Etiket Arşivleri: experiment
BACTERIOLOGICAL MEDIA AND STERILIZATION
The purpose of this experiment is that to prepare media from nutrient agar for cultivating microorganisms. Bacteria have special requirements to grow. In order to see bacterial growth, a medium is needed. A MEDIUM is a nutritional environment for bacteria to grow. There are two primary forms of media: Liquid (Broth) and Solid (Agar). The most common solid medium used to grow bacteria in a microbiology lab is Nutrient Agar. The most common liquid medium used in the lab is Nutrient Broth. AGAR is derived from the extract of seaweed (an alga).Agar contains two main components: agarose and agaropectin. Agar contains solidifying factors and therefore has a gelatin like nature. Some agars are used in cooking and preparation of ice cream. The most familiar food is Jell-O. Nutrients such as peptone, tryptone, soy, and beef extract, salt, calcium, magnesium, water, and manganese are added to the agar or broth providing the bacterium with a proper growing environment.
All bacteriological media must first be sterilized before it can be used. Distilled water is added to agar, heated to a boil then autoclaved. An AUTOCLAVE sterilizes the bacterial growth medium so that a pure culture can be obtained. The autoclave sterilizes the medium by subjected it to a temperature of 121° C for 15 to 20 minutes. It uses steam under pressure to obtain this temperature. This will kill any heat resistant bacteria that have contaminated the medium. Once autoclaved, the agar can then be poured into a PETRI PLATE or test tube. When placed in a test tube, it can either be tilted on a slanted board so that it will solidify at a SLANT or remain upright to solidify into a DEEP. Broths are dissolved with water, added to test tubes, capped then autoclaved.
Media must include source of C, N, P, S, 4 of the 6 major nutrients (CHNOPS), as well as micronutrients. These are usually present as trace contaminants in water, on glassware, or in chemicals used to make media.
Media can be liquid or solid. Use for different purposes:
Liquid media: easiest to prepare and use. Good for growing quantities of microbes needed for analysis or experiments. Unless inoculated with pure culture, cannot separate different organisms.
Solid media: usually made by adding agar, seaweed extract, to appropriate liquid. 1.5% agar is standard for plates. Agar melts at 80-90 deg. C, will remain liquid until temperature cools to 40-42 deg. C. Very few microbes can degrade agar, so it is normally not a source of C, and acts as inert gelling medium.
Synthetic of Defined Media: usually relatively simple media, all components are known. Useful for photoautotrophs, also in some experimental situations where want to select mutants unable to use certain compounds, or for radioisotope labeling. Example: you want to select a microbe that can obtain all its nitrogen from atmospheric N2. You would prepare synthetic medium with sources of C, P, and S, but no N source. Organisms would be unable to grow unless they can fix nitrogen from air.
Complex Media: composition of media not completely known. Often made from inexpensive organic materials such as slaughterhouse wastes (tryptic digests called tryptone, trypticase, etc.), soybeans, yeast wastes from brewing (rich source of vitamins), animal blood, etc. All our standard laboratory media in MCB 229 are complex media, such as Tryptone agar, TSA (trypticase soy agar), Nutrient agar, etc.
Selective Media: media favors the growth of one or more microbes. Example: bile salts inhibit growth of most gram-positive bacteria and some gram-negative bacteria, but enteric bacteria adapted to life in animal gut can grow well. Include bile salts in some media such as EMB, MacConkey agar (will use later in this course) to select for enteric.
Differential Media: media allows distinguishing between different bacteria that grow. Ex: MacConkey agar has color indicator that distinguishes presence of acid. Bacteria that ferment a particular sugar (e.g., glucose in culture media) will produce acid wastes on plates, turn pH indicator red. Bacteria that cannot ferment the same sugar will grow but not affect pH, so colonies remain white.
Note that it is possible to design a medium that is both selective and differential.
Sterilization of Media and Equipment:
Sterilization denotes the use of physical or chemical agents to eliminate all viable microbes from a material, whereas disinfection generally refers to the use of germicidal chemical agents to destroy the potential infectivity of a material and need not imply elimination of all viable microbes. Sanitizing refers to procedures used to lower the bacterial content of utensils used for food without necessarily sterilizing them. Antisepsis usually refers to the topical application of chemicals to a body surface to kill or inhibit pathogenic microbes.
Heat is generally preferred for sterilizing materials excepts those that it would damage. The agent penetrates clumps and reaches sites that might be protected from a chemical disinfectant. Fungi, most viruses and vegetative cells of various pathogenic bacteria are sterilized within a few minutes at 50 to 70 °C and the spores of various pathogens at 100 °C. The spore of some saprophytes, however, can survive boiling for hours. Because absolute sterility is essential for culture media and for the instruments used in major surgical procedures, it has become standard practice to sterilize such materials by steam in an autoclave at a temperature of 121 °C (250 °F) for 15 to 20minutes.
In using an autoclave, it is important that flowing steam be allowed to displace the air before building up pressure , for in steam mixed with air, the temperature is determined by the partial pressure of the water vapour. Thus, if air at 1 atm (15 psi) remains in the chamber and steam is added to provide an additional gauge pressure at 1 atm, the average temperature will be only 100° (that of steam at 1 atm). More over, heating will be uneven because the air will tend to remain at the bottom of the chamber.
Pasteurization is now used primarily for milk. It consists of heating at 62 °C for 30 minutes or in “flash” pasteurization, at a higher temperature for a fraction of a minute. The total bacterial count is generally reduced by 97% to 99%. Pasteurization is effective because the common milk borne pathogens (tubercle bacillus, Salmonella, Streptococcus, and Brucella) do not form spores.
