Are you researching on experiments on ecology ? You are in the right place. The below mentioned article includes a collection of nineteen experiments on ecology: 1. Community Structure Study 2. Biomass Study 3. Soil Science 4. Aquatic Ecosystem 5. Physico-Chemical Analysis of Water.
Contents:
- Experiments on Community Structure Study
- Experiments on Biomass Study
- Experiments on Soil Science
- Experiments on Aquatic Ecosystem
- Experiments on Physico-Chemical Analysis of Water
1. Experiment on Community Structure Study: (8 Experiments)
1. Aim of the Experiment:
To determine the minimum size of the quadrat by species area-curve method.
Requirements:
Nails, cord or string, metre scale, hammer, pencil, notebook.
Method:
i. Prepare a L-shaped structure of 1 × 1 metre size in the given area by using 3 nails and tying them with a cord or string.
ii. Measure 10 cm on one side of the arm L and the same on the other side of L, and prepare 10 x 10 sq. cm area using another set of nails and string. Note the number of species in this area of 10 x 10 sq. cm.
iii. Increase this area to 20 × 20 sq. cm and note the additional species growing in this area.
iv. Repeat the same procedure for 30 × 30 sq. cm, 40 × 40 sq. cm and so on till 1 × 1 sq. metre area is covered (Fig. 67) and note the number of additional species every time.
Record your data in the following table:
v. Prepare a graph using the data recorded in the above table. Size of the quadrats is plotted on X- axis and the number of species on Y-axis (Fig. 67 B).
Observations:
The curve starts flattening or shows only a steady increase (Fig. 67 B) at one point in the graph.
Results:
The point of the graph, at which the curve starts flattening or shows only a steady or gradual increase, indicates the minimum size or minimum area of the quadrat suitable for study.
2. Aim of the Experiment:
To study communities by quadrat method and to determine % Frequency, Density and Abundance.
Requirements:
Metre scale, string, four nails or quadrat, notebook.
(i) Frequency:
Frequency is the number of sampling units or quadrats in which a given species occurs.
Percentage frequency (%F) can be estimated by the following formula:
(ii) Density:
Density is the number of individuals per unit area and can be calculated by the following formula:
(iii) Abundance:
Abundance is described as the number of individuals per quadrat of occurrence.
Abundance for each species can be calculated by the following formula:
Method:
Lay a quadrat (Fig. 68) in the field or specific area to be studied. Note carefully the plants occurring there. Write the names and number of individuals of plant species in the note-book, which are present in the limits of your quadrat. Lay at random at least 10 quadrats (Fig. 69) in the same way and record your data in the form of Table 4.1.
In Table 4.1, % frequency, density and abundance of Cyperus have been determined. Readings of the other six plants, occurred in the quadrats studied, are also filled in the table. Calculate the frequency, density and abundance of these six plants for practice. (For the practical class take your own readings. The readings in Table 4.1 are only to give an explanation of the matter).
Observations:
See Table 4.1.
Results:
Calculate the frequency, density and abundance of all the plant species with the help of the formulae given earlier and note the following results:
(i) In terms of % Frequency (F), the field is being dominated by…
(ii) In terms of Density (D), the field is being dominated by…
(iii) In terms of Abundance (A), the field is being dominated by…
Observations:
Table 4.1: Size of quadrat: 50cm × 50cm = 2500 cm2
3. Aim of the Experiment:
To determine minimum number of quadrats required for reliable estimate of biomass in grasslands.
Requirements:
Metre scale, string, four nails (or quadrat), note book, graph paper, herbarium sheet, cello tape.
Method:
i. Lay down 20-50 quadrats of definite size at random in the grassland to be studied, make a list of different plant species (e.g., A-J) present in each quadrat and note down their botanical names or hypothetic numbers (e.g., A, B, C,…, J) as shown in Table 42. u
ii. With the help of the data available in Table 4.2, find out the accumulating total of the number of species for each quadrat.
iii. Now take a graph paper sheet and plot the number of quadrats on X-axis and the accumulating total number of species on Y-axis of the graph paper.
Observations and results:
A curve would be obtained. Note carefully that this curve also starts flattening. The point at which this curve starts flattening up would give us the minimum number of quadrats required to be laid down in the grassland.
