Module 4: Liquid Characteristics - Concentration

Features

Units of Measurement

: Parts Per Million; Milligrams Per Liter; Molarity and Normality

Practice Problems

Objective

Calculate the concentration of solutions on a parts per million, milligram per liter, molar, and normal basis.

Units of Measurement

Since there are so many different units of measurement for expressing concentrations of solutions in the field of air pollution control, this lesson only presents those that are most commonly used.

A solution is a liquid mixture, which consists of a solid and a liquid where the solid is dispersed in the liquid at the molecular level. The solid is referred to as a solute and the liquid is referred to as the solvent.

Concentration of liquids plays an important role in the air pollution control field. For example, it is often necessary to calculate the concentrations of solutions that enter wet scrubbers. In emissions testing often the concentrations of impinger solutions and recovery solutions must be known.

The concentrations of pollutants in solvents are usually low and are often expressed as parts per million (ppm) or milligrams per liter (mg/L). Concentrations of solutions with higher concentrations of solutes than normally occur with pollutants (e.g. scrubbing liquids used in control equipment or liquids used in emission sampling trains) are usually expressed as molarity or normality.

Parts Per Million

Specific gravity is not an especially useful means of expressing the concentration of pollutants in aqueous liquids, slurries, and ionic solutions. The concentrations are simply too low in most industrial applications to affect the liquid specific gravity. Liquid concentrations are often expressed as a ratio of masses as shown in Equation 1.

The concentrations of the pollutants in liquids are often expressed on a parts per million (ppm) basis. Equation 1 can be expressed on a parts per million basis as shown below.

Rearranging the terms in Equation 2 and solving for the pollutant concentration in ppm yields Equation 3.

In aqueous solutions, parts per million is a ratio of the number of grams (or pounds mass) of solute per million grams (or million pounds mass) of solvent and solute.

Note: Parts per million for liquids is different than parts per million for gases. For liquids, ppm is a ratio of two masses [often designated as ppm(w/w)]. While for gases, it is a ratio of two volumes [often designated as ppmv or ppm(v/v)]. Throughout these modules, the term ppm when applied to liquids means ppm(w/w).

Example Problem 1.
Converting from Mass Concentration to ppm

If 15 mg of sodium hydroxide (NaOH) is added to water and the mass of the resulting solution is 1500 grams, what is the concentration of the solution in ppm?

Solution:

Convert the mass of NaOH from milligrams to grams.
Calculate the concentration of the solution in ppm.

Milligrams Per Liter

Liquid concentrations are often expressed on a milligram per liter (mg/L) basis. This is a convenient way to represent low concentrations as mass per volume.

Example Problem 2.
Converting a Concentration from ppm to mg/L

If the concentration of sugar in water is 10 ppm at 20°C, what is its concentration in mg/L at the same temperature? The density of sugar is 1.59 gm/cm³ at 20°C.

Answer: i. 10 mg/L

Solution:

Assuming a basis of 1,000,000 gm of solution, convert 10 ppm of sugar to its equivalent mass ratio (in gm).
Calculate the volume of the water in liters. Recall that the density of pure water is 1 gm/cm³.
Using the density data, calculate the volume of sugar in liters.
Calculate the concentration of the sugar and water solution in mg/L.

Note: This example demonstrates three important points:

For solutions in which the solvent is water the units of ppm and mg/L are equivalent.
Usually the mass of the solute is negligible compared to the mass of the solvent and can be ignored in the denominator of Equation 1 and 2.
For city water, 1 gm = 1 mL.

Molarity and Normality

At higher concentration levels than normally occur with pollutants, molarity and normality are convenient ways to express concentrations of solutions.

Molarity is the concentration of a solution given in gram moles of solute per liter of solution. A solution containing exactly one gram mole per liter of solution is referred to as "molar."

The Normality expresses the concentration of a solution based on the number of gram equivalent weights of solute per liter of solution. A solution containing one gram equivalent weight per liter is referred to as "normal."

The number of gram equivalent weights is calculated as follows:

The equivalent weight is defined as the molecular weight (MW) divided by the change in the oxidation number experienced by a molecule in a chemical reaction.

#1

How do you abbreviate normality and molarity in the following two cases?

0.1 normal solution of H₂SO₄
0.25 molar solution of HCl

Answer to Question

Example Problem 3.
Creating a 0.1 Normal Solution

Suppose you must determine the amount of hydrogen halides (HCl, HBr and HF) in the flue gas leaving a chemical reactor. The emission sampling train for hydrogen halide determination calls for a total of 200 mL of 0.1 N H₂SO₄ as an absorbing solution. The absorbing solution will be located on the impingers of the sampling train. How many grams of H₂SO₄ should be added to water to create 200 mL of a 0.1 N H₂SO₄ solution?

The chemical reaction of H₂O and H₂SO₄ is as follows:

Approximate atomic weights of elements:
Example Equation 3b

Solution:
Use the following equations to solve this problem.

Step 1. Calculate the equivalent weight of the solute, H₂SO₄.

Determine the approximate molecular weight of H₂SO₄.
Determine the change in oxidation number.
The chemical reaction of H₂O and H₂SO₄ is as follows:

The acid (H₂SO₄) changes from a neutral molecule to ions with 2 charges each ( and ). Therefore the change in oxidation number is 2.
Calculate the equivalent weight of H₂SO₄.

Step 2. Determine the amount of H₂SO₄ to be added to 200 mL of H₂O.

Calculate the number of gram equivalent weights of H₂SO₄ required.
Calculate the mass of H₂SO₄ that must be added to 200 mL of water to create a 0.1 N solution. The solute volume can be neglected in the following equation.