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CHM 1020--Chemistry for Liberal Studies--Spring 1999

Chemistry 1020—Lecture 16—Notes

Chapter 6—Acid Rain

Recall in Chapter 1 we discussed tropospheric air quality. This problem was primarily a local problem.In Chapter 2 and 3 we discussed ozone and global warming. These problems were global problems.In this chapter on acid rain, we are dealing with a regional problem.

You should all have read the chapter. It describes effects, or supposed effects, of acid rain on:

Lakes and Streams (aquatic life)
Marble and limestone
Human health

The sources of acid rain are primarily two classes species: oxides of nitrogen (NOx) and oxides of sulfur (SOx). Table 6.2 summarizes the sources of these gaseous oxides, pointing out that there are both natural and anthropogenic sources. The anthropogenic (i.e. related to human activity) sources of NOx include primarily the burning of gasoline in automobiles as well as combustion in coal burning electrical plants. The nitrogen comes from the air, and it is the high temperature of automobile engines that favors the combination of the nitrogen and oxygen of the air. The natural sources include lightning (where temperatures are also elevated) and the action of plants and microorganisms in the soil. About 41% of the NOx production is natural.

The anthropogenic sources of SOx are primarily from the burning of coal in electric generation plants and other industrial plants. Natural sources include oceans and volcanoes. About 21% of SOx production is from natural sources.

Are there solutions to these problems? The book discusses several.

Remediation, at least as far as lakes go, is possible to some extent by adding lime (Ca(OH)2) to lakes. Some lakes are already partly protected from the problem by existing in contact with limestone (CaCO3) which can also neutralize the acid.
Reduction of the oxides is also a potential solution.
For NOx, that involves use of catalytic converters in automobiles. The catalysts in these devices promote the dissociation of the NO formed in the high temperatures of automobiles back to nitrogen and oxygen. (The favored direction of this reaction changes with temperature).
For sulfur oxides, there are three methods of attack. One is using low sulfur coal (anthracite) instead of more plentiful high sulfur coal (bituminous). A second is the removal of the sulfur from the coal by grinding the coal into small particles and washing it. A third is the removal of the sulfur dioxide from the gaseous output of coal burning plants, a process called scrubbing.

$$$$ These solutions all cost money. $$$$

Another solution, common to solutions of both local air pollution and global warming, is the decrease our energy use. Most of the problems are generated in the production of energy, primarily by burning coal or petroleum. Of course there is a cost here, one of convenience.

Alternative energy sources provide another possible solution. We will discuss some of these in a later chapter. For the most part they are also more expensive.

 So a basic question is what is it worth to us to try to avoid these problems? Answers to this question not clear cut. Nothing is black and white, and there are trade-offs in the risks and benefits of various courses of action.

These are questions for political debate; perhaps for an economics class. The goal in this class is to try to get you to understand the chemistry underlying the problem.

So the basic chemical question has to do with what is an acid?

The term comes from the Latin acidus, meaning sour or tart. The taste of lemon juice, for example, comes from the acidity there due to a natural acid called citric acid.

In the 1880’s the Swedish chemist Svante Arrhenius defined an acid as something produces H+ ions in water. A base, which is sort of the opposite of an acid, produces OH- ions in water. If the amount of H+ > OH-, the solution is said to be acidic, and if the amount of H+ < OH-, the solution is said to be basic.

Acids and bases can react to neutralize each other to form water:

H+ + OH- ---> H2O

Nature of the H+ in an acid

Recall that H+ is simply the nucleus of a hydrogen atom, a proton. We don’t expect bare protons to exist other than in specialized particle beams. In aqueous solution, the proton attaches itself to one of the lone pairs of a water molecule to form a hydronium ion (H3O+). We might also think of the proton as being solvated not by one but by two or more water molecules, therefore representing species such as H5O2+ or H7O3+. For convenience we often simply refer to this species either as H+(aq) or H3O+(aq), where the (aq) refers to an aqueous solution (a solution in water).

The Arrhenius definition is sufficient for our discussion in this chapter. For sake of completeness, though, I should mention that another, more general definition, is probably a better one—a definition attributed to Bronsted and Lowry. In this definition, an acid is a proton donor and a base is a proton acceptor.