4. Aim of the Experiment:
To study frequency of herbaceous species in grassland and to compare the frequency distribution with Raunkiaer’s standard frequency diagram.
Requirements:
Quadrat, pencil, note-book, graph paper.
Method:
i. Lay 10 quadrats in the given area and calculate the percentage frequency of different plant species by the method and formula given above in Exercise No. 2.
ii. Arrange your data in the form of following Table 4.3:
Raunkiaer (1934) classified the species in a community into following five classes as shown in Table 4.4:
Arrange percentage frequency of different species of the above Table 4.3 in the five frequency classes (A-E) as formulated by Raunkiaer (1934) in Table 4.4.
Draw a histogram (Fig. 70) with the percentage of total number of species plotted on Y-axis and the frequency classes (A-E) on X-axis.
This is the frequency diagram (Fig. 70):
Observations and results:
The histogram takes a “J- shaped” curve as suggested by Raunkiaer (1934), and this shows the normal distribution of frequency percentage. If the vegetation in the area is uniform, class ‘E’ is always larger than class ‘D’. And in case class ‘E’ is smaller than class ‘D, the community or vegetation in the area shows considerable disturbance.
5. Aim of the Experiment:
To estimate Importance Value Index for grassland species on the basis of relative frequency, relative density and relative dominance in protected and grazed grassland.
Requirements:
Wooden quadrat of 1×1 metre, pencil, notebook.
What is Importance Value Index?
The Importance Value Index (IVI) shows the complete or overall picture of ecological importance of the species in a community. Community structure study is made by studying frequency, density, abundance and basal cover of species. But these data do not provide an overall picture of importance of a species, e.g., frequency gives us an idea about dispersion of a species in the area but does not give any idea about its number or the area covered.
Density gives the numerical strength and nothing about the spread or cover. A total picture of the ecological importance of a species in a community is obtained by IVI. For finding IVI, the percentage values of relative frequency, relative density and relative dominance are added together, and this value out of 300 is called Importance Value Index or IVI of a species.
Relative frequency (RF) of a species is calculated by the following formula:
Relative density (RD) of a species is calculated by the following formula:
Relative dominance of a species is calculated by the following formula:
Basal area of a plant species is calculated by the following formula:
Basals area of a species = p r2
where p = 3.142, and r = radius of the stem
Method:
i. Find out the values of relative frequency, relative density and relative dominance by the above-mentioned formulae.
ii. Calculate the IVI by adding these three values:
IVI = relative frequency + relative density + relative dominance.
Results:
Arrange the species in order of decreasing importance, i.e., the species having highest IVI is of most ecological importance and the one having the lowest IVI is of least ecological importance.
6. Aim of the Experiment:
To determine the basal cover, or vegetational cover of one herbaceous community by quadrat method.
Requirements:
Wooden quadrat of 1×1 m, Verniercalliper, pencil, notebook.
Method:
i. Lay a wooden-framed quadrat of 1 x 1 metre randomly in a selected plot of vegetation and count the total number of individuals of the selected species inside the quadrat.
ii. Cut a few stems of some plants of this individual species and measure the diameter of the stem with the help of Verniercalliper.
iii. Calculate the basal area of the individuals by the formula:
Average basal area = π r2 where r is the radius of the stem.
iv. Take 5 readings, arrange them in tabular form and find out the average basal area by the above formula.
v. Lay the quadrat again randomly at another place and note the same observations in the table.
vi. Lay about 10 quadrats in the same fashion and each time note the total number of the species and average basal area of the single individual.
Observations and results:
(a) For finding the average basal area, divide the sum of average basal area in all quadrats with the total number of quadrats studied.
(b) For finding the total basal cover of a particular species multiply the average basal area of all observations with the density of that particular species as under:
Basal cover of a particular species = Average basal area x Density (D) of that species.
The basal cover of a particular species is expressed in… sq. cm/sq. metre.
7. Aim of the Experiment:
To measure the vegetation cover of grassland through point-frame method.
Requirements:
Point-frame apparatus, graph paper sheet, herbarium sheet, cello tape, note-book.
Point-frame apparatus:
A point-frame apparatus is a simple wooden frame of about 50 cm long and 50 cm high in which 10 movable pins are inserted at 45° angle. Each movable pin is about 50 cm long.
