Development Team

Paper No: 16 Environmental Chemistry Module: 01 Environmental Concentration Units

Development Team

Prof. R.K. Kohli Principal Investigator & Prof. V.K. Garg & Prof. Ashok Dhawan Co- Principal Investigator Central University of Punjab, Bathinda

Prof. K.S. Gupta Paper Coordinator University of Rajasthan, Jaipur Prof. K.S. Gupta Content Writer University of Rajasthan, Jaipur Content Reviewer Dr. V.K. Garg Central University of Punjab, Bathinda

Anchor Institute Central University of Punjab 1

Environmental Chemistry Environmental Sciences Environmental Concentration Units

Description of Module

Subject Name Environmental Sciences

Paper Name Environmental Chemistry

Module Name/Title Environmental Concentration Units

Module Id EVS/EC-XVI/01

Pre-requisites A basic knowledge of concentration units

1. To define exponents, prefixes and symbols based on SI units 2. To define molarity and molality 3. To define number density and mixing ratio 4. To define parts –per notation by volume Objectives 5. To define parts-per notation by mass by mass 6. To define mass by volume unit for trace gases in air 7. To define mass by volume unit for aqueous media 8. To convert one unit into another

Keywords Environmental concentrations, parts- per notations, ppm, ppb, ppt, partial pressure

Environmental Chemistry Environmental Sciences Environmental Concentration Units

Module 1: Environmental Concentration Units

Contents 1. Introduction 2. Exponents 3. Environmental Concentration Units 4. Molarity, mol/L 5. Molality, mol/kg 6. Number Density (n) 7. Mixing Ratio 8. Parts-Per Notation by Volume 9. ppmv, ppbv and pptv 10. Parts-Per Notation by Mass by Mass. 11. Mass by Volume Unit for Trace Gases in Air: Microgram per Cubic Meter, µg/m3 12. Conversion from µg m-3 to ppbv 13. Mass by Volume Unit for Aqueous Media: mg per liter or ppm 14. Conversion from One Unit to Another 15. Pressure, Partial Pressure and Units 16. Suggesting Reading

Introduction

A large number of chemical species are found in the environment. To express their amounts, it is necessary to specify the unit of measurement. In general, SI units (Système international d'unités) are used, although some other units may also be found occasionally. SI unit for length is meter, m, for mass is kilogram, kg, and for time is second, s. These basic units are utilized to obtain derived units for

Environmental Chemistry Environmental Sciences Environmental Concentration Units

physical parameters. It must be pointed out that environmental chemistry is concerned mainly with air and aquatic chemistry and therefore the discussion on units is concerned mainly with these systems.

Exponents

The amounts of different substances present in environment differ by several orders of magnitude, say from 0.00000001 to 1,00,00,00,000, i. e., from 1x10-8 to 1x 108. For the sake of brevity, ease in writing and to avoid writing several zeros, the exponents, as given in Table 1, are used. One may come across the use of billion, million etc., hence following equalities may be remembered: tera = trillion; giga = billion; mega = million; kilo = thousand, hecto = hundred and deca = ten. Environmental chemistry deals mainly with trace amounts of contaminants and pollutants in air and aqueous systems, therefore, the entire discussion is based on this...

Table 1. Exponents, prefixes and symbols based on SI units.

Multiple Prefix Symbol Multiple Prefix Symbol

1012 tera T 10-2 centi c

109 giga G 10-3 milli m

106 mega M 10-6 micro µ

103 kilo k 10-9 nano n

102 hecto h 10-12 pico p

10 deca da 10-15 femto f

10-1 deci d 10-18 atto a

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Environmental Concentration Units

The concentrations of the gaseous substances can be expressed using absolute and relative scales. An example of the former are molarity, molality and number density and of the latter are mixing ratio and mole fraction.

Molarity, mol/L The most preferred unit of expressing concentrations of the substances in chemistry is molarity. It is defined as the number of moles of a solute dissolved in one liter of the solution. Its unit is mol/L or mol L-1. It is defined by Eq. 1.

molarity = mass of solute in g/(M. W. of solute × volume of solution in L) (1)

Problem 1.. A 100 mL solution was prepared by dissolving 0.5845 g NaCl (M. W. = 58.45 g/mol) in water. Calculate molarity of NaCl.

Solution. Convert volume100 mL in to 0.1 L and then the molarity of NaCl using Eq. 1. is:

molarity of NaCl = 0.5845 g/58.45 g mol-1x 100 mL = 0.5845 g/ 58.45 g mol-1x 0.1L = 0.1 mol L-1.