Moist heat and Dry heat:
Sterilization by heat involves protein denaturation and the melting of membrane lipids as a consequence of disruption of multiple weak bonds. Among these, hydrogen bonds between a >C=O and an HN< group are more readily broken if they can be replaced with hydrogen bonds. Accordingly, sterilization requires a higher temperature for dry than for wet material. Reliable sterilization of glassware and instruments in a dry oven requires 160 °C for 1 to2 hours. In addition, bacteria and viruses, like isolated enzymes, are more stable in an aqueous medium when the water concentration is reduced by the presence of a high concentration of glycerol or glucose.
The role of water in heat denaturation of proteins is illustrated by the usefulness of steam in pressing woolen fabrics (e.i. in shifting the multiple weak bonds between fibrous molecules of keratin).
When a suspension of bacteria is frozen, the crystallization of the water results in the formation of tiny pockets of concentrated solution of salts, which do not themselves crystallize unless the temperature is lowered below the eutectic point (about -20°C for NaCl); at this temperature, the solution becomes saturated, and the salt also crystallizes. The localized high concentrations of salt, and possibly the ice crystals, damage the bacteria, as shown by their increased sensitivity to lysozyme. Only some of the cells are killed, but repeated cycles of freezing and thawing result in the progressive decrease in the viable count.
With radiation of decreasing wavelength, the killing of bacteria first becomes appreciable at 330nm and then increases rapidly. The sterilizing effect of sunlight is attributable mainly to its content of UV light (300-400 nm). Most of the UV light approaching the earth from the sun, and all of that shorter than 290 nm, is screened out by the ozone in the outer regions of the atmosphere; otherwise, organisms could not survive on the earth’s surface.
This mechanism is not convenient for routine laboratory use, but intense source of radioactivity are now being used to sterilize food. Public fear of danger from the irradiation is unwarranted, as the activated mutagenic molecules produced by the irradiation are extremely short-lived.
Ultrasonic and sonic waves:
In the supersonic (ultrasonic) range, with a frequency of 15,000 to several hundred thousand per second, sound waves denature proteins, disperse a variety of materials, and sterilize and fragment bacteria. The effect has not been of practical value as a means of sterilization, but it is useful for disrupting cells for experimental purposes (sonication).
Bacteria- free filtrates may be obtained by the use of filters with a maximum pore size not exceeding 400nm.This procedure is used for solutions that cannot tolerate sterilization by heat (e.g. sera an media containing proteins or labile metabolites). The early, rather absorptive filters of asbestos or diatomaceous earth were replaced by unglazed porcelain or sintered glass, and these in turn have been replaced by nitrocellulose membrane filters of graded porosity. Membrane filters can also used to recover bacteria quantitatively for chemical and microbiological analysis.
Chemical methods of microbial control: Most of the chemical agents are used for disinfecting and cannot achieve sterility. The term disinfectant is restricted to those that are rapidly bactericidal at low concentrations. The activity of a disinfectant depends upon the pathogen/nature of material being disinfected.
Salts: Pickling in brine or thermal treatment with solid NaCl has been used for many centuries as a means of preserving perishable meats and fish. Bacteria differ widely in susceptibility.
Heavy Metals: The various metallic ions can be arranged in a series of decreasing antibacterial activity. With small inocula, Hg + and Ag +. At the head of the list, are effective at less than 1 ppm because of their high affinity for –SH groups. The antibacterial action of Hg 2+ can be reversed readily by sulfhydryl compounds.
Phenol and Phenol derivatives: Phenol (carbolic acid) has an anaesthetic effect at concentration below 1% but at concentrations above 1% it has significant antibacterial effect. Phenol and its derivatives (phenolics), exert antimicrobial effect by damaging plasma membrane, denaturing proteins and inactivating enzymes. The phenolics contain a molecule of phenol that has been chemically altered to reduce its irritating qualities and increase its antibacterial activity in combination with soap or detergent. They are often used as disinfectants. Cresols are a group of phenolic found in Lysol. Another phenolic derivative a bisphenol -hexachlorophene is ingredient in soaps and lotions used as disinfectants. It is effective against gram positive streptococci, and staphylococci which cause skin infections.
Halogens: Particularly chlorine and iodine are antimicrobial agents effective against all kinds of bacteria, many endospores, fungi, and viruses. Iodine inhibits the microbial protein synthesis by combining with the amino acid tyrosine. The germicidal action of chlorine is due to the formation of hypochlorous acid, which forms when chlorine is mixed with water. A liquid form of compressed chlorine is used for disinfecting drinking water, swimming pools, and sewage. Also hypochlorite solutions (200ppm Cl2) are used to sanitize clean surfaces in the food and the dairy industries and in restaurants.
Alcohols: Effective against bacteria and fungi but not endospores and viruses. Alcohol denatures proteins and dissolves lipid component of cell membranes. Alcohol has the advantage of acting rapidly and evaporating without leaving any residue. Ethanol and isopropanol are the two important alcohols most commonly used.
600 mL Nutrient agar
600 mL distilled water
a piece of folio
a piece of cotton
Firstly, required amount of nutrient agar was calculated. This amount was 12gr. for 600 mL, then 12gr. of nutrient agar was weighed and was put in Florence flask. Next, 600 mL of distilled water was added above 12gr. of nutrient agar. Afterwards, solution was heated until it completely dissolved and solution had clear appearance, then the prepared solution in Florence flask was put in the autoclave at 121 °C for 2 hours for sterilisation. Solution was cooled to about 45 °C after than 2 hours. Finally, about 20 mL of solution was poured into each Petri plate and solutions in Petri plate were waited until it solidified.
