Revised May 19, 1998
Chemicals are applied to ponds and lakes to control weeds; to control fish diseases; to eliminate undesirable fish; to control undesirable insects and aquatic invertebrates and to correct undesirable water quality problems. Pond owners are often confused by terminology, units of measure, and formulations. This confusion makes it difficult to select the right chemical, to calculate the proper amount to be applied, and to apply it to the pond in a correct and safe manner.
Once you have accurately identified the problem, then select the most effective control measure. This does not mean that a chemical can or should be used to correct every pond management problem. The best approach is to consider preventive measures first. If they are not practical or do not produce the desired results, then other control methods should be considered. It is always easier and more economical to prevent a problem than to cure one. Even when preventive measures are only partially successful, they quite often facilitate the effectiveness of other control measures. Preventive measures may or may not include the use of chemicals.
Matching the management problem with an effective chemical is not enough. You must also consider the effect that chemicals may have on nontarget organisms. For example, some chemicals used to treat diseases in fish are also toxic to plants. Use of these chemicals during the summer months may cause oxygen depletion. Also, the water chemistry and its effect on the chemical may need to be considered. Some chemicals break down rapidly in the presence of sunlight, high pH, and high temperature and are less likely to be effective during the hot summer months. Be sure to consider other water uses and effects the chemical may have on them. For example, aquatic herbicides applied to a pond used for irrigation may have a disastrous effect upon the irrigated crops. Also, consider the effects the chemical may have downstream from your pond.
Whenever you use a chemical in a pond, it must be applied properly and all warnings and precautions concerning use must be understood and observed. Fortunately, all of this information is on the label for most chemicals approved for use in ponds. Anyone who uses a chemical in a pond should always thoroughly read and understand the chemical label before purchasing and applying it.
Obviously the effectiveness of some chemical treatments can be quite variable. If you are not certain of the identification of the aquatic pest or the best control method, consult your county Extension agent or state fisheries biologist. Assuming you have selected the most effective chemical for use, the following information should be used to determine the proper amount to apply and to determine the best and safest way to apply it.
An acrefoot is one surface acre one foot deep. Acrefeet are computed by multiplying the area (in acres) by the average depth (in feet). In the example above, eight surface acres times the average depth of four feet equals 32 acrefeet of water.
Most county Natural Resources Conservation Service offices can assist pond owners in determining the water volume of their ponds. The surface acreage of most ponds can also be determined by county Farm Service offices. Assuming the surface acreage of a pond is known, the following method can be used to determine the average depth of a pond. Average depth can be determined by use of a sounding line at regular intervals along several transects of the pond. Both deep and shallow area of the pond should be included in the transects.
Average depth is computed by adding all of the depth measurements and dividing by the number measurements. The average depth multiplied by the surface area should give an accurate estimate of the pond water volume.
Know the water volume of your pond before a treatment is needed. You can lose valuable time if the determination must be made after a problem has arisen. Table 1 can be used to convert acre feet into other measures of water volume.
Table 1. Equivalents of 1 acre foot of water
1 acrefoot  = 43,560 cubic feet
= 4,840 cubic yards = 326,000 gallons (approximately) = 2,780,000 pounds (approximately) 
For a particular chemical, the application rate is based upon the amount of active ingredient in the chemical formulation. Fortunately, the amount of active ingredients contained in the chemical formulation and the application rate are printed on most product labels. This is one reason why it is important to read the information printed on the label.
One part per million is equivalent to the ratio of one pound of chemical to 999,999 pounds of water or one gram of chemical to 999,999 grams of water. In other words, one part per million equals one pound or one gram in one million pounds or grams of a solution or mixture, respectively.
Notice that parts per million is a weighttoweight relation. Units of volume cannot be used directly. This is because an equal volume of two different chemicals may have considerably different weights. For example, one cubicfoot of lead weighs much more than one cubicfoot of water.
Amt of Chemical = Volume x CF x ECC x Al Needed
Where:
Volume = Volume of water to be treated. Although the unit of measure can be in gallons, liters, cubic feet, cubic yards, etc., when treating ponds, the more common and easier to use expression of volume is acrefeet.
