EXPERIMENT 5 TITLE
: Preparation and Precipitation of Lyophobic Colloid
OBJECTIVE
:
i. To prepare prepare mangan manganum um dioxid dioxidee (MnO (MnO2) sol. ii. To test test the Schulz Schulzee – Hardy Hardy rule. rule. THEORY
The term “colloid” was first introduced by Thomas Graham (1861) A colloid system consists of a dispersion phase and a dispersion medium. If the diameter of the dispersed particles is in a range between 20 A to 2000 A, then a colloid is formed. There are eight classes of common colloidal systems. These colloids are made-up of solids, liquid and gases. A colloid formed by the dispersion of a solid in a liquid is called a sol. Sols exhibit several physical properties such as Tyndall effect, electrical effect and molecular-collision effect. In general, sol can be classified into two types; The terms lyophilic (liquid-loving) and lyophobic (liquid-hating) are frequently used to describe the tendency of a surface or functional group to become wetted or solvated. If the liquid medium is aqueous, the terms hydrophilic and hydrophobic are used. Lyophilic surfaces can be made lyophobic, and vice versa. For example, clean glass surfa surface ces, s, whic whichh are hydro hydrophi phili lic, c, can be made made hydro hydropho phobi bicc by a coat coatin ingg of wax; wax; conversely, the droplets in a hydrocarbon oil-in-water emulsion, which are hydrophobic, can be made hydrophilic by the addition of protein to the emulsion, the protein molecules adsorbing on to the droplet surfaces. Lyophilic sol can be prepared by dispersion (normally by vigorous stirring of a solid in the form of gel or resin in a liquid), and condensation. condensation. During dispersion, solids solids are disintegrated into particles of suitable sizes to form a colloid in a dispersion medium. During condensation however, molecules or atoms are condensed into clusters / particles of suitable sizes to form a colloid. Lyophobic colloids are very sensitive to the presence of electrolytes, in which a small amount of added electrolytes resulted in precipitation. The precipitation of sol by electrolytes can be predicted using Schulze – Hardy rule, which stated that “The effect of precipitation by a given ion on a (lyophobic) dispersed dispers ed phase of opposite charge increases increa ses by the increase of the valency of the ion”.
METHODOLOGY Part A: Preparation of manganum dioxide, MnO 2
3 ml of 30% H2O2 + 50 ml of distilled water
Solution prepared filled into the burette 50 ml of 0.02M KMnO4 solution introduced into a clean and dry conical flask The diluted H2O2 solution added into the 0.02M KMnO4 solution drop-by-drop while swirling the conical flask. H2O2 solution added until the color of the KMnO4 solution turns brown. A drop of the brown-colored solution taken and tested using 1.0ml of BaCl 2 solution. If a pinkish solution is observed, some more of H2O2 added into the KMnO4 solution until a pinkish solution CANNOT be further observed.
DO NOT ADD too much H2O2 to avoid precipitate formed.
Part B: Verification of the Schulze – Hardy Rule.
1
2 3 4 Clean test tubes labeled
10 ml of 0.10M NaCl solution added into test tube number 1. 1 ml of the same solution and 9 ml of distilled water added into test tube number 2, and then homogenize the solution by swirling.
1 ml of the NaCl solution from the test tube number 2 transferred into the test tube number 3, and then 9 ml of distilled water added.
Further dilution performed on the NaCl solution in the test tube number 4. 1 ml of the MnO2 sol added into each test tube containing the NaCl solutions. The test tubes mildly swirled and in 2 minutes time, the one which form a precipitate observed. The concentration of the NaCl solution recorded.
After the results obtained, further tests performed using NaCl solutions with concentrations lower than the lowest concentration that yielded a precipitate.
Steps repeated using CaCl2 and AlCl3 solutions.
DATA / RESULTS A) Preparation of manganum dioxide (MnO 2) sol.
Initial reading (cm 3) final reading (cm 3) Volume of H2O2 solution used (cm 3)
1.9 5.1 3.2
B) Verification of the Schulze - Hardy rule.
