1. CONCEPTUAL FRAMEWORK & PHASE-SPECIFIC DIMENSIONS

A. Liquid Phase Concentration Units

In water and wastewater engineering, pollutant concentration describes the amount of a substance present in a given volume or mass of water.

1. Milligrams per Litre — mg/L

A mass-per-volume concentration unit.

Example:

50 mg/L means 50 mg of pollutant is present in 1 litre of water.

2. Parts per Million — ppm

For water, ppm is commonly expressed as a mass-per-mass ratio.

1 ppm = 1 mg pollutant per kg solution

For dilute aqueous solutions, ppm and mg/L are numerically equivalent because the density of water is approximately 1 kg/L.

3. Relationship Between mg/L and ppm

For dilute water:

1 mg/L ≈ 1 ppm

Also:

1 mg/L = 1,000 µg/L = 1,000 ppb

And:

1 g/m³ = 1 mg/L

Why?

For water:

Density ≈ 1 kg/L

Therefore:

1 L water ≈ 1 kg = 1,000,000 mg

So:

1 mg/L = 1 mg / 1,000,000 mg = 1 part per million

Therefore:

1 mg/L ≈ 1 ppm

Exam point: The equality between mg/L and ppm is an approximation based on water density being approximately 1 kg/L. It is not universally valid for concentrated or non-aqueous liquids.

4. Specific Gravity Correction

When the liquid density differs significantly from water:

ppm(w/w) = Concentration (mg/L) / Specific Gravity

Therefore, for a liquid with S.G. > 1:

ppm(w/w) < mg/L

Liquid vs Gas Phase

LIQUID / AQUEOUS PHASE
────────────────────────────────────
1 mg/L ≈ 1 ppm
1 mg/L = 1,000 µg/L
1 mg/L = 1,000 ppb
1 g/m³ = 1 mg/L

Basis: mass of pollutant / volume of water


GAS / AIR PHASE
────────────────────────────────────
1 ppmᵥ = 1 volume of pollutant / 10⁶
        volumes of air

ppmᵥ and mg/m³ are NOT numerically equal.

Conversion depends on:
• Molecular weight
• Temperature
• Pressure
• Molar volume

B. GAS / AIR PHASE CONCENTRATION UNITS

Air pollutants are commonly expressed in two different ways.

1. Volumetric Concentration — ppmᵥ or ppbᵥ

ppmᵥ means parts per million by volume.

For example:

1 ppmᵥ SO₂

means approximately:

1 volume of SO₂ per 1,000,000 volumes of air

Similarly:

1 ppbᵥ = 1 volume of pollutant per 1,000,000,000 volumes of air

2. Mass Concentration — µg/m³ or mg/m³

This expresses the actual mass of pollutant present in a cubic metre of air.

Example:

100 µg/m³ SO₂

means 100 micrograms of SO₂ are present in each cubic metre of air.

Important Difference

ppmᵥ
│
├── Volume ratio
├── Depends on pollutant molecular weight when converted to mass
└── Does not directly give mg/m³

mg/m³ or µg/m³
│
├── Mass per volume
├── Depends on temperature and pressure
└── Changes with gas density/volume conditions

For a fixed gas mixture, the ppmᵥ ratio is essentially unaffected by expansion or contraction when temperature and pressure change uniformly.

2. CORE MATHEMATICAL FORMULAS

Formula 1 — Liquid Phase Conversion

For dilute aqueous solutions:

1 mg/L ≈ 1 ppm

1 mg/L = 1,000 µg/L

1 mg/L = 1,000 ppb

1 g/m³ = 1 mg/L

Specific Gravity Correction

ppm(w/w) = mg/L ÷ S.G.

Where:

• mg/L = concentration by volume

• S.G. = specific gravity of the liquid

• ppm(w/w) = concentration by mass

Formula 2 — Gas Phase: ppmᵥ → µg/m³

From the ideal gas law:

PV = nRT

The general conversion is:

µg/m³ = (ppmᵥ × MW × 1,000) / Vₘ

Where:

• ppmᵥ = concentration by volume

• MW = molecular weight in g/mol

• Vₘ = molar volume in L/mol

Gas Phase: ppmᵥ → mg/m³

mg/m³ = (ppmᵥ × MW) / Vₘ

Therefore:

µg/m³ = 1,000 × mg/m³

Reverse Conversion

To convert µg/m³ to ppmᵥ:

ppmᵥ = (µg/m³ × Vₘ) / (MW × 1,000)

Formula 3 — Temperature & Pressure Correction

Molar volume changes with temperature and pressure.