Method:
i. Put the point-frame apparatus (with 10 pins) at a place in the vegetation of grassland (Fig. 71) and note down various plant species hit by one or more of 10 pins of the apparatus. Treat this as one sampling unit.
ii. Now put the apparatus at random at 10-25 or more places and note down each time the various plants species in a similar fashion. In case three plants of any species touch three pins in one sampling unit put at a place, the numerical strength of that particular species in this sampling unit will be three individuals. Write this value against the species below this sampling unit.
Observations and results:
Note down the details in the form of following Table 4.5:
Now calculate the percentage frequency of each species as already done in Exercise No. 2. Allocate the various species among five frequency classes (A, B, C, D, E) mentioned in Exercise No. 4, find out the percentage value of each frequency class and prepare a frequency diagram as done in Exercise No. 4. Compare the thus-developed frequency diagram with normal frequency diagram.
Results:
Find out the three most frequently occurring species in the area studied. Also find out whether the vegetation is homogeneous or heterogeneous. Also try to determine the density values of individual species. Also find out at each place the total number of individuals of each species being hit by 10 pins of the point-frame apparatus.
8. Aim of the Experiment:
To prepare a list of plants occurring in a grassland and also to prepare chart along the line transect.
Requirements:
Two nails, 25 feet cord.
Method:
i. Prepare a 25 feet long line transect in a selected grassland by tying each end of a 25 feet cord to the upper knobs of two nails.
ii. Note down the names of the plant species whose projection touches one edge of the cord along the line transect, and assign all of them a definite number (e.g., 1,2,3,4, …etc.).
iii. Take several such samples at regular or irregular intervals in the grassland along the line transect.
iv. Also record the plant species from different grassland types in the similar fashion.
Observations:
Record your data in the following Table 4.6 in the form of the following manner:
Result:
Table 4.6 gives the complete list of plants occurring in the selected grassland. Also find out the name of the species represented in maximum number in each locality.
These data will also provide a clear picture of the dominant species of the grassland in a particular area.
2. Experiment on Biomass Study: (2 Experiments)
What is Biomass?
It is usually expressed as dry weight of all the living materials (plants as w ell as animals) in an area. Under biomass we include plants (their aboveground and underground parts) as well as animals. Fallen leaves and twigs of the plants are also taken in consideration at the time of studying biomass.
In the forests, the humus is in different stages of decomposition. The floor of the forest remains covered by organic matter which is slightly or not at all decomposed. This is called litter. A partially decomposed matter is present below this layer. It is called duff. Further decomposed matter, which has lost its original form, is present below duff and called leaf mould.
9. Aim of the Experiment:
To measure the above-ground plant biomass in a grassland.
Or
To determine the biomass of a particular area.
Requirements:
Nails (4), metre scale, string, khurpa (a weeding instrument), polythene bags, oven.
Method:
i. Make a quadrat of the size of 50 cm × 50 cm in the field by digging the nails and connecting them with the string. Weed out all the above-ground parts of the plants growing in that limit with the help of weeding instrument. Collect all of them in a polythene bag.
ii. Collect the fallen leaves and other parts of the plants in the second polythene bag.
iii. Collect all the animals such as ants, larvae, earthworms, insects, etc., in the third polythene bag.
iv. By digging the soil to about 20 to 25 cm., take out all the underground parts of the plants and collect them in a separate bag after washing.
In the same way lay some more quadrats in the area under study and collect all the materials in polythene bags.
Calculations:
Suppose:
Dry weight of aboveground parts = 15 gm.
Dry weight of underground parts = 4 gm.
Dry weight of animals = 1 gm.
... Total dry weight = 20 gm.
50 × 50 cm field area contains = 20 gm. total dry biomass
...100 × 100 cm field area will contain
20×100×100/50×50 = 80gm
Results:
80 gm. is the biomass of 100 × 100 cm. field area.
Results of Different Parts:
(i) 50 x 50 cm. field area contains 15 gm. of aboveground parts.
100 × 100 cm. field area will contain
= 15×100×100/50×50 = 60 gm. biomass.
(ii) 50 x 50 cm. field area contains 4 gm. of underground parts.
100 × 100 cm. field area will contain
= 4×100×100/50×50 = 16 gm. biomass
(iii) 50 × 50 cm. field area contains 1 gm. of animals
100 × 100 cm. field area will contain
1×100×100/50×50 = 4 gm. biomass.