In this experiment, we prepared Nutrient Agar Media for Petri plate. Previously, a certain amount of nutrient agar was mixed with distilled water and then solution was heated, and it was put in autoclave for sterilization, after then two hours as far as about 45 °C itwas cooled. Afterwards, approximately 32 Petri plates it was poured.
During experiment, we learned to prepare different media according to a certain microorganisms and a certain purposes. These media are anaerobic, synthetic, transport, enriched, selective, differential and microbiological assay media. For example; in anaerobic media; oxygen is removed from media with reducing agent, in synthetic media; their chemicals and concentration is known and identified, in transport media; microorganism is transferred from place to other place temporarily, in enriched media; number of scarce microorganisms are increased, but during this process growth of other microorganisms are not prevented, in selective media; growth of a special microorganisms is supplied and growth of other microorganisms are prevented, in differential media; appearance and size of microorganisms are determined with indicator, in microbiological assay media; concentration of some substances are measured. These media are prepared from nutrient broth and nutrient agar that broth represents liquid media and agar represents solid media.
Some media are slant media shape some media also are vertical shape. Slant media provide a large surface area so slant media is convenient for aerobic organisms, in vertical (deep) media, microorganism is cultivated toward the penetration of media and Petri plate also supplies a large surface area and because of this it supplies growth of microorganisms in a short time.
Finally; sterilization of media and equipment were learned during experiment. Moreover, required criteria one by one was told us. For sterilization, applied processes are classified as physical and chemical methods. Physical methods are heat, wet sterilization, tyndallization dry heat sterilization, radiation, freezing and bacteriological filtration. Chemical methods are salt, phenol and phenol derivatives, halogens, and alcohols. Chemical and physical methods and their effects on the microorganisms and equipment were learned and examined.
To learn what the effects of pH (Bacillus subtilis) and ultraviolet light on the bacterial (E.coli) growth are. To be able to should learn how determine the oxygen requirement of a particular microorganism (Bacillus subtilis).
Effect of pH
Firstly, prepare Glucose-phenol red broth (pH 3, 5, 7, 9) containing 0.5%w/v glucose+basal medium. Then, under aseptic conditions inoculate 4 GPRB tubes with a microorganism obtained from the stock culture. Finally, incubate at 37°C for 24 hr. Following week check bacterial growth by observing turbidity by comparing turbidity observations of 4 tubes relative to each other
Effect of UV
First of all, agar plates should be labeled as in the figure on the bottom before inoculation as shown below:
Study under aseptic conditions inoculate 0.1 ml of culture and open a plate and expose two halves to UV for indicated time periods by shielding the other half with a card. Then, incubate at 37°C for 24 hr. Following week, check bacterial growth by counting colonies via comparing colony numbers of the parts exposed to UV for different time periods and comparing you results with other results of other groups
Study under aseptic conditions, inoculate thioglycollate medium by the culture (B.subtilis). Then, rotate the tube between your palms gently to get good dispersion of inoculation. Continue with incubation at 37°C for 24 hr. Following week, check bacterial growth and decide oxygen requirement
Results & Observations
Table: pH Effect (Bacillus subtilis)
+++(light color change, yellow become darker)
Effect of UV
-Lawn formation due to lack of UV effect in all plates
-No growth is observed, no color change occurred
In this experiment our purpose was that to investigate the effects of irradiation with ultraviolet light on the growth of the bacteria (E.coli), effects of pH of the medium to growth of Bacillus subtilis, and lastly we tried to determine experimentally the oxygen requirements of it. The results are summarized in the results part. As you can see, Bacillus subtilis growth the best at alkaline conditions especially at pH=9. This indicates that the microorganism is alkaliophile but it also can grow at neutral pH and pH=5 meaning that it can behave like neutrophile and to some extent acidophile. Theoretically it is known that Bacillus spp are classiied as some of them as acidophile and alkaliphile. But we know that B.subtilis is an alkaliphile but not obligatory as we see from our experiment. However, to confirm our results, we should have done some further inoculations into tubes with more refined pH values.
In the following study, we tried to observe the lethal effect of ultraviolet light on the growth of E.coli. However, we could not obtain any results due to not working UV lamb or ineffective UV lamb. Therefore, we obtained lawn on the plate in all plates.
Finally, we tried to determine the oxygen requirements of B. subtilis. We know that it is a aerobic bacteria, therefore, it requires oxygen to live. Thus we expected no change in the color of thioglycollate medium and no growth as we could observe in our experiment.
To sum up, except UV exercise, all of our studies resulted as expected. However, UV experiments should be done with new UV lamb or UV apparatus to prove the effect of UV on the growth of organisms because as scientist candidate I expect to see theoretical events in experiment results to be satisfied.
Information retrieved December 21, 2005 from
In this experiment , we were cleaned flasks and we marked 1,2,3 and 4. We added indicator and 20 mL of distilled water into the each flasks. We added NaOH from burette into the each flasks and we determined volume of NaOH for each flasks.When the color of solution change we closed the burette. We found the volume of spilled NaOH for each flasks and we reached the equilibrium.
In this experiment we learned that how mass and volume carried out. This mass and volume measurements then be used to determine the density of a salt solution and liquid water. During this experiment also we learned that how the density changes some conditions such as temperature , pressure ext. Personal errors are on a small scale during the experiment and we try to minimized this errors.