Using the ideal gas relationship:

Vₘ = 22.414 × [(T + 273.15) / 273.15] × (1 / P)

Where:

• Vₘ = molar volume, L/mol

• T = temperature in °C

• P = pressure in atm

This equation uses:

22.414 L/mol at 0°C and 1 atm

Standard Temperature and Pressure — STP

For the convention used here:

T = 0°C = 273.15 K

P = 1 atm

Therefore:

Vₘ = 22.414 L/mol

25°C and 1 atm

At:

T = 25°C

P = 1 atm

Molar volume:

Vₘ ≈ 24.45 L/mol

Therefore:

µg/m³ = (ppmᵥ × MW × 1,000) / 24.45

This is a very useful environmental-engineering conversion.

Memory Logic

LOWER TEMPERATURE
        ↓
Gas occupies less volume
        ↓
Higher mg/m³ for the same ppmᵥ


HIGHER TEMPERATURE
        ↓
Gas expands
        ↓
Lower mg/m³ for the same ppmᵥ


HIGHER PRESSURE
        ↓
Gas occupies less volume
        ↓
Higher mg/m³

Formula 4 — Chemical Equivalency as CaCO₃

Hardness and alkalinity are often reported as:

mg/L as CaCO₃

The conversion is:

mg/L as CaCO₃
= mg/L of ion × (Equivalent weight of CaCO₃ / Equivalent weight of ion)

Since:

Equivalent weight = Molecular weight / Valence

For CaCO₃:

Molecular weight ≈ 100.09 g/mol

Valence = 2

Therefore:

Equivalent weight of CaCO₃ ≈ 50.04 g/eq

For most environmental calculations:

Equivalent weight of CaCO₃ ≈ 50

Hence:

mg/L as CaCO₃
= mg/L of ion × (50 / Equivalent weight of ion)

3. KEY MOLECULAR WEIGHTS

Exam Focus

For gas-phase ppm → mass concentration, the most important values are usually:

SO₂ = 64

NO₂ = 46

CO = 28

O₃ = 48

CH₄ = 16

For hardness calculations:

Ca²⁺ = 40.08; Eq. wt. = 20.04

Mg²⁺ = 24.31; Eq. wt. = 12.15

CaCO₃ Eq. wt. ≈ 50

4. SOLVED NUMERICAL EXAMPLES

Problem 1 — Specific Gravity Correction

Question

Industrial wastewater contains 45 mg/L of a toxic organic solvent.

Specific gravity of wastewater:

S.G. = 1.15

Calculate:

1. Concentration in ppm(w/w)

2. Concentration in µg/mL

3. Concentration in g/m³

Solution

Step 1 — Convert mg/L to ppm(w/w)

Formula:

ppm(w/w) = mg/L ÷ S.G.

Therefore:

ppm = 45 / 1.15

ppm = 39.13 ppm(w/w)

Step 2 — Convert mg/L to µg/mL

We know:

1 mg = 1,000 µg

1 L = 1,000 mL

Therefore:

45 mg/L = 45 µg/mL

Step 3 — Convert mg/L to g/m³

We know:

1 m³ = 1,000 L

1 g = 1,000 mg

Therefore:

45 mg/L × 1,000 L/m³ ÷ 1,000 mg/g

= 45 g/m³

Answer

39.13 ppm(w/w)

45 µg/mL

45 g/m³

Problem 2 — SO₂: ppmᵥ to µg/m³

Question

Ambient SO₂ concentration:

0.05 ppmᵥ

Molecular weight of SO₂:

64 g/mol

Conditions:

25°C and 1 atm

Calculate the concentration in µg/m³.

Step 1 — Identify molar volume

At 25°C and 1 atm:

Vₘ = 24.45 L/mol

Step 2 — Apply the conversion formula

µg/m³ = (ppmᵥ × MW × 1,000) / Vₘ

Substitute:

µg/m³ = (0.05 × 64 × 1,000) / 24.45

= 3,200 / 24.45

= 130.88 µg/m³

Answer

SO₂ = 130.88 µg/m³

Problem 3 — CO at Stack Conditions

Question

A stack gas contains:

CO = 15 ppmᵥ

Molecular weight:

MW = 28 g/mol

Stack conditions:

T = 150°C

P = 0.90 atm

Calculate CO concentration in mg/m³ at the actual stack conditions.