Results:
One square metre (100×100 cm.) field area contains 80 gm. biomass in terms of dry weight of the total plant and animal parts.
10. Aim of the Experiment:
To determine diversity indices (richness, Simpson, Shannon-Wiener) in grazed and protected grassland.
Or,
To study species diversity (richness and evenness), Index of dominance, Similarity index, Dissimilarity index and Species diversity index in grazed and protected grassland.
Requirements:
Khurpa, note book, pen.
Method and results:
(a) Species diversity:
Species diversity is a statistical abstraction with two components.
These two components are:
(i) Richness (or number of species), and
(ii) Evenness or equitability.
In any grassland, to be studied, if there are seventy species in a stand, then its richness is seventy. Pick out individual plants of different species with the help of khurpa, count the number of species in a stand of the area provided, and calculate the richness. On the other hand, if all the species in the grassland have equal number of individuals, then its evenness or equitability is high and if some species have only a few individuals then the evenness is low.
(b) Index of dominance:
The species which have strongest control over energy flow and environment in given habitat are called ecological dominant. According to Simpson (1949), the Index of dominance (C) is calculated by the formula
C= ∑ (ni/N)2
where∑ (sigma) refers to summation, ni refers to the importance value of the species in terms of number of individuals or biomass or productivity of each species over a unit area, and N refers to the total of corresponding importance values of all the component species in the same unit area and period. Count the Index of Dominance by the above-mentioned formula.
(c) Similarity index:
Similarity Index between two stands of vegetation can be worked out by the formula S = 2 C/(A+B), where S is the Similarity Index, C is the number of species common to both the stands, and A and B are number of species on stand A and stand B. For example, if there are 20 species on site A and 20 on site B and 14 species are common in both sites, the Similarity Index (S) will be;
S = 28/40 or S = 0.7.
(d) Dissimilarity index:
The Dissimilarity Index is counted by the formula D = 1 – S, where D is the dissimilarity index and S is the similarity index. For example, if there are 20 species on site A and 20 species on site B and 14 species are common in both sites the similarity index (S) comes to 0.7 as calculated above in case of similarity index. Therefore, dissimilarity index (D) can be counted, as
D= 1-S or D= 1-0.7 = 0.3.
(e) Diversity index:
Species diversity index (d) is calculated by the following formula given by Menhinick (1964):
d=S/√N
where d = diversity index, S = number of species, and N= number of individuals of that particular species.
3. Experiment on Soil Science: (1 Experiment)
11. Aim of the Experiment:
To study the characteristics of different types of soils.
Requirements:
Samples of different types of soils (e.g., clay soil, sand or alluvial soil, humus, black soil, yellow soil, red soil, laterite or lateritic soil).
Method and observations:
Different soil samples are taken and studied individually.
Some of their major characteristics are under mentioned:
I. Clay soil:
i. It is a compact and heavy-textured soil.
ii. The size of its particles is less than 0.002 mm.
iii. It has very minute spaces in between its particles.
iv. It is quite sticky when wet but becomes hard and develops cracks on drying.
v. It has higher water-holding capacity and poor aeration.
vi. It gets waterlogged easily.
vii. Its particles are negatively charged and have the ability to absorb cations of Mg, Ca, K, P, Fe and Na.
viii. This soil is made up of hydrated alumino silicate.
ix. It is quite rich in calcium carbonate and magnesium carbonate.
x. The pore space between its particles is greater than sand.
xi. This soil has high degree of fertility.
Identification:
On the basis of above characteristics, the given sample belongs to clay soil.
II. Fine sand or alluvial soil:
i. This soil is loose, light-textured and silver-grey in colour.
ii. The size of its particles is between 0.02 mm to 0.2 mm.
iii. It has poor water-holding capacity.
iv. This soil shows quite rapid rate of water infiltration.
v. It is poor in humus.
vi. The carbonate content of this soil is very low.
vii. It does not get waterlogged easily.
viii. It shows good aeration.
ix. It is non-sticky and non-plastic when wet.
x. It has very low contents of phosphate, nitrogen and organic matter.
xi. It has shiny particles of aluminium silicate or mica.
xii. Some amounts of iron, magnesium, sodium, aluminium, silicon and calcium are present in this soil.
xiii. Its particles become warm on long exposure to the sun.