CF = Conversion factor, a figure that equals the weight of a chemical to be used to give one part per million (ppm) in a given unit volume of water. Table 2 lists conversion factors (CF) for various measures of volume. For example, select the CF that corresponds to the unit of measure used for pond volume. For example, if the pond volume is measured in acrefeet, the appropriate CF is 2.72 if the chemical weight is measured in pounds or 1,233 if weight is measured in grams.
Table 2. Conversion Factors (CF)  Weight of Chemical in One Unit Volume of Water to Give One Part Per Million ppm.
2.72 pounds per acrefoot
1,233 grams per acrefoot 0.0283 grams per cubic foot 0.0000624 pounds per cubic foot 0.0038 grams per gallon 0.0584 grains per gallon 1 milligram per liter 0.001 gram per liter 8.34 pounds per million gallons of water 
= 1 ppm
= 1 ppm = 1 ppm = 1 ppm = 1 ppm = 1 ppm = 1 ppm = 1 ppm = 1 ppm 
Al = The total amount of active and inert ingredients divided by the amount of active ingredients. Products, which are liquid formulations, usually list the amount of active ingredients as pounds active per gallon. For such products Al = 1 gallon divided by the pounds per gallon of active ingredients. A few chemicals are liquids in their pure form and their specific gravity must be known to calculate Al. See Example 4 to calculate Al using specific gravity. Nonliquid formulations usually list active ingredients as a percentage of the total formulation. For nonliquid formulations, Al = 100% divided by the percentage of active ingredients.
The following examples illustrate how the equation previously mentioned can be used in calculating pond water treatments.
Example 1. How much Chemical A is needed to treat a pond that has
4 surface acres and an average depth of 3 feet with 2 ppm active ingredient?
Chemical A is 100% active.
Volume
CF

= 4 acres x 3 feet
= 12 acrefeet = 2.72 pounds (from Table 2) = 2 ppm (active ingredient needed in the water) = 100% 
100% (Chemical A is 100% active) 
Volume x CF x ECC x Al
Thus: (12 acrefeet x 2.72 pounds x 2 ppm x 100) / 100 = 65.3 pounds of Chemical A are needed to treat the pond.
Example 2. How much Chemical B (80 percent active) is needed to treat a pond measuring 1,000 feet long by 500 feet wide by 5 feet deep with a concentration of 0.25 ppm active ingredient?
Volume
CF

= 100 feet x 50 feet x 5 feet
= 25,000 cubic feet = 0.0000624 pounds/cubic foot (from Table 2) = 0.25 ppm (active ingredient needed in the water) = 100% 
80% 
Volume x CF x ECC x Al
Thus: (25,000 cu. ft. x 0.0000624 pounds/cu.ft. x 0.25 ppm x 100) / 80 = 0.49 pounds of Chemical B (80 percent) are needed to treat the pond.
Example 3. How much Chemical C (2 pounds active per gallon) is needed to treat a pond that has 6 surface acres and an average depth of 4 feet with 0.5 ppm active ingredient?
Volume
CF

= 6 acres x 4 feet
= 24 acrefeet = 2.72 pounds/acrefoot (From Table 2) = 0.5 ppm (active ingredient needed in water) = 1 gal. 
2 lbs. 
Volume x CF x ECC x Al
Thus: (24 acrefeet x 2.72 pounds/acrefoot x 0.5 ppmx 1 gal) / 2 lbs = 16.3 gallons of Chemical C (2 lbs active/gallon) are needed to treat the pond.
Example 4. How much Chemical D is needed to treat a pond measuring 180 feet long by 90 feet wide by 4 feet deep with a concentration of 25 ppm active ingredient. Chemical D is a liquid and is 100 percent active.