Concentration of solutions

The concept of acidity refers to the quantity of H+. The amount of H+ in a solution can be determined in a number of ways, on is by using a indicator:


A universal indicator is added to solutions with different amounts of acid, and one can see that the color of the solutions is different. As an illustration of how dissolved substances can change this amount of acid, dry ice is placed in the beaker with the lowest acididty, and the color gradually changes to increasing acidity, illustrating that carbon dioxide does form acid in solution. It undergoes the following reaction:

CO2 + H2--->  H2CO3 --->  H+ + HCO3-

We speak here of the amount or quantity of acid in a solution. How do we express this quantity? Before tackling that question we need to understand in general how we talk about the concentration of substances in general.

Recall that solutions are homogeneous mixtures, with no fixed composition. In Chapter 1, we considered pollutants as components of gaseous solutions, and we had ways of expressing the amount of those pollutants in terms of parts of the whole, such as parts per thousand, parts per million, etc, generally referring to relative masses or relative numbers of particles or moles.

In other solutions, the principle is the same. The concentration refers to the amount of one component (usually the solute—the component in lower amount) either relative to the solvent—the component in greater amount, (i.e. water for aqueous solutions) or relative to the total solution.

There are several concentration units used in chemistry to express this concept. Each has its specialized use. In the case we are discussing, the convention is to express the ratio of the solute, expressed in moles to the amount of total solution, expressed in liters. Hence the concentration unit is moles/liter, abbreviated molar or M.

This unit makes it convenient to relate volumes of a solution to moles of a solute:

V x M = moles

For example:

Q: 0.75 L of 0.12 M sucrose solution contains how many moles of sucrose?
A: 0.75 L x 0.12 moles/L = 0.90 moles
Q: What volume of a 1.2 M solution of HCl is necessary to have 3.5 moles of HCl?
A: 3.5 moles x 1 L/1.2 moles = 2.9 L

Preparing a solution of a given molarity usually involves the use of a volumetric flask. The proper quantity of the substance is added to the flask, and then water (or whatever solvent is being used) is added until the total volume of the solution comes to the volume mark on the flask. See figure 6.2, page 190 of the text, which illustrates this procedure. This figure describes addition of 0.250 moles of acetic acid to a 1.000 L volumetric flask in order to make a solution of 0.250 M acetic acid.

The interconversion of volume, molarity, and number of moles is pretty simple and straightforward. For some practice, go to the web site link at: http://proton.csudh.edu/lecture_help/molarity.html.

Of course there is no balance that reads in moles. For real world problems, the calculation is slightly more complicated by the need to convert the moles to grams (or vice versa) when considering quantities of a material. For example, in order to measure 0.250 moles of acetic acid, one must calculate the molecular (or formula) weight of C2H4O2 (which is 60.0 g/mole) and then convert to grams as follows:

0.250 moles x 60 g/mole = 15.0 g of acetic acid.

Returning to the 0.75 L of 0.12 M sucrose solution above, one calculates the molecular weight of sucrose (C12H22O11) as 342.0 g/mole, and calculates the mass of 0.090 moles of sucrose :

0.090 moles x 342 g/mole = 30.78 g (rounded to 3.1 g)

Lets work a couple of the exercises in the book:

Question 14, page 217:

Calculate the molarity of:

(a) 0.30 moles HCl in 3.0 L of solution.
     0.30 moles/3.0 L = 0.10 moles/L = 0.10 M
(b) 11.2 g KOH in 1.0 L of solution.
     FW of KOH = 56.1 g/mol
     11.2 g KOH x 1 mol/56.1 g = 0.1996 mol
     0.1991 mol/1.0 L = 0.1996 M (round to 0.20 M)
(c) 42.6 g glucose (C6H12O6) in 0.40 L of solution.
     MW of glucose = 180.0 g/mol
     42.6 g x 1 mol/180.0 g = 0.2366 mol
     0.2366 mol/0.40 L = 0.59166 M (round to 0.59 M)

Question 16, page 217

How many liters of the following solutions can be prepared from 100 g of the solute?

(a) 1.25 M NH4NO3 (FW = 80.0 g/mol)
     100 g x 1 mol/80.0 g = 1.25 mol
     volume = 1.25 mol/1.25 mol/L = 1.00 L
(b) 0.50 M Al2(SO4)3 (FW = 342.3 g/mol)
     100 g x 1 mol/342.3 g = 0.292 mol
     volume = 0.292 mol/0.50 mol/L = 0.58 L
(c) 0.154 M NaCl (FW = 58.5 g/mol)
     100 g x 1 mol/58.5 g = 1.709 mol
     volume = 1.709 mol/0.154 mol/L = 11.1 L

You can get some practice with more problems of this type by going to the web site http://proton.csudh.edu/lecture_help/startmolarity.htmll.

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