CE 510 Hazardous Waste Engineering

CE 510Hazardous Waste Engineering Department of Civil Engineering Southern Illinois University Carbondale Instructors: Jemil Yesuf and Dr. L.R. Chevalier Lecture Series 3: Hazardous Waste Properties and Classification

Course Goals • Review the history and impact of environmental laws in the United States • Understand the terminology, nomenclature, and significance of properties of hazardous wastes and hazardous materials • Develop strategies to find physical and chemical properties, i.e., water solubility, density, flammability, and toxicity for hazardous compounds • Elucidate procedures for describing, assessing, and sampling hazardous wastes at industrial facilities and contaminated sites • Predict the behavior of hazardous chemicals in surface impoundments, soils, groundwater and treatment systems • Assess the toxicity and risk associated with exposure to hazardous chemicals • Apply scientific principles and process designs of hazardous wastes management, remediation and treatment

Concentration - mg/L to ppm In environmental engineering, we are sometimes dealing with dilute aqueous solutions. In such a case, we assume that the substance does not change the density of water. Based on this assumption, we can relate mg/L to ppm.

Concentration - mg/L to ppm From the density of water, we know that 10-3 ml of water weighs 1 mg.

Concentration - mg/L to ppm Substituting the volume of water for the mass of water in 1 mg/L we obtain the following: THEREFORE, IN OUR DILUTE AQUEOUS SOLUTIONS WE COMMONLY INTERCHANGE mg/L WITH ppm

Class Example A 500-L tank contains 255 mg/L 2,4,6-trichlorophenol dissolved in hexane. What is the trichlorophenol concentration in ppm? Assume the temperature of the system is 20°C. Specific gravity of Hexane is 0.6603 (Appendix C)

Solution Mass of trichlorophenol dissolved in hexane is: (255 mg/L)(500L) = 127,500 mg = 0.128 kg Mass of hexane as a solvent is: (500 L)(0.6603)(1000kg/1000L) = 330 kg Therefore, concentration of trichlorophenol in hexane is: (127,500 mg)/(330 kg) = 386 mg/kg = 386 ppm ......end of example

Concentration - air • In dilute aqueous solutions, ppm is a mass-to-mass ratio • In air, ppm is a volume-to-volume ratio • Changes in temperature and pressure do not change the volume of the pollutant gas to the volume of the air that contains it • i.e. we can compare ppm from Washington D.C. to Denver without further conversion • We will define the Ideal Gas Law to further understand this…

Ideal Gas Law

Example How much volume in liters will one mole of any ideal gas occupy at standard temperature and pressure (STP). Note: STP is 273.13°K and 101.325 kPa (0°C and 1 atm).

Solution Use the ideal gas law to solve for volume. Note: J = N. m Pa = N/m2 ......end of example

Solution Similarly, if we consider the volume at 25°C ......end of example

Example The chemical para-dichlorobenzene (p-DCB) is used in an enclosed area. At 20C (68F) the saturated vapor pressure of 1,4-DCB is 5.3 x 10-4 atm. What would be the concentration in the air of the enclosed area (units of g/m3) at 20C ?

Solution Rearrange the Ideal Gas Law to solve for the concentration of 1,4-DCB in the air

Example Anaerobic microorganisms metabolize organic matter to carbon dioxide and methane gases. Estimate the volume of gas produced (at atmospheric pressure and 25° C) from the anaerobic decomposition of 2 moles of glucose. The reaction is:

Solution Each mole of glucose produces 3 moles of methane and 3 moles of carbon dioxide gases, for a total of 6 moles. Therefore, 2 moles of glucose produces a total of 12 moles. The entire volume is then Note: The volume of 1 mole of any gas is the same. Thus, 1 mole of carbon dioxide gas is the same volume of 1 mole of methane gas.