Guideline: √ = precipitate exist Х = precipitate do not exist
Concentration of NaCl (M) 0.1 0.08 0.06 0.04 0.02 0.01 0.001 0.0001
Volume of NaCl (ml) 10.0 8.0 7.5 6.7 5.0 1.0 1.0 1.0
Volume of distilled water (ml) 0.0 2.0 2.5 3.3 5.0 9.0 9.0 9.0
Observation of precipitate (√ / Х) √ √ √ √ Х Х Х Х
Volume of distilled water (ml) 0.0 9.0 9.0 2.0 2.5 3.3 5.0 9.0
Observation of precipitate (√ / Х) √ √ √ √ √ √ Х Х
Minimum concentration: 0.04 M
Concentration of CaCl2 (M) 0.1 0.01 0.001 0.0008 0.0006 0.0004 0.0002 0.0001
Volume of CaCl2 (ml) 10.0 1.0 1.0 8.0 7.5 6.7 5.0 1.0
Minimum concentration: 0.0004 M
Concentration of AlCl3 (M) 0.1 0.008 0.006 0.004 0.002 0.01 0.001 0.0001
Volume of AlCl3 (ml) 10.0 8.0 7.5 6.7 5.0 1.0 1.0 1.0
Volume of distilled water (ml) 0.0 2.0 2.5 3.3 5.0 9.0 9.0 9.0
Minimum concentration: 0.01 M
CALCULATION Dilution of NaCl solution:
Used the equation: M1V1 = M2V2 From the equation, M1V1 = M2V2 (0.1 M) V1 = (0.04 M) (10.0 ml) V1 = 4.0 ml Volume of 0.1 M NaCl solution been used = 4.0 ml Volume of distilled water been used = 6.0 ml So, the minimum concentration of NaCl solution = 0.04 M
Observation of precipitate (√ / Х) √ Х Х Х Х √ Х Х
By using the same method, we can determine the volume of NaCl (ml) to use for preparing the concentration of NaCl (M). 1) For 0.08 M NaCl M1V1 = M2V2 (0.1 M) V1 = (0.08 M) (10.0 ml) V1 = 8.0 ml 2) For 0.06 M NaCl M1V1 = M2V2 (0.08 M) V1 = (0.06 M) (10.0 ml) V1 = 7.5 ml 3) For 0.04 M NaCl M1V1 = M2V2 (0.06 M) V1 = (0.04 M) (10.0 ml) V1 = 6.7 m 4) For 0.02 M NaCl M1V1 = M2V2 (0.04 M) V1 = (0.02 M) (10.0 ml) V1 = 5.0 ml 5) For 0.01 M NaCl M1V1 = M2V2 (0.1 M) V1 = (0.01 M) (10.0 ml) V1 = 1.0 ml 6) For 0.001 M NaCl M1V1 = M2V2 (0.01 M) V1 = (0.001 M) (10.0 ml) V1 = 1.0 ml 7) For 0.0001 M NaCl M1V1 = M2V2 (0.001 M) V1 = (0.0001 M) (10.0 ml) V1 = 1.0 ml
Dilution of CaCl 2 solution:
Used the equation: M1V1 = M2V2 From the equation, M1V1 = M2V2 (0.001 M) V1 = (0.0004 M) (10.0 ml) V1 = 4.0 ml Volume of 0.001 M CaCl2 solution been used = 4.0 ml Volume of distilled water been used = 6.0 ml So, the minimum concentration of CaCl2 solution = 0.0004 M By using the same method, we can determine the volume of CaCl 2 (ml) to use for preparing the concentration of CaCl2 (M). 1) For 0.0008 M CaCl2 M1V1 = M2V2 (0.001 M) V1 = (0.0008 M) (10.0 ml) V1 = 8.0 ml 2) For 0.0006 M CaCl2 M1V1 = M2V2 (0.0008 M) V1 = (0.0006 M) (10.0 ml) V1 = 7.5 ml 3) For 0.0004 M CaCl2 M1V1 = M2V2 (0.0006 M) V1 = (0.0004 M) (10.0 ml) V1 = 6.7 ml
4) For 0.0002 M CaCl2 M1V1 = M2V2 (0.0004 M) V1 = (0.0002 M) (10.0 ml) V1 = 5.0 ml 5) For 0.01 M CaCl2 M1V1 = M2V2 (0.1 M) V1 = (0.01 M) (10.0 ml) V1 = 1.0 ml 6) For 0.001 M CaCl2 M1V1 = M2V2 (0.01 M) V1 = (0.001 M) (10.0 ml)
V1 = 1.0 ml 7) For 0.0001 M CaCl2 M1V1 = M2V2 (0.001 M) V1 = (0.0001 M) (10.0 ml) V1 = 1.0 ml
Dilution of AlCl 3 solution:
Used the equation: M1V1 = M2V2 From the equation, M1V1 = M2V2 (0.1 M) V1 = (0.01 M) (10.0 ml) V1 = 1.0 ml Volume of 0.001 M AlCl3 solution been used = 1.0 ml Volume of distilled water been used = 9.0 ml So, the minimum concentration of AlCl3 solution = 0.01 M By using the same method, we can determine the volume of AlCl3 (ml) to use for preparing the concentration of AlCl3 (M). 1) For 0.08 M AlCl3 M1V1 = M2V2 (0.1 M) V1 = (0.08 M) (10.0 ml) V1 = 8.0 ml 2) For 0.06 M AlCl3 M1V1 = M2V2 (0.08 M) V1 = (0.06 M) (10.0 ml) V1 = 7.5 ml 3) For 0.04 M AlCl3 M1V1 = M2V2 (0.06 M) V1 = (0.04 M) (10.0 ml) V1 = 6.7 ml
4) For 0.02 M AlCl3 M1V1 = M2V2 (0.04 M) V1 = (0.02 M) (10.0 ml) V1 = 5.0 ml 5) For 0.01 M AlCl3 M1V1 = M2V2 (0.1 M) V1 = (0.01 M) (10.0 ml) V1 = 1.0 ml 6) For 0.001 M AlCl3 M1V1 = M2V2 (0.01 M) V1 = (0.001 M) (10.0 ml) V1 = 1.0 ml 7) For 0.0001 M AlCl3 M1V1 = M2V2 (0.001 M) V1 = (0.0001 M) (10.0 ml) V1 = 1.0 ml The minimum concentrations for the salt solution (NaCl, CaCl2 and AlCl3) are needed to precipitate MnO2 sol in 2 minutes. [NaCl] : [CaCl2] : [AlCl3] 0.04 M : 0.0004 M : 0.01 M 100 : 1 : 25
DISCUSSION