The solubility of any substance is a function of temperature. Most substances are endothermic, absorbing heat in the process of dissolution. For these substances, an increase in temperature results in an increase in solubility. A few substances, such as calcium hydroxide and sodium carbenicillin, are exothermic and give off heat in the process of dissolution. The solubility of such substances would decrease with an increase in temperature. The application of this aspect of solubility is of limited use to us, since pharmaceutical solutions must be administered at or near room or body temperature. It is more a factor to be considered for product storage than for formulation.
FE 106 GENERAL CHEMISTRY
Experiment No : 9
Experiment Name : TOTAL HARDNESS WİTH
Submitted by : Mutlu DEMIREL
Group No : 2
Group Members : Sevgi COLAKOGLU
Determine the hardness in a water sample.
The quality of natural waters is important to the health and survival of all living things. A part of water quality deals with the hardness of water. Water hardness is defined as the concentration of dissolved divalent cations such as Ca2+ and Mg2+. The contributing factors to water hardness are municipal and individual waste, chemicals, fertilizers, herbicides, and pesticides that leach into the groundwater supply. Another factor is the fact that water is the universal solvent, which means it dissolves minerals as it percolates through the soil and rocks.1 Different types of rock give up Ca2+ and Mg2+ ions. Sedimentary rocks such as limestone (calcium carbonate CaCO3) are soluble in acidic water. Calcium also comes from the dissolution of igneous rock revealed by the decomposition of anorthite (CaAl2Si2O8), which occurs in the higher acidity of groundwater. An example of magnesium from the dissolution of rock is the decomposition of fosterite (Mg2SiO4) into serpentite (Mg6(OH)8Si4O10).1
There are no national standards for water hardness published by the Environmental Protection Agency (EPA), because the minerals, calcium and magnesium, are not considered health threats. Levels of water hardness are based the amount of calcium carbonate found in the water sample if the water were evaporated. The water is considered soft if it contains 0 to 60 mg/L of hardness, moderately hard from 61 to 120 mg/L, hard between 121 and 180 mg/L, and very hard if it contains if more than 180 mg/L.2 Commonly accepted level range for calcium is 100-300mg/L and the levels for magnesium are lower, based on how the water tastes.
There are various methods to measure water hardness. These include EDTA titration, Atomic Absorption Spectrometry, TDS (total dissolved solids) test, fluorescence fibre-optic sensor, and test strips.
The EDTA titration process uses EDTA (ethylenadiaminetetracetic acid) as the chelating agent.1 The EDTA attaches itself to the minerals with a bond called a chelate. The EDTA molecule forms complexes with nearly all metals. It forms particularly strong complexes, because it has six points of attachment that allow the molecule to wrap itself around the metal ion. Eriochrome black T (EBT) is the indicator dye used in EDTA titration, which is water soluble due to the sulfonate group’s (-SO3-) negative charge and also a pH indicator, so a buffer is added to make the blue form (HD2-) of EBT predominant. EBT forms a colored chelate with Mg2+, but not Ca2+.1 The initial wine red color is caused by the dye attaching itself to the metal ions. The endpoint of the titration is when all of the ions are coated by the EDTA and there are none left to complex with the dye, so the red color is replaced with sky blue.6 The values gained from the EDTA titration after some calulations are moles of Ca2+ and Mg2+, which can be converted into parts per a million (ppm) hardness of CaCO3.
Applications of EDTA chetation of Ca2+ and Mg2+ include bathroom cleaners in the form of tetrasodium salt because it dissolves lime and scum deposites. CaEDTA (calcium chelate) removes iron/copper that would spoil oil so it is used as a preservative in salad dressings.1 The calcium chelate is also effective in medical treatment for lead toxicity and blood vessel disease.5 EDTA is used as a titrant in industry because it is inexpensive; however, the downsides of it are that it is time consuming, requires a certain level of skill, and is subject to operational errors.
Atomic Absorbtion Spectrometry (AAS) is an instrumental technique that is used to determine the amount of trace elements, metals, dissolved in a solution. AAS is used in modern analytical chemistry, biochemistry, and ecology for inorganic trace analysis. It is used because of its high sensitivity, selectivity, broad scope, low cost, and reliability. Atoms have unique electronic energy levels. The light falling on the atom must match the energy separation between two electronic energy levels to be excited. In other words, absorption will occur if the wavelength of the light is the same as the change of energy.1 The atomic absorbtion spectrophotometer is based on the principle of excited atoms. A monochromatic light that has the energy correponding to the delta energy of the atoms is projected through the sample being analyzed.1 The wavelength used for calcium is 422.7 nanometers (nm), 202.5 nm for magnesium.8 The Beer’s law, A=abc, is used to calculate the metal concentration in the sample because absorbance is proportional to the concentration of the atoms in the sample.1
There are three main chambers of the atomic absorption spectrophotometer. The first chamber is where the source of the light is contained in cathode lamps.1 The fuels / oxidents used for analyzing calcium and magnesium are C2H2 and air.8 The monochromatic light is produced when the atoms relax, which is the opposite of absorption. The light is then absorbed by the Mg or Ca in the water sample in the second chamber. The second chamber is where the water sample enters the process by being aspirated into the sample chamber where it is converted into an aerosol, only about 10% of which is introduced to the flame from the burner. The 2300°C flame is capable of atomizing all of the components of the water sample.1 The light is absorbed by only the Mg or Ca atoms which produces a rich yellow flame.8 The third chamber of the atomic absorption spectrophotometer contains a monochrometer, a detector, and an internal computing system. Only the wavelength of the light corresponding to delta energy, which was produced from the excited metal emissions can pass through the narrow slit for the monochrometer. The detector is a photomultiplier tube (PMT), which detects a decrease in the initial signal from the lamp. The intensity transmitted (It) from the lamp through the flame is regulated by the Beer-Lambert law : It= I0(10-abc). The value given by the internal computing system is the light absorbance, which can be converted to concentration of the metal ion in ppm by the use of a linear regression graph produced from the calibration curve.1
The dissolved total solids test is a visual observation that compares the amount of total dissolved solids in a water sample to that in distilled water. It compares the ring of residue left on aluminium foil when the water is evaporated. The disadvantages of this are that it is only an observation and includes all of the dissolved minerals, not just Mg and Ca.