Molality, mol/kg It is defined as the number of moles of a solute dissolved in one kg of the solvent. Its unit is mol/kg as in Eq 2. molality = mass of solute in g/(M. W. x mass of solvent in kg) (2) It does not depend on temperature.

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Problem 2. For preparing 0.01 molal solution of NaCl, how many grams of NaCl should be dissolved in 500 g of water?

Solution. After converting 500 g of water in to 0.5 kg, use Eq. 2 as follows:

0.01 mol kg-1= mass of NaCl in g /(58.45 g mol-1 x 0.5 kg) mass of NaCl = 0.01 mol kg-1x 58.45 g mol -1x 0.5 kg = 0.29225 g

Number Density (n)

The number density, n, is defined as the number of molecules per unit volume as in Eq. 1.

nx = total number of molecules of X / total volume of air (3)

where nx is the number density of gaseous substance X. The unit of number density is molecules /cm3 or molecules cm-3. Number density is widely used in measuring reaction rates and optical properties of atmosphere. While reporting number density, it is necessary to report environmental conditions such as temperature and pressure because the value of volume depends on these parameters and so the number density will also depend on these parameters.

-6 6 Problem 3. Then number of CO2 molecules in 4 x 10 L air is 2x10 . Calculate the number density of

CO2.

Solution. The number density, nCO2, can be calculated as follows:

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nCO2 = total number of molecules of CO2/Total volume of air. 6 -6 nCO2 = 2x10 / 4 x 10 11 3 nCO2 = 5 x10 molecules/ cm

Mixing Ratio

Mixing ratio, Cx, is equal to the number of moles per mole of air as in Eq. 2.

Cx = total number of moles of X/ total number of moles of air (4)

The mixing ratio is a dimensionless quantity because the amounts of X and air are in the same unit of mole, which cancels out.

The environmental science, in general, deals with trace amounts of pollutants in gas phase and with trace impurities in aqueous systems. It is, therefore, preferred to express the concentrations of trace substances in the form of parts-per notation, which can be expressed as volume by volume (V/V) or mass by mass (m/m). ppm, ppb and ppt are calculated in volume-per-volume ratio and so the correct representation is ppmv, ppbv and pptv. In an environment if the amount CO be 1 ppmv, it means 1mL of CO is present in 1 million mL of air.

Parts-Per Notation by Volume

The concentrations of gaseous substances in air determined as volume by volume ratio should be expressed as parts per million by volume (ppmv), parts per billion by volume (ppbv) and parts per trillion by volume (pptv). Unfortunately, most often ‘v’ is dropped and the values are written as ppm, ppb and ppt.

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ppmv, ppbv and pptv

One ppm means one unit volume of the trace substance to be present in one million volume. For example if in air reported CO be 2 ppmv, it means 2mL CO is present in one million mL of air, or 2mL CO is present in one m3 of air. ppbv and pptv have similar meaning. According to parts-per notation the values of ppmv, ppbv and pptv can be expressed as in Eqs.5-7.

푛푢푚푏푒푟 표푓 푝푎푟푡푠 표푓 푎 표푛푠푡𝑖푡푢푒푛푡 푝푝푚푣 = ×106 (5) 푡표푡푎푙 푛푢푚푏푒푟 표푓 푝푎푟푡푠 표푓 푎푙푙 푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

푛푢푚푏푒푟 표푓 푝푎푟푡푠 표푓 푎 푐표푛푠푡𝑖푡푢푒푛푡 푝푝푏푣 = ×109 (6) 푡표푡푎푙 푛푢푚푏푒푟 표푓 푝푎푟푡푠 표푓 푎푙푙푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

푛푢푚푏푒푟 표푓 푝푎푟푡푠 표푓 푎 푐표푛푠푡𝑖푡푢푒푛푡 푝푝푡푣 = ×1012 (7) 푡표푡푎푙 푛푢푚푏푒푟 표푓 푝푎푟푡푠 표푓 푎푙푙 푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

The quantities in numerator and denominator must be in the same unit and hence values of ppmv, ppbv and pptv are dimensionless.

The amounts of the components are expressed in several units, which all are proportional to number of moles, for example: number of moles of a gaseous constituent are proportional to its volume, number of molecules of a gaseous constituent are proportional to number of its moles and number of moles of a gaseous constituent are proportional to its partial pressure

Based on these considerations, the formulas in Eq. 5 become:

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푛푢푚푏푒푟 표푓 푚표푙푒푠 표푓 푎 푐표푛푠푡𝑖푡푢푒푛푡 푝푝푚푣 = ×106 (8) 푡표푡푎푙 푛푢푚푏푒푟 표푓 푚표푙푒푠 표푓 푎푙푙 푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