Colloid and interface science deals with multi-phase systems in which one or more phases are dispersed in a continuous phase of different composition or state. Classical colloid science deals with dispersions for which at least one dimension of a dispersed phase falls within about 1 and 1000 nm. In applied colloid science the upper size limit is commonly extended to at least 10 000 to 100 000 nm. Interface science deals with dispersions in which there is an extremely large interfacial area between two of the phases. The dispersed phases may be particles, droplets or bubbles. Colloidal particles are larger than molecules but too small to be observed directly with a microscope; however, their shape and size can be determined by electron microscopy. In a true solution the particles of dissolved substance are of molecular size and are thus smaller than colloidal particles; in a course mixture the particles are much larger than colloidal particles. Although there are no precise boundaries of size between the particles in mixtures, colloids or solutions, colloidal particles are usually on the order of 10-7 to 10-5 cm in size. The tendency of colloidal dispersions in a fluid state not to separate is aided by the collisions that the dispersed particles experience from the constantly moving molecules of the ‘solvent’. The erratic movement of colloidally dispersed particles caused by such uneven buffeting is called Brownian movement after Scottish botanist, Robert Brown (1773 – 1858). Colloidal dispersions that do eventually separate are those in which the dispersed particles, over time, grow too large. Evidently, to prepare a stable colloidal dispersion, we must not only make the dispersed particles initially small enough but must also keep them from joining together. The dispersed particles will not coalesce if they carry the same kind of electrical charge, either all positive or all negative. Ions of the opposite charge are in the solvent, keeping the whole system electrically neutral. Some of the most stable dispersions form when the surfaces of their colloidal particles have preferentially attracted ions of just one kind of charge from a dissolved salt. The dispersed particles of most sols, which are colloidal dispersions of solids in a fluid. Alternatively, dispersions may form when extremely large, like-charged ions, such as those of proteins, are involved. Colloidal dispersions of one liquid in another are called emulsions. They are often relatively stable provided that a third component called an emulsifying agent is also present. Its molecules act to give an electrically charged surface to each microdroplet of the oil, which keeps the microdroplets from coalescing. Water-in-oil emulsions are also possible. Even when a beam of light is focused on a starch dispersion so dilute as to look as clear as water, the path of the beam is revealed by the light scattered to the side. Light scattering by colloidal dispersions is called the Tyndall effect, after John Tyndall (1820 –
1893), a British scientist. Solutes in true solutions, however, involve species too small to scatter light, so solutions do not give the Tyndall effect. From the experiment, there are some errors that may effects the results. Errors while using apparatus such as the measuring cylinder and pipette. KMnO4 is dark in colours that may cause to inaccurate value while measuring it. We also did not observe the changes within 2 minutes. Some of the precipitate formed dissolve after 2 minutes. Precaution Steps; 1. Use dry an clean conical flask. 2. Avoid overtitration with H2O2 to avoid precipitate form. 3. Observe the changes occurred within 2 minutes only. 4. Calculate the volume using the concentrations.
CONCLUSION
NaCl = Minimum concentration: 0.04 M CaCl2 = Minimum concentration: 0.0004 M AlCl3 = Minimum concentration: 0.01 M The minimum concentrations for the salt solution (NaCl, CaCl2 and AlCl3) are needed to precipitate MnO2 sol in 2 minutes. [NaCl] : [CaCl2] : [AlCl3] 0.04 M : 0.0004 M : 0.01 M 100 : 1 : 25
REFERENCES
1. Chemistry matter and Its Changes, 3rd Edition Brady, Russel and Holum Pages 558 – 560 2. Atkins’ Physical Chemistry, 7th Edition Atkins de Paula Page 752
QUESTIONS
1. From the results of this experiment, determine whether the MnO2 sol is of positively or negatively charged. The MnO2 sol is negatively charged because the solution used contains some cation such as Cl-, and other anion Al3+, Ca2+ and Na+. 2. By what method the MnO2 sol was prepared? 2MnO4- + 5H2O2 + 6H+ 2Mn2+ + 8H2O + MnO2The MnO2 sol was prepared by condensation method. It refers to the formation of particles by precipitation reaction chemistry. 3. Supposed a soluble lyophilic sol was added into an MnO2 sol prior to the addition of an electrolyte, what would be the effect? Explain your answer. By adding electrolyte after lyophilic are added to MnO2. so the precipitation are MnO2 sol as we expect. Using H2O2 30% lyophilic sol charges positive. It can not occur between MnO2 sols that are in negative charge with lyophilic sol that is in charge positive. Complete precipitation will done during isoelectronic point are reach.