Identification:
The above characteristics show that the given soil sample belongs to fine sand or alluvial soil.
III. Humus:
i. It is decomposed matter of plant and animal remains.
ii. This organic matter is amorphous and dark brown to black in colour.
iii. It is soluble in dilute alkali solution like KOH and NaOH but insoluble in water.
iv. It is actually a layer of organic matter at the top of a soil profile. It is the habitat of most decomposers. The main decomposers are bacteria and fungi.
v. It is made up of nitrogen-rich proteins, lignin and polysaccharides.
vi. A large amount of carbon and small amounts of sulphur, phosphorus and some other elements are also present.
vii. It is colloidal in nature.
Identification:
On the basis of the above characteristics, it can be concluded that the given material is humus.
IV. Black soil:
i. This is black-coloured soil. The black colour is due to the presence of iron in this soil.
ii. High percentage of iron oxides, calcium carbonates, magnesium carbonates and alumina are present in this soil.
iii. It also contains large amount of nitrogen and organic matter. It, however, contains very low amount of phosphorus.
iv. If made wet by adding some water, this soil is sticky. On drying, it contracts and shows cracks.
v. It has high water-retaining capacity.
vi. It is highly productive and suits most for the crops like cotton.
Identification:
Based on the above-mentioned characteristics, it can be concluded that the given sample belongs to black soil.
V. Yellow soil:
i. The yellow colour of this soil is due to the enhanced hydration of ferric oxide.
ii. It is a porous soil with nearly neutral pH.
iii. Size of the particles of this soil is between 0.002 mm to 0.02 mm.
iv. It is a granite-derived soil with moderately rich humus.
v. It contains very low amount of oxides of phosphorus, nitrogen and potassium.
vi. It contains large amount of silicon oxide and alminosilicate.
Identification:
The above characteristics show that this is a sample of yellow soil.
VI. Red soil:
i. This is the sample of red-coloured soil.
ii. The red colour is due to the diffusion of large amount of iron compounds such as ferrous oxide and ferric oxide.
iii. It is a slightly acidic type of soil. Its pH varies between 5 and 8.
iv. Some amount of silicon oxide and aluminium oxide are also present in this soil.
v. This soil is not good for agriculture because it is poor in nitrogen, phosphorus and humus.
Identification:
Because of the above characteristics, the given sample is of red soil.
VII. Laterite soil:
i. This is yellowish or red-coloured soil. On exposure to sun it turns black.
ii. It is produced from aluminium-rich rocks.
iii. It is quite compact type of soil made up of hydrated oxides of iron and aluminium.
iv. It also contains small amounts of compounds like magnesium oxide and titanium oxide.
v. Small amounts of nitrogen, phosphorus, magnesium, potash and lime are also present in this soil.
vi. It is also quite rich in humus.
vii. Because of the above characteristics, this soil is good for the purpose of agriculture.
Identification:
Above-mentioned characteristics show that the given sample is of laterite or lateritic soil.
4. Experiment on Aquatic Ecosystem: (1 Experiment)
What is ecosystem?
Living organisms are structurally and functionally inter-related with the external world or the environment, and this functional and structural relationship of communities and the environment is called ecological system or ecosystem.
Ecosystem normally contains:
(i) Biotic components and
(ii) Abiotic components.
Taking in view the organisms and their habitat conditions, the ecosystem can be classified as follows:
Pond ecosystem can be studied as follows:
12. Aim of the Experiment:
To study the biotic components of a pond. Make diagram of a pond ecosystem.
Requirements:
Hand lens, collection net, meshes of different sizes, collection tubes, iron hook, scissor, forceps, and centrifuge.
Method:
Biotic components of a pond can be studied exactly according to the classification of a pond ecosystem given above. Hydrophytes can be picked by hand and collected in polythene bags. Other submerged plants may also be taken out by iron hooks (Fig. 74).
Phytoplankton and zooplankton can be collected in plankton bottles.
With the help of plankton nets, microorganisms can be collected in tubes.
Macro-producers and macro-consumers can be estimated in gm./cubic metre by the quadrat method used in the exercise of biomass.
Micro-producers and micro-consumers can be estimated in gm./litre of water collected as sample from an undisturbed part of the pond. They can be separated by centrifuging a little amount of pond water (containing micro-producers and micro-consumers) in test tube.