푣표푙푢푚푒 표푓 푎 푐표푛푠푡𝑖푡푢푒푛푡 푝푝푚푣 = ×106 (9) 푡표푡푎푙 푣표푙푢푚푒 표푓 푝푎푟푡푠 표푓 푎푙푙 푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

푛푢푚푏푒푟 표푓 푚표푙푒푐푢푙푒푠 표푓 푎 푐표푛푠푡𝑖푡푢푒푛푡 푝푝푚푣 = ×106 (10) 푡표푡푎푙 푛푢푚푏푒푟 표푓 푚표푙푒푐푢푙푒푠 표푓 푎푙푙 푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

푝푎푟푡𝑖푎푙 푝푟푒푠푠푢푟푒 표푓 푎 푐표푛푠푡𝑖푡푢푒푛푡 푝푝푚푣 = ×106 (11) 푡표푡푎푙 푝푟푒푠푠푢푟푒 표푓 푝푎푟푡푠 표푓 푎푙푙 푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

Likewise, the equations for ppbv and pptv can be written based on Eqs. 6 and 7.

Mixing ratios are independent of pressure and temperature.

Problem 4. Ten molecules of NO2 were found in 100 million molecules of air. What is the concentration of NO2 in ppbv?

Solution. The NO2 concentration I ppbv is given by following Eq.

푛푢푚푏푒푟 표푓 푚표푙푒푐푢푙푒푠 표푓 푎 푁푂2 푵푶ퟐ 풊풏 푝푝푚푣 = ×109 푛푢푚푏푒푟 표푓 푚표푙푒푐푢푙푒푠 표푓 푎𝑖푟푠

10 10 푵푶ퟐ 풊풏 푝푝푚푣 = ×109 = ×109 = 100 ppbv 100 푚𝑖푙푙𝑖표푛 100푥10 6

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Parts-Per Notation by Mass by Mass.

The concentration of gaseous components can be expressed in mass by mass (or weight by weight) units also. For example parts per million mass by mass, ppmm(or ppmw), is defined in Eq.10.

푚푎푠푠 표푓 푎 표푛푠푡𝑖푡푢푒푛푡 푝푝푚푚 = ×106 (12) 푡표푡푎푙 푚푎푠푠 표푓 푚표푙푒푠 표푓 푎푙푙 푡ℎ푒 푐표푛푠푡𝑖푡푢푒푛푡푠

ppbm and pptm may be defined similarly.

Mass by mass values are independent of temperature.

Problem. 5.2 ng ammonia is present 1.0 mole of air. Calculate the amount of ammonia in ppmm. Assume molecular weight of air to be 28.9 g/mol.

Solution. NH3 in ppmm can be calculated by Eq. 12.

6 NH3 in ppmm = mass of NH3×10 / total mass of one mole of air (13)

6 -9 6 -4 NH3 in ppmm = 3.2 ng ×10 / 28.9 = 3.2×10 ×10 / 28.9 = 1.107×10

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Mass by Volume Unit for Trace Gases in Air: Microgram per Cubic Meter, µg/m3

In air, pollutants are found in small amounts only. And a frequently used unit to express such low concentration is µg/m3. Volume of air depends on temperature, while pressure remains almost unchanged. It is necessary to mention environmental conditions while reporting values in µg/m3.

-6 Problem 5. In an air analysis experiment, NO2 was found to be 1×10 g/ L of air. Calculate the 3 concentration of NO2 in µg/m .

-3 Solution. NO2 in µg m is given by Eq.:

-3 3 NO2 in µg m = (mass of NO2 in µg)/ (volume of air in m )

Since 1x10-6 g is equal to 1.0 µg and 1.0 L volume is equal to 10-3 m3 so on using these values in above Eq. we get, -3 -3 3 3 -3 NO2 in µg m = (1.0 µg)/ (10 m ) = 1×10 µg m

Conversion from µg m-3 to ppbv

Such conversions become necessary, when it is required to compare the data given in µg m-3 and ppbv. The procedure is described in the following problem.

-3 o Problem 6. Convert 32.05 µg m SO2 present in air in to ppbv at 1.0 atm pressure and 27 C.

Solution. (i) Convert 32.05 µg SO2 into moles: -6 -7 SO2 in moles = 32.05 µg/M. W. of SO2 = 32.05×10 /64.1 = 5×10 mol -5 (ii) Convert 5x10 mol into number of SO2 molecules.