Volume
CF

= 180 feet x 90 feet x 5 feet
= 81, 000 cubic feet = 0.0000624 pounds per cubic foot = 25 ppm = 100% 
100% 
Volume x CF x ECC x Al
Thus: (81,000 cu. ft. x 0.0000624 pounds/cu.ft. x 25 ppm x 100) / 100 =126.4 pounds of Chemical D
However, Chemical D is a liquid and 126.4 pounds must be converted to a unit of volume. Since (ppm) parts per million is a weighttoweight relation, it is necessary to know how Chemical D compares in weight with water. Chemical D is heavier than water, thus a smaller amount of Chemical D is needed to equal 250 ppm in water on a Chemical D to water weighttoweight ratio. Chemical D weighs about 9 pounds per gallon and water 8.34 pounds per gallon; or Chemical D is 1.08 times as heavy as water (9 divided by 8.34). This figure is called the specific gravity (SG) of Chemical D. If the weight of Chemical D is computed in grams, the weight divided by the specific gravity equals the number of cubic centimeters required. If the weight (as in this example = 126.4 pounds) is computed in pounds, divide by 8.34 times the specific gravity to convert it to gallons. In this example the amount of Chemical D needed is:
(126.4 pounds) / (8.34 Ibs/gal x 1.08 SG) = 140 gallons
From  To  
CM^{3}  liter  M^{3}  IN^{3}  ft^{3}  fl.oz.  fl. pt.  fl.qt.  gal.  
CM^{3}
liter M^{3} IN^{3} ft^{3} fl. oz. fl. pt. fl.qt. gal. 
1
1000 1x10^{6} 16.39 2.83x10^{4} 29.57 473.2 946.2 3785 
0.001^{3}
1 1000^{3} 0.0164 ^{3} 28.32 0.0296 0.4732 0.9463 3.785 
1x10^{6}
0.001 1 1.64x10^{5} 0.0283 2.96 x 10 ^{5 } 4.73x10^{4} 9.46x10^{4 } 0.0038 
0.0610
60.98 6.1x10^{4} 1 1728 1.805 28.88 57.75 231.0 
3.53x10^{5}
0.353 35.31 5.79x10^{4} 1 0.00104 0.0167 0.0334 0.1337 
0.0338
33.81 3.38x10^{4} 0.5541 957.5 1 16 32 128 
0.00211
2.113 2113 0.0346 59.84 0.0625 1 2 8 
0.00106
1.057 1057 0.0173 29.92 0.0313 0.5 1 4 
2.64x10^{4}
0.2642 264.2 0.0043 7.481 0.0078 0.125 0.25 1 
From  To  
cm  m  in.  ft.  yd  
cm
m in. ft. yd. 
1
100 2.54 30.48 91.44 
0.01
1 0.0254 0.3048 0.9144 
0.3937
39.37 1 12 36 
0.0328
3.281 0.0833 1 3 
0.0109
1.0936 0.0278 0.3333 1 
From  To  
g  kg  gr.  oz.  lb.  
g
kg gr. oz. lb. 
1
1000 0.0648 28.35 453 6 
0.001
1 6.48 x 10^{5} 0.0284 0.4536 
15.43
1.54 x 10^{5} 1 437.5 7000 
0.0353
35.27 0.0023 1 6 
0.0022
2.205 1.43 x 10 ^{4} 0.0625 1 
1 acrefoot
1 acrefoot 1 acrefoot of water 1 cubicfoot of water 1 gallon of water 1 gallon of water 1 liter of water 1 fluid ounce 1 fluid ounce 
43,560
325,580 2,718,144 62.4 8.34 3,785 1,000 29.57 1,043 
cubic feet
gallons pounds pounds pounds grams grams grams ounces 
cm
cm^{3} fl oz fl pt fl qt ft ft^{3} gal g gr in in^{3} kg lb m m^{3} oz yd 
=
=^{3} = = = = =^{3} = = = = =^{3} = = = =^{3} = = 
centimeter
cubic centimeter^{3} fluid ounce fluid pint fluid quart foot cubic foot ^{3} gallon gram grain inch cubic inch ^{3} kilogram pound meter cubic meter^{3} ounce yard 
WWellborn T.L. 1978 Calculation of Treatment Levels for Control of Fish Diseases and Aquatic Weeds. Information Sheet 673, Mississippi State University, Cooperative Extension Service.
Wellborn, T.L. 1979. Control and Therapy in Principal Diseases of FarmRaised Catfish. Southern Cooperative Series No. 225. Southern Regional Research Project S83. Pages 61  89.