185 = 0.815 j. XW10 = 1 – XP10 = 1 – 0.155 = 0.845 k. XW11 = 1 – XP11 = 1 – 0.134 = 0.866 l. XW12 = 1 – XP12 = 1 – 0.117 = 0.883 m. XW13= 1 – XP13 = 1 – 0.104 = 0.896 n. XW14 = 1 – XP14 = 1 – 0.09 = 0.91 o. XW15 = 1 – XP15 = 1 – 0.077 = 0.934 7. Mass percent of phenol % WP =
ρ water =
a. b. c. d. e. f. g.
1 g/mL % WP1 = 70.37 % % WP2 = 68.44 % % WP3 = 66.61 % % WP4 = 65.74 % % WP5 = 63.24 % % WP6 = 60.92 % % WP7 = 59.47 % 9
h. % WP8 = 56.76% i. % WP9 = 54.29 % j. % WP10 = 48.96 % k. % WP11 = 44.58 % l. % WP12 = 40.93 % m. % WP13 = 37.82 % n. % WP14 = 33.96 % o. % WP15 = 30.26 % 8. Mass percent of water a. % WW1 = 100 % - % WP1 = 100 % - 70.37 % = 29.62 % b. % WW2 = 100 % - % WP2 = 100 % - 68.44 % = 31.55 % c. % WW3 = 100 % - % WP3 = 100 % - 66.61 % = 33.38 % d. % WW4 = 100 % - % WP4 = 100 % - 65.73 % = 34.26 % e. % WW5 = 100 % - % WP5 = 100 % - 63.24 % = 36.76 % f. % WW6 = 100 % - % WP6 = 100 % - 60,92 % = 39.07 % g. % WW7 = 100 % - % WP7 = 100 % - 59.47 % = 40.52 % h. % WW8 = 100 % - % WP8 = 100 % - 56.76 % = 43.23 % i. % WW9 = 100 % - % WP9 = 100 % - 54.29 % = 45.7 % j. % WW10 = 100 % - % WP10 = 100 % - 48.96 % = 51.03 % k. % WW11 = 100 % - % WP11 = 100 % - 44.58 % = 55.41 % l. % WW12 = 100 % - % WP12 = 100 % - 40.93 % = 59.06 % m. % WW13 = 100 % - % WP13 = 100 % - 37.82 % = 62.17 % n. % WW14 = 100 % - % WP14 = 100 % - 33.96 % = 66.03 % o. % WW15 = 100 % - % WP15 = 100 % - 30.25 % = 73.02 % B. ANSWER TASKS AND QUESTIONS A. Tasks 1. Write the chemical formula of phenol and the molecular mass (Mr)! Answer : Phenol has the chemical formula C 6H5OH, The molecular mass is : 12(6) + 6(1) + 1(16) = 72 + 6 + 16 = 94 The formula is as follows : OH
2. If phenol content used levels of 95% (w/w) and a weighed mass of phenol is 5,140 grams, calculate the number of moles of phenol! Answer : mass of phenol : moles of phenol :
x 4,989 = 4,739 grams
=
= 0,052 moles
3. Briefly, what is meant by phase ? Is there any difference with the form ? Answer : Phase is homogeneous part of a substance that can be separated by mechanical and uniform in physical and chemical properties, while a form is a 10
form of substance at a particular temperature. Substances at different temperatures may have different manifestations. Suppose the water at a temperature of-10 ºC its form solid, while at a temperature of 10ºC mol ten his form B. Questions 1. What is the composition of a mixture of phenol and water in % (w/w) at the critical temperature of the solution ? Answer : Mass of phenol = 4.939 g ; mole fraction of phenol = 0,165 Mass of water = 5.2 g ; mole fraction of water = 0,835 Composition of a mixture % Phenol = Water =
2.
x 100 % = 48,71%
x 100% = 51.29%
What is the composition of a mixture of phenol and water in units of mole fraction at 50oC in which the system is in single phase and two phase ? Answer : The composition of a mixture at temperature 50 oC (at average temperature is 52,5oC) XP = 0,276 XW = 0,724 Two phase at below temperature : 51oC Single phase at on temperature : 50oC
11