Concentration - Air Correction factor for conditions other than STP Incorporate this factor in order to report as ppm

Concentration - Air Based on these two equations, we can convert mg/L (g/m3) to ppm using M is the molecular weight in g/mole Concentration is g/m3

Example Convert 80 mg/m3 of SO2 in 1 m3 of air, 25° C, 101.325 kPa to ppm

Solution

Water Solubility • the maximum concentration of a substance that will dissolve in water at equilibrium at a given temperature and pressure • Controls • Concentration of a chemical in groundwater • Proportion that exists as a free product • Partitioning into solids • Inversely related to sorptivity, bioaccumulation, and volatilization • Affects a number of other pathways such as biodegradation, photolysis, chemical oxidation • Together with another chemical parameter, Kow, determine the fate of chemicals in the environment

Equilibrium Most chemical reactions are to some extent reversible when the rates of reaction are the same, i.e. the products are formed on the left at the same rate at which they are formed on the right. In such a case, the reaction is said to have reached equilibrium.

Generalized Reversible Reaction

Generalized Reversible Reaction Here, K is the equilibrium constant Provides the ratio of the concentration of individual reactants and products for any reaction at equilibrium

Use this equation with a degree of caution • Valid only when chemical equilibrium is established • Natural systems are often subject to constantly changing inputs • Some reactions occur very slowly • Equilibrium may never be established

Important Equilibrium Processes in Environmental Systems • Acids and bases • Dissolved and precipitated chemicals • Pure compounds and air (volatilization) • Chemicals dissolved in water and air (Henry’s Law) • Chemicals dissolved in water and adsorbed on a solid (adsorption and ion exchange)

Acid-Base Chemistry • Among the most important in environmental engineering • Waste must be neutralized • Aquatic life sensitive to changes • The concentration of acid/base is given by the pH • By controlling pH, unwanted substances may be driven out of solution as gases or precipitates

Acid-Base Chemistry We’ll evaluate this equation further later. First, lets consider the dissociation of water and the definition of pH

Water dissociates slightly intoHydrogen Ions (H+)Hydroxide Ions (OH-) Acid-Base Chemistry

Acid-Base Chemistry pH= -log [H+] [H+] = 10-pH

Example What percentage of total ammonia (i.e. NH3 + NH4) is present as NH3 at a pH of 7? The pKa for NH4+ is 9.3.

Solution The problem is asking:

Solution The problem is asking: However, this expression has two unknowns. Therefore, we need a second equation. ?????

Solution The problem is asking:

Solution Recall, pH=7 means [H] = 10-7

Solution Therefore:

Solution

Acid-Base Chemistry Relationship between pH and pKa Consider a generalized acid reaction

Acid-Base Chemistry Take the logarithm Henderson-Hasselbalch equation

Example How is pH related to pKa when the acid is 50% ionized? (i.e. [A-] = [HA])

Solution ......end of example

Solubility Product Solid Chemical  Dissolved Chemical Equilibrium Constant Notation: Solubility Product (Ksp)

Solubility Product • All complexes are soluble in water to a certain extent • Likewise, all complexes are limited by how much can be dissolved in water • Example • NaCl very soluble • AgCl only a small amount will go into solution

Solubility Product • Visualize a solid compound being placed in distilled water • Some of the compound will go into solution • At some time, no more of the compound will dissolve • At this point, equilibrium is reached • may take seconds • may take centuries

Solubility Product Consider the following The solid form, NaCl, may be dissociating into its ionic components (dissolution) OR The ionic components may be recombining into the solid form (precipitation)

The conditions of equilibrium can be expressed in an equilibrium or mass action equation for a generalized reaction. Solubility Product Solid Compound Ionic Components

Solubility Product • The brackets indicate molar concentrations • K is an equilibrium constant for a given substance in pure water at a given temperature • At equilibrium, the solid phase does not change concentration because dissolution and precipitation are equal.

Solubility Product • Supersaturated Solution • Created by dissolving a solid at an elevated temperature and allowing it to cool • Once cooled, precipitation may not occur, although the solution is not at equilibrium • Precipitation will occur if the reaction vessel is shaken or otherwise disturbed

Solubility Product • Unsaturated Solution • Not at equilibrium • Can dissolve more solid • Saturated Solution • At equilibrium • Cannot dissolve more solid unless the temperature or pressure is increased

CE 510 Hazardous Waste Engineering