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Number of SO2 molecules = number of mole × Avogadrow’s No. = 5x10-7× 6.02×1023 = 3.1× 1017 (iii) Convert 1.0 m3 air into number of moles using ideal gas Eq.: PV = n RT Number of moles of air, n = PV/RT = 1.0 atm × 1.0 m3/0.082×300K = 1.0 ×1.0x103/ 0.082 mol-1×300 = 40.65 mol (iv) Convert moles of air into number of molecules of air. Number of molecules of air = number of mole × Avogadrow’s No = 40.65 × 6.02×1023 = 244.7×1023 = 2.44x1025

(v) From data in (ii) and (iii) calculate SO2 in ppbv using Eq. 10. 17 9 25 SO2 in ppbv = 3.1× 10 × 10 /2.44×10 = 12.7ppb

Mass by Volume Unit for Aqueous Media: mg per liter or ppm

In aqueous systems, ppm is a preferred and widely unit for representing concentrations of contaminants. In environmental aqueous systems, the concentrations of contaminants are generally and the solution can be treated as dilute. In dilute solutions, the concentration(s) of the dissolved solute(s) is(are) so small that the density of dilute solution can be taken as the density of pure water. Thus, the density of dilute solution can be taken as 1 g /mL. Let there be x g of a contaminant in 1 million g of solution. By definition, one-ppm means one part of the solute to be in one million parts of water as defined by Eq. 15. Using gram as unit, for a dilute aqueous solution Eq. 15 modifies to Eq. 16:

Contaminant in ppm = x g / million g of dilute solution of water (15)

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As discussed earlier, the volume of 1 million g of water solution is equal to 1 million mL of water. The Eq. 15 then modifies to Eq. 16.

Contaminant in ppm = x g solute/ 1 million mL of dilute solution of water (16) = x g/106 mL

Since 1 g = 1000 mg = 103mg, and 1 million L = 106 mL = 103L, we get ,

Contaminant in ppm = x 103mg/103 L Contaminant in ppm = x mg/L

In aquatic chemistry, ppm means mg L-1. So the dissolved amounts of solids, liquids and gases in -1 water are expressed in mg L , which is invariably expressed as ppm. For example, 6.7 ppm O2 in

water means 6.7 mg O2 to be dissolved in 1 L of water.

Problem 7. 1×10-4 g copper was determined in 1 L of water. Calculate the concentration of copper( atomic weight = 63.55) in ppm and in mmol/L.

Solution. The concentration of copper in ppm = copper in mg/ volume in L

Remembering, 1000 mg = 1 g, copper in 1×10-4 g = 1×10-4 × 103 mg = 0.1 mg So from above Eq. we get, Cu in ppm = 0.1 mg/ 1L = 0.1mg/L = 0.1 ppm

Cu in mol /L = mass of Cu in g/(M. W. of Cu× volume in L) = 1×10-4/63.55×1= 0.0157 mol/ L Remembering , 1 mol = 1 millimol = 1 mmol = 1×103mmol, we get, Cu in mmol/L = 0.0157× 103mmol/L = 15.7 mmol/L

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Conversion from One Unit to Another

Conversions from one unit to another are frequently required. To aid, conversion factors at atmospheric pressure are given below:

mg/m3 to ppmv = ={(mg/m3)(273+toC)}/{12.187}×M.W.} (17)

ppmv to mg/m3 = {ppmv×12.187× M.W.}/(273+toC) (18)

Pressure, Partial Pressure and Units

Partial pressure, pi, of a constituent, i, is related to total pressure, P, by Eqs.19 and 20.

pi = xi P = ni(n1+ n2+------+ni) P (19)

where xi is the mole fraction of i and is given by Eq.: xi = ni(n1+ n2+------ni), where n1, n2 and ------

-ni, are the number of moles of the constituents. Based on Eq. 19, it can be shown that

(p1+p2+------+pi ) = P (20) The atmospheric pressure can be expressed in several units, which are interrelated to each other as given below.

1 atmosphere = 101.325 Pa = 1.01325 bar = 760 mm Hg = 1.033228 kg/cm2

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Suggesting Reading:

http://web.viu.ca/krogh/chem311/units%20of%20concentration.pdf

http://how-it-looks.blogspot.in/2010/07/how-to-convert-to-and-from-parts-per.html

http://www.rapidtables.com/math/number/PPM.htm

http://chem3400.blogspot.in/2012/03/mathematically-measuring-atmospheric.html

http://en.wikipedia.org/wiki/Parts-per_notation

http://www.lenntech.com/calculators/ppm/converter-parts-per-million.htm

1. James E. Girard(2011), Principles of Environmental Chemistry, James and Bartlett, New Delhi. 2. Chemistry Part I Text Book for XII Class(2007), NCERT, New Delhi.

3. Colin Baird(2008), Environmental Chemistry, W. H. Freeman, New York

4. D.J. Jacob (1999), Introduction to Atmospheric Chemistry, Princeton University Press, Princeton

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