Water sample, conical flask, , Erioch,
1. We took 100 mL of sample into conical flask and add tree drops of phenol phatalein when the color is pink titrate the solution with 0.04 M and the recorded the used amount of acid.
2. Into same sample added three drops of methyl orange when the yellow titrate the sample with same acid until the color turns to red.
1. Total Hardness : The combined concentration of earth-alkali metals, predominantly magnesium (Mg 2+ ) and calcium (Ca 2+ ), and some strontium (Sr 2+ ). Most natural waters have a more or less high level of hardness. The source of this hardness is limestone dissolved by water rich in carbon dioxide. Total Hardness is expressed in mg/L of calcium carbonate (CaCO 3 ), though calcium carbonate is not water soluble, but calcium bicarbonate is. Hardness levels range from <50 mg/L (soft) to >500 mg/L (very hard).
Calcium Hardness : The calcium portion of total hardness expressed as calcium carbonate. Typically 65-75% of total hardness. Low calcium hardness can cause damage to pool surface and components.
Carbonate Hardness : The part of the total hardness that is formed by the ions of carbonates(Co3) and hydrogen carbonate(HCo3). It is symbolized by dCH. It is important to know the dCH of your water, as it affects both the ph. and Carbon Dioxide amounts in your water. It is also commonly called “buffering capability”. a dCH of 4 to 8 is fine for most fish.
Temporary Hardness : Hardness is produced by calcium and magnesium in water. These ions cause a precipitate with soap. Temporary hardness is caused by bicarbonate and can be removed by boiling which converts the bicarbonate to carbon dioxide and calcium carbonate.
Permanent Hardness : Permanent hardness can be removed by
· the addition of washing soda (sodium carbonate)
· ion exchange
· the use of polyphosphate water softeners
All the methods that remove permanent hardness also remove temporary hardness at the same time.
Alkalimetric Titer: the term can reflect the nature of the titrant, such as acidimetric, alkalimetric, iodimetric titrations as well as coulometric titrations, in which the titrant is generated electrolytically rather than being added as a standard solution.
German hardness: 1 °DH = 10 mg/L CaO = 0.178 mmole/L Ca (M(CaO) = 56 g/mole)
English hardness: 1 °EH = 1 grain/UK gallon CaCO3 = 0.142 mmole/L Ca (1 grain = 64.79891 mg; 1 UK gallon = 4.54596 L)
French hardness: 1 °FH = 10 mg/L CaCO3 = 0.1 mmole/L Ca (M(CaCO3) = 100 g/mole)
American hardness: 1 °AH = 1 ppm CaCO3 = 1mg/L CaCO3 = 0.01 mmole/L Ca
2. The following hardness classes can be distinguished
0-4 °DH (0-0.7 mmole/L Ca): very soft water
4-8 °DH (0.7-1.4 mmole/L Ca): soft water
8-12 °DH (1.4- 2.1 mmole/L Ca): normal water
12-18 °DH (2.1-3.2 mmole/L Ca): hard water
>18 °DH (>3.2 mmole/L Ca): very hard water
In The Netherlands the hardness of drinking water should be in the range of 1-2.5 mmole/L Ca.
Hard water is not unhealthy. Calcium en magnesium are essential minerals for our bones.
3. Water is used a lot of process in food industry. So water is very importatnt for food industry.
Effects of hard water :
a) cause the heat loses ( energy loses )
b) reduce the cleaning ability of soap
c) reduce the foaming ability of soap
d) changes the water boiling scala
4. 1. We took 100 mL of sample into conical flask and add tree drops of phenol phatalein when the color is pink titrate the solution with 0.04 M and the recorded the used amount of acid.
2. Into same sample added three drops of methyl orange when the yellow titrate the sample with same acid until the color turns to red.
Water is referred to as the “universal solvent” because, over the course of time, water will dissolve or erode almost any material that it is in contact with. It is this natural occurrence that contributes to the hardness of water.
Water hardness is normally referred to as a measure of the soap or detergent consuming power of water. Technically, hard water is water having a high concentration of calcium and magnesium ions. These, along with other minerals, are commonly present in all natural water.
Water is used a lot of process in food industry. So water is very importatnt for food industry.
