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Other Questions

Test Methods:  Frequently Asked Questions

pH of Organic Liquids

Question: Is Method 9045C (pH of solids) applicable to organic liquid waste matrices?

Answer: Section 1.1 of Method 9045C states that the pH of non-aqueous liquids can be determined by the procedure. Water should be added to the non-aqueous liquid and mixed, because pH cannot be determined without the presence of water. The non-aqueous liquid should be separated from the water [aqueous phase] and the water analyzed. The water will contain compounds that leached out of the non-aqueous liquid waste so that the waste is not directly measured. Liquid organic wastes, such as oily wastes, may damage the pH electrode if the electrode is exposed directly to the oil.

Note: This does not mean that you must use Method 9045C to evaluate non-aqueous liquids to see if they exhibit the RCRA characteristic of corrosivity (40 CFR 261.22).

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Total Organic Carbon Method for Soils

Question: Where in SW-846 can I find a method for total organic carbon (TOC) in soil and other solid matrices?

Answer: There is no SW-846 method for TOC in soils. EPA does not have any regulatory requirements that govern the total organic carbon content of soils or solid matrices. Therefore, without a regulatory driver, there is no need for a method for TOC in soil in SW-846. Soil scientists have developed a variety of methods for TOC over the years. Some of them are based on titration of "easily oxidizable organic carbon," while others involve the high temperature combustion of any organic carbon in the soil and measurement of the resulting carbon dioxide. If you need a method for TOC in soil, you should either consult the open soil science literature or check with the environmental regulatory agency in your state to see if they have a method of choice.

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Equation for Surface Area in Method 1110

Question: The equation for the surface area of the stainless steel coupon in Method 1110 is not clear to us. We have received widely differing answers from different laboratories for the analysis of the same waste and we believe it is because they calculated the area differently.

Answer: The apparent cause of the confusion is due, in part, to the age of Method 1110, for corrosivity towards steel. It was written at a time when word processing programs did not have the ability to display equations as clearly as they do at present. However, going back to first principles, the equation is used to determine the surface area of the coupon. The area of a circle is determined from the familiar product of the radius of the circle, squared, and the constant Pi. Since the radius (r) is half of the diameter (d) of the circle, the area can expressed in the three ways shown below.

Area = II (r) 2 = II (d/2) 2 = II (d 2/4)

The total surface area of the coupon is the sum of the areas of the two faces of the coupon, plus the area around the outside edge, plus the area around the edge of the hole in the center of the coupon. Sec. 4.5 of Method 1110 contains an equation that illustrates adding these areas together to determine the total surface area.

Some laboratories are apparently misreading the equation in Sec. 4.5 of Method 1110 and placing the two diameters in the first term of the expression in the denominator of the equation, not the numerator. Using the mathematical rules for the order of precedence of the operations of addition, subtraction, multiplication, and division, the equation shown in the method is correct. However, it is shown below in a format that is less subject to misunderstanding:

A = [3.14 (D 2 - d 2)]/2  + (t)(3.14)(D)  + (t)(3.14)(d)


t = thickness
D = diameter of the coupon
d = diameter of the mounting hole
and 3.14 is a simple approximation of the constant Pi.

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