Unit 4 Chemistry in the Environment
Are You Ready Page 262 - 265   Answers
Activity Unit 4 Getting Started - Simulated Water Treatment
Read Section 4.1 Water: Essential for Life Page 268 through 274
Activity 4.1 How Much Water is Essential?
Section 4.1 Questions  Page 274


Read Section 4.2 What's in Clean Water?  Page 275 through 279
Activity 4.2 Testing For Hard Water
Section 4.2 Questions  Page 279
Read Section 4.3 What's in Polluted Water?  Page 280 through 284
Section 4.3 Questions  Page 284
Extension Exercise 4.3 Case Study
Investigation 4.4 Testing for Ions in Water Page 285

Self Quiz 4.1 - 4.6

Acids and Bases
Arrhenius Model of Acids and Bases
The classical, or Arrhenius, model was developed by Svante Arrhenius in the nineteenth century. He defined an acid as any substance that liberates or yields hydrogen ions (H+) or protons in water. An example would be hydrogen chloride, HCl, gas, which when put in water ionizes to yield hydrogen ions, H+, and chloride ions. The resulting water solution of ionized H+ and Cl- is known as hydrochloric acid.

HCl(g) + H2O  <------------->   H3O+(aq) + Cl-(aq)

This process involving the breakdown of a substance into ions is known as ionization.
An Arrhenius base