Effects of hard water :
a) cause the heat loses ( energy loses )
b) reduce the cleaning ability of soap
c) reduce the foaming ability of soap
d) changes the water boiling scala
FE 111 GENERAL CHEMISTRY LAB SHEET
Experiment: 1 Density of liquids INTRODUCTION The density, mass per unit volume, of liquids can be determined like that of solids by measuring both the mass and the volume of a given sample. The density is a characteristic property of a substance; it remains fixed unless the temperature or pressure is changed. For liquids, a relatively small change in temperature can affect the density appreciably, but a pressure change must be quite great to have a measurable effect. As an intensive property the density is independent of the quantity of material measured. In this experiment, you will practice using a balance to measure mass. In addition, you will learn how to measure volume using a graduated cylinder and then you will determine the density of a) liquid water and b) a salt solution with different concentrations. Once the density of liquid water and its change with temperature are known, you can use the information to find out what volume should be occupied by a known mass of water at a given temperature. Once a data such as concentration vs. density of the salt solution is obtained, it is possible to plot a curve what we called a calibration curve. When such a plot is obtained we can determine the density of a substance with a known concentration, or the concentration of a substance with a known density. PROCEDURE A) – Weigh an empty graduated cylinder – Add about 20 mL of distilled water in this graduated cylinder. – Record the volume to the nearest 0.1 mL. – Weigh the graduated cylinder plus water on an analytical balance and record. – Measure the temperature of the water – Calculate the density. – Compare the calculated density of the water with the values given in the Table 1. and calculate the percent error.
B) – Rinse your graduated cylinder with about 5 mL of the 4 % NaCl solution, and then pour about 20 mL of the 4 % NaCl solution into the graduated cylinder. – Determine the density of this solution as in Part A. – Determine the density of 8, 12, and 16 % NaCl solutions similarly. – Obtain a salt solution of unknown concentration of NaCl from your instructor and determine its density. – Plot a graph showing the density (on the vertical axis) vs. concentration of NaCl (horizontal axis). – Using the graph, obtain a value for the concentration of NaCl for your unknown and report it to your instructor. Table 1. Temp°C Density(g/mL) Temp°C Density (g/mL) 15 0.9979 26 0.9959 17 0.9977 28 0.9955 18 0.9975 29 0.9952 20 0.9972 31 0.9946 22 0.9968 33 0.9941 24 0.9964 35 0.9935
PRELAB QUESTIONS 1. Write the definition of the density and a) explain the effect of the temperature and pressure on the density of the liquids.How does the density of liquids changes with the temperature? b) why do we need a great pressure change to observe a change in the density of the liquids? c) if we have a gas sample instead of liquids, do we need great pressure change to have a measurable change in the density? Explain briefly. d) if we have a solid sample instead of liquid or gas, does the density of this solid change if we increase or decrease pressure too much? 2. a) Suppose that, you measure the temperature of 20 mL of the water as 30°C when it is actually 20°C. Find the minimum percent error in the calculation of the density by using the table given above. b) Suppose that you determine the mass of 41.2 mL of water as 41.052 g. at 28°C. Calculate the density from the experimental values and match it with the value given in the table. Calculate the percent error. 3. Plot the temperature vs. density values given in the table. Determine the density of the liquid water at 19°C, 27°C, 30°C and 32°C. 4. What is the calibration? Explain briefly.
FE 111 GENERAL CHEMISTRY DATA SHEET Density of liquids Experiment : 1 Date: a) Mass of empty graduated cylinder : ……………….. Volume of water in the cylinder : ……………….. Mass of graduated cylinder plus water : ……………….. Temperature of water : ………………… b) NaCl solution 4 % 8 % 12 % 16 % ……….. ……….. ……….. ……….. Mass of grd.cyl.+ known NaCl soln. ……….. ……….. ……….. ………… Volume of NaCl in cylinder Mass of grd. cyl+ unknown soln : ………… Volume of unknown soln : ………… Name of the student: Submitted to :
FE 111 GENERAL CHEMISTRY LAB SHEET Experiment: 2 Heat and Temperature INTRODUCTION Heat differs from temperature in that heat is a quantity of energy whereas temperature is a measure of the hotness or coldness of an object. Indeed, temperature is a physical property that determines the direction of heat flow. Heat always transfers from hot objects to cold ones. For example, if you touch an ice cube, heat flows out of your hand to the ice, cooling your hand. The specific heat capacity is the amount of heat required to raise the temperature of 1g or 1mol of substance by o 1 C. In this experiment you will investigate a) the calibration of a thermometer as a temperature-measuring device b) the determination of the specific heat capacity of a given metal Table 2-1 Boiling point of water at various pressures Pressure mm-Hg atm o Bp, C 700 0.921 97.7 The thermometer can be calibrated by 705 0.928 97.9 o immersing it first in melting ice (0 C) and than in 710 0.934 98.1 o 715 0.941 98.3 boiling water (100 C at 1 atm pressure). Since 720 0.947 98.5 the boiling point of water varies with changing barometric pressure. Table 2-1 will enable you to 725 0.954 98.7 find the t rue boiling point of water under the 730 0.961 98.9 conditions of the experiment. To determine the 735 0.967 99.1 specific heat capacity of a metal, you will heat a 740 0.974 99.3 weighed amount of metal to a known 745 0.980 99.4 temperature (that of boiling water), and then 750 0.987 99.6 place the hot metal in a known amount of water. 755 0.993 99.8 From the temperature rise of the water, you can 760 1.000 100.0 calculate the amount of heat transferred from the 765 1.007 100.2 metal to the water. This can be done directly 770 1.013 100.4 because it takes 4.2 joules (or 1 calorie) to raise o 1 g of water by 1 C. Since the density of water is almost exactly 1 g per mL. 1 g of water is very nearly 1 mL. The specific heat capacity of the metal is equal to the amount of heat liberated by the the metal divided by its temperature drop times its mass.