Observations:
On the basis of their trophic position in the ecosystem different organisms may be grouped as follows:
(a) Producers:
(i) Submerged:
Vallisneria, Ceratophyllum, Hydrilla, Potamogeton, Chara, etc.
(ii) Free-floating:
Azolla, Eichhornia, Lemna, Pistia, Spirodella, Salvinia, etc.
(iii) Rooted floating:
Trapa, Jussiaea, Nymphaea, Potamogeton, Nelumbium, etc.
(iv) Rooted Amphibious:
Marsilea, Typha, Ranunculus, Polygonum, Cyperus, etc.
(v) Phytoplankton:
Agal members of Chlorophyceae, Xanthophyceae, Bacillariophyceae, Myxophyceae, etc.
(b) Consumers:
(i) Consumers of the 1st order (Primary consumers):
e.g., Zooplankton, some insects.
(ii) Consumers of the 2nd order (Secondary consumers):
e.g., Fishes, frogs and some insects.
(iii) Consumers of the 3rd order (Tertiary consumers):
e.g., Birds, big fishes, etc.
(c) Decomposers:
Fungi and bacteria.
5. Experiment on Physico-Chemical Analysis of Water: (7 Experiments)
13. Aim of the Experiment:
To measure temperature and pH of different water bodies.
Requirements:
Maximum-minimum thermometer or thermometer or thermo flask.
Method:
(i) Temperature:
The temperature of the pond can be determined by any of the following apparatuses:
(a) Maximum-minimum thermometer:
It contains two indicators (Fig. 73). With the help of a magnet these indicators are set to the present atmospheric temperature. Quickly lower down the thermometer to the desired depth in the pond. Keep it there for 10 minutes.
Bring out the thermometer quickly and note the readings of both the indicators. Out of the two indicators, one remains at the point and other moves to some extent giving the reading of temperature at that particular depth of the pond.
(b) Thermistor:
It is an instrument which gives correct reading of temperature in centigrade. It contains a long cable. At the end of the cable is attached a thermocouple (Fig. 75).
A milliammeter is present which is calibrated in C° and gives direct reading. Quickly lower the thermocouple upto a desired depth and note the temperature.
(c) Thermo flask:
It is also one of the good apparatuses for measuring the temperature of a pond. After lowering to the desired depth, bring it out when it is filled completely with water. With the help of a good sensitive thermometer, note the temperature of the water.
(ii) pH:
pH of the pond water can be tested by pH meter, pH paper or B.D.H. Universal Indicator.
14. Aim of the Experiment:
To determine transparency or turbidity of different water bodies.
Transparency (clarity of pond):
It is directly related to and mainly depends upon the presence of microorganisms and microscopic soil particles in the pond water. If the quantity of soil particles and microscopic organisms will be more, the pond water will be less transparent. It also depends upon the depth of the pond water. The turbidity value will be very low in the deep water.
The instrument for knowing the turbidity value is called Secchi disc (Fig. 76). It is a circular disc with black and white or other contrasting colours. The disc is lowered down in the water. Note the depth of the water where there is no colour contrast on the disc.
15. Aim of the Experiment:
To find out the light intensity available to pond.
Light intensity:
Light intensity available to the pond is measured with the help of ‘photometer’ (Fig. 77).
A ‘photometer’ consists of a photoelectric cell and a micro-ammeter. Photometer for pond is specially sealed in water-tight containers fitted with a glass window. Photoelectric cell is sensitive to light and generates current when light falls on it. Light intensity is proportional to the current generated in the photoelectric cell by the light falling on it. Readings can be noted in micro-ammeter.
Calculate the light intensity by the following formula:
Light Intensity = r × 100 / a
where r – Reading of lux-meter or photometer
a = Reflected light from the cardboard.
16. Aim of the Experiment:
To measure amount of dissolved oxygen content in polluted and unpolluted water bodies.
Or
To measure amount of dissolved oxygen in pond water.
Requirements:
Water sample, glass stoppered conical flask, manganous-sulphate, potassium iodide solution (alkaline), pipettes, sulphuric acid (conc.), sodium thiosulphate solution, starch solution, reagent bottles.