Step 1 — Convert temperature to Kelvin

T = 150 + 273.15

T = 423.15 K

Step 2 — Calculate molar volume

Formula:

Vₘ = 22.414 × (423.15 / 273.15) × (1 / 0.90)

Therefore:

Vₘ ≈ 38.58 L/mol

Step 3 — Convert ppmᵥ to mg/m³

Formula:

mg/m³ = (ppmᵥ × MW) / Vₘ

Substitute:

mg/m³ = (15 × 28) / 38.58

= 420 / 38.58

≈ 10.89 mg/m³

Answer

CO ≈ 10.89 mg/m³

at 150°C and 0.90 atm.

Problem 4 — Total Hardness as CaCO₃

Question

A groundwater sample contains:

• Ca²⁺ = 52.0 mg/L

• Mg²⁺ = 18.2 mg/L

Equivalent weights:

• Ca²⁺ = 20.04

• Mg²⁺ = 12.15

• CaCO₃ = 50

Calculate total hardness as CaCO₃.

Step 1 — Calcium hardness

Formula:

Ca hardness = Ca²⁺ concentration × (50 / Eq. wt. of Ca²⁺)

= 52.0 × (50 / 20.04)

≈ 129.74 mg/L as CaCO₃

Step 2 — Magnesium hardness

Mg hardness = Mg²⁺ concentration × (50 / Eq. wt. of Mg²⁺)

= 18.2 × (50 / 12.15)

≈ 74.89 mg/L as CaCO₃

Step 3 — Total hardness

Total hardness = Ca hardness + Mg hardness

= 129.74 + 74.89

= 204.63 mg/L as CaCO₃

Answer

Total hardness ≈ 204.63 mg/L as CaCO₃

Problem 5 — O₃ and NO₂ Conversion

Question

At 25°C and 1 atm:

• O₃ = 180 µg/m³

• NO₂ = 0.04 ppmᵥ

Molecular weights:

• O₃ = 48 g/mol

• NO₂ = 46 g/mol

Calculate:

1. O₃ in ppmᵥ

2. NO₂ in µg/m³

Part 1 — O₃: µg/m³ → ppmᵥ

At 25°C and 1 atm:

Vₘ = 24.45 L/mol

Formula:

ppmᵥ = (µg/m³ × Vₘ) / (MW × 1,000)

Substitute:

ppmᵥ = (180 × 24.45) / (48 × 1,000)

= 4,401 / 48,000

≈ 0.0917 ppmᵥ

O₃ Answer

0.0917 ppmᵥ

Part 2 — NO₂: ppmᵥ → µg/m³

Formula:

µg/m³ = (ppmᵥ × MW × 1,000) / Vₘ

Substitute:

= (0.04 × 46 × 1,000) / 24.45

= 1,840 / 24.45

≈ 75.26 µg/m³

NO₂ Answer

75.26 µg/m³

5. FORMULA SUMMARY — QUICK REFERENCE

6. RAPID EXAM RECALL

LIQUID
────────────────────────────────
1 mg/L ≈ 1 ppm
1 mg/L = 1,000 µg/L
1 mg/L = 1,000 ppb
1 g/m³ = 1 mg/L

If S.G. ≠ 1:
ppm(w/w) = mg/L ÷ S.G.


GAS
────────────────────────────────
ppmᵥ = volume ratio
mg/m³ = mass / volume

ppmᵥ → mg/m³:
ppmᵥ × MW ÷ Vₘ

ppmᵥ → µg/m³:
ppmᵥ × MW × 1,000 ÷ Vₘ


MOLAR VOLUME
────────────────────────────────
0°C, 1 atm → 22.414 L/mol
25°C, 1 atm → 24.45 L/mol


CaCO₃ BASIS
────────────────────────────────
Eq. wt. = MW / valence

CaCO₃ Eq. wt. ≈ 50

mg/L as CaCO₃
= mg/L ion × 50/Eq. wt.

Most important conceptual distinction

Water: concentration is usually handled using mg/L.

Air: concentration may be reported as either ppmᵥ or µg/m³.

ppmᵥ → mass concentration: you need molecular weight + temperature + pressure.

Hardness/alkalinity: convert different ions to a common CaCO₃ equivalent basis before adding them.