PROCEDURE a) Wash off some crushed ice, and place it in a small beaker. Add distilled water until the ice is nearly covered. Immerse your thermometer in the ice-water mixture as deep as possible; stir gently; record the thermometer reading when it has become constant. Place about 100 mL of distilled water in your 500 mL flask. Drop in a boiling chip to prevent bumping during boiling. Insert your thermometer into the water. Heat the water to boiling and record the thermometer reading when it has become constant. Record the barometric pressure. b) Obtain a piece of metal (about 10 g) for which the specific heat is to be found. Weigh it to the nearest 0.1 g. Tie it with a string and suspend it above boiling water in a flask as the thermometer is set in Figure. Do not let the metal get wet. (If water condenses on the metal, pull the metal out, dry, and replace.) Allow sufficient time for the metal to reach the temperature of the vapor from boiling water. Record the barometric pressure. With a graduated cylinder, measure 10.0 mL of distilled water into a test tube. Prop the test tube in the mouth of your 300 mL flask. o Measure the temperature of the water to the nearest 0.1 . Withdraw the thermometer, and hold it directly over the test tube so that any drops of water will drain back. Quickly remove the heated metal, and carefully lower it into the water in the test tube. Stir by raising and lowering the metal. Return the thermometer, and note the maximum temperature registered. PRELAB QUESTIONS 1- Define Boiling point, Normal Boiling Point, Normal Freezing point 2- What is the difference between heat capacity and specific heat capacity ?
FE 111 GENERAL CHEMISTRY DATA SHEET Heat and Temperature Experiment : 2 Date: A) Thermometer reading in ice-water mixture………………. Thermometer reading at water boiling point………………. Barometric pressure………………. B) Weight of metal ……………. Temperature of the heated metal………………. Temperature of the water in the test tube………………. Temperature of the water and metal mixture………………. Name of the student: Submitted to :
FE 111 GENERAL CHEMISTRY LAB SHEET Experiment 3 Charles’s Law INTRODUCTION Charles’s law gives the relation between the volume and temperature of a gas at constant pressure. It states that the volume of a given mass of gas is directly proportional to the absolute temperature when pressure is constant. Mathematically we may state this in either of two ways: V = kT (P constant) or V /T = V /T (P constant) 1 1 2 2 In this experiment the student will heat the air inside a flask to the boiling temperature of water. Then he will cool the flask by placing it in a cold water bath. As the air inside the flask cools, its volume will decrease, and water will be drawn into the flask. The volume of this water is the difference between the initial and final air volumes. One correction must be made. The water drawn into the flask has a vapor pressure (about 20 mm). Therefore, the air in the flask at the end of the experiment will be mixed with water vapor. The pressure of air: Pair = Patm — PH 2O Now we can use Boyle’s law to calculate the volume the air would occupy if P = P . Let’s call this volume V . Then air atm 2 Patm V2 = (Patm – P 2 ) ( V1 – V 2 ) H O H O where V 2 is the volume of water drawn into the flask and V is the volume of the H O 1 flask. Now it should be possible to compare V /T (the volume and temperature of the 1 1 air at the boiling temperature of water) with V /T (where T is room temperature). 2 2 2 Since in both cases P = Patm, if Charles’s law is obeyed, we will find that V /T = V /T 1 1 2 2 PROCEDURE Take a clean 250 mL flask and make sure it contains no moisture by passing it back and forth through a (luminous) bunsen burner flame, until it is completely dry. (You have to hold the flask in your hand and heat the area that contains moisture). Take a one-holed rubber stopper that fits the flask and insert a 7-8 cm length of fire- polished glass tubing so that one end is even with the bottom of the stopper. Fit the stopper firmly into the flask and mark with a pencil where it comes in contact with the flask. Place the flask in a 1000 mL beaker and set up apparatus as shown below. The flask should be almost completely immersed in water.
Gently boil the water for about 10 minutes. Record the temperature of the boiling water (T 1) and the atmospheric pressure (Patm). Turn the burner off and cover the open end of the glass tube with your finger. Loosen the clamp and, keeping your finger over the glass tube, turn the flask upside down and immerse it in a room temperature water bath. Remove your finger and keep the flask completely submerged for 10 minutes. At the end of this time, raise the inverted flask until the water levels inside and outside are the same. Cover the end of the glass tube with your finger, remove the flask from the bath, and restore it to an upright position. Record the temperature of the water bath (T ). Pour the water from the flask into a 2 graduated cylinder and record its volume ( V 2 ). H O Fill the flask with water and place the rubber stopper in its original position (using the pencil mark). The glass tube should be completely filled with water, and no air bubbles should be trapped around the bottom of the stopper. Dry the outside of the stopper, then remove the stopper and carefully drain the contents of the glass tube into the flask. Pour the water from the flask into a graduated cylinder and record its volume (V ). 1 From the appendix of your textbook obtain the vapor pressure of water ( PH 2O ) at room temperature T . 2 CALCULATIONS Remember to convert T and T to the absolute temperature scale. Now calculate 1 2 V , as shown in the introductory discussion. Calculate and compare V /T and V /T . 2 1 1 2 2
QUESTIONS 1- Assuming your values of V , T , and T were correct, what is the theoretical value 1 1 2 of V ? (Calculate from Charles’s law). Calculate your percent error. 2 % error = difference between V (theoretically) and V (experimentally) x 100 2 2 V (theoretically) 2 2- How would the experimental results be affected if the flask contained moisture at the beginning of the experiment ? 3- Why was it necessary to match the water levels inside and outside the cooled flask before removing it from the water bath ?