Preparation of reagents:
(a) Starch solution:
Add 1 gm. starch in 100 ml distilled water, warm and dissolve it.
(b) Potassium iodide solution (alkaline):
Heat 200 ml of distilled water and dissolve in it 100 gm. KOH and 50 gm. KI.
(c) Manganous sulphate solution:
Add 200 gmMnSO4 . 4H2O in 200 ml distilled water. Heat it to dissolve maximum salt. Cool it and then filter it.
(d) Sodium shiosulphate solution:
Dissolve 24.82 gmNa2SO4 . 5H2O in some amount of distilled water and make up the volume to 1 litre by adding more distilled water. To stabilize the solution add a small pellet of sodium hydroxide. Thus prepared solution is 0.1N stock solution. In 250 ml of this stock solution add 750 ml of distilled water to dilute the solution of sodium thiosulphate.
Method:
i. Take 100 ml of water sample in a 250 ml glass-stoppered conical flask and add 1 ml of manganous sulphate solution and 1 ml alkaline potassium iodide solution by separate pipettes. Appearance of brown precipitate indicates the presence of oxygen in the water sample.
ii. Shake it well and then allow the precipitate to settle down.
iii. Add 2 ml sulphuric acid (conc.) and again shake it well. The precipitate will be dissolved.
iv. Decant the liquid and titrate it with sodium thiosulphate solution. Starch solution is used as an indicator. The blue black colour disappears when the end point is reached.
Calculations and results:
Dissolved oxygen in mg/ liter is calculated by the following formula:
Dissolved O2 (mg/l)
= (ml × N of sodium thiosulphate) × 8 ×1000/V1 – V2
where V1 = Volume of water sample titrated,
V2 = Volume of MnSO4 and KI solution added.
17. Aim of the Experiment:
To determine the total dissolved solids (TDS) in water.
Requirements:
Water, sample, evaporating dish, Whatman filter paper, oven, desiccator, balance, beakers.
Method:
i. Weigh a dry and clean evaporating dish of200 ml capacity.
ii. Shake the water sample well and filter it through Whatman filter paper.
iii. Take 100 ml of filtrate in a pre-weighed evaporating dish and keep it in an oven at 180°C for some time. The water will be evaporated and the sample will become dry.
iv. Cool it in a desiccator and weigh. Calculate the total dissolved solids using following formula:
Total Dissolved Solids (mg/1) = (a-b) × 10/V
where
a = Weight of dish and dried filtered sample (in gm.)
b = Weight of empty evaporating dish (in gm).
V= Volume of sample evaporated (in ml).
18. Aim of the Experiment:
To count phytoplankton by haemocytometer method.
Requirements:
Haemocytometer, water sample, cover slip, microscope, dropper.
Method:
Haemocytometer is a special type of glass slide having more than 500 small grooved chambers or counting chambers (1 × 1 × 0.5 mm) in the middle portion. This specially designed slide is used for counting the microorganisms or plankton present in a water drop.
i. Take a haemocytometer and put a drop of concentrated water sample on its counting chambers.
ii. Put a cover slip, wait for about 2-5 minutes and examine under the high power of microscope. Count the plankton present in each chamber.
19. Aim of the Experiment:
To determine plankton biomass of a pond.
Requirements:
Pond water, shallow water bottle, chemical balance, oven, beaker, funnel, Whatman filter paper.
Method:
i. Collect 1000 ml of surface water of pond with the help of a shallow water bottle. This water contains phytoplankton and zooplankton.
ii. Weigh a dry filter paper. Suppose it is A1 gm.
iii. Take another filter paper. Make it wet and weigh it. Suppose it is A2 gm.
iv. Filter the water sample through a Whatman filter paper and weigh this filter paper containing plankton. Suppose it is A3 gm.
v. Now put this plankton-containing filter paper in oven for 24 hours at 85°C. Weigh this dry filter paper with plankton. Suppose it is A4 gm.
Calculations and result:
Calculate the biomass (fresh weight or dry weight of organisms) in mg/litre as follows:
(i) Fresh weight of plankton/1000 ml = A3– A2gm.
(ii) Fresh weight of plankton/ml = A3 – A2gm/1000
(iii) Dry weight of plankton/1000 ml = A4– A1 gm.
(iv) Dry weight of plankton/ml = A4 – A1/1000 gm.