FE 111 GENERAL CHEMISTRY DATA SHEET Charles’s Law Experiment : 3 Date: Temperature of the Boiling water (T ) : ……………… 1 Atmospheric Pressure (Patm ) : ………………. Temperature of Water Bath (T ) : ………………. 2 Volume of Water in the Flask ( V 2 ) : ………………. H O Vapor Pressure of Water at T ( P ) : ………………. 2 2 H O Volume of the Flask (V ) : ………………. 1 Name of the student: Submitted to :
FE 111 GENERAL CHEMISTRY LAB SHEET Experimet: 4 Chemical Equilibrium INTRODUCTION Chemical reactions occur so as to approach a state of chemical equilibrium. The equilibrium state can be characterized by specifying its equilibrium constant, i.e., by indicating the numerical value of the equilibrium constant expression. In this experiment you will determine the value of the equilibrium constant for the reaction 3+ – between ferric ion, Fe and isothiocyanate ion, SCN . 3+ – 2+ Fe + SCN ══ FeSCN for which the equilibrium condition is 2+ [FeSCN ] ———————- = K 3+ – [Fe ][SCN ] To find the value of K, it is necessary to determine the concentration of each of 3+ – 2+ the species Fe , SCN , and FeSCN in the system at equilibrium. This will be done 2+ colorimetrically, taking advantage of the fact that FeSCN is the only highly colored species in the solution. The color intensity of a solution depends on the concentration of the colored species and on the depth of solution viewed. Thus 2 cm of a 0.1 M solution of a colored species appears to have the same color intensity as 1 cm of a 0.2 M solution. Consequently, if the depths of two solutions of unequal concentrations are chosen so that the solutions appear equally colored, then the ratio of concentrations is simply the inverse of the ratio of the two depths ( h M = h M ). It should be noted that this 1 1 2 2 procedure permits only a comparison between concentrations. It does not give an absolute value of either one of the concentrations. To know absolute values it is necessary to compare with a standard of known concentration. 2+ For color determination of FeSCN concentration, you must have a standard 2+ solution in which the concentration of FeSCN is known. Such a solution can be – prepared by starting with a small known concentration of SCN and adding such a 3+ – 2+ large excess of Fe that essentially all the SCN is converted to FeSCN . Then, 2+ the concentration of FeSCN may be calculated from the equation:
PROCEDURE – Obtain five clean test tubes. Rinse with distilled water and let drain. – Give numbers to each test tube. – Add 5 mL of 0.0020 M NaSCN solution to each. – To the first test tube, which will serve as your standard, add 5 mL of 0.20 M Fe(NO ) . 3 3 – For the remaining four tubes proceed as described below. – Add 10 mL of 0.20 M Fe(NO ) to your graduated cylinder. 3 3 – Add 15 mL distilled water to make the volume exactly 25 mL. Stir this solution thoroughly. – Pour 5 mL of the this solution into the second test tube. – Pour out the half of the remaining 20 mL of the solution in graduated cylinder so that the volume will be exactly 10 mL. – Make the volume again 25 mL with distilled water and mix thoroughly. – Pour 5 mL of this solution into the third test tube. – Pour out half of the solution in the graduated cylinder until exactly 10 mL remain. – Add 15 mL of distilled water and mix thoroughly. – Pour 5 mL of this solution into the fourth test tube. – Again pour out half of the solution in the graduated cylinder until exactly 10 mL solution remain. – Add 15 mL of distilled water and mix thoroughly. – Pour 5 mL of this solution into the fifth test tube. 2+ The next step is to determine the relative FeSCN concentration in each test tube. To do this you will compare the color intensity in test tube 1 with that in each of other four test tubes. – Take test tube 1 and test tube 2. – Hold them side by side. – Wrap a piece of white paper around them.
– Look down through the solution towards the white background that your table makes. – If color intensities appear identical record this fact. – If not, pour out some of the standard in the test tube 1 until the colour intensities are the same. – Measure the heights of solutions in the two tubes being compared. – Repeat this comparison for the remaining three test tubes and record the heihts of both the standard solution and solution being compared. RESULTS AND CALCULATIONS 3+ – 1- Calculate the concentration of Fe and SCN in each of the five tubes, assuming 2 3+ – that no FeSCN + had been formed as [ Fe ] and [ SCN ] . ( Use M V = M V ). o o 1 1 2 2 – 2+ 2- Assume that all the SCN in the test tube 1 was converted to FeSCN . Then 2+ 2+ calculate the concentration of FeSCN , [ FeSCN ], in each of the other test tubes from (h ) M = (h ) M 1 1 2 2 3+ – 3- Calculate [ Fe ] and [ SCN ] from 3+ 3+ 2+ – – 2+ [ Fe ] = [ Fe ] – [ FeSCN ] and [ SCN ] = [ SCN ] – [ FeSCN ] o o 4- Determine the K value separately for each tube by using the equilibrium constant expression. PRELAB QUESTIONS 1- Define the following terms. Equilibrium Constant, Le Chatelier’s Principle, Chemical Equilibrium, Homogeneous Equilibria and Heterogeneous Equilibria 2- Explain the effect of concentration, temperature, pressure and catalyst according to Le Chatelier’s principle on chemical equilibrium. 3- Write a brief procedure for the experiment in your own words. QUESTIONS 1- What are possible sources of error in this experiment ? 2- Why are the values of K determined for test tubes 3, 4, and 5 probably more reliable than that determined for tube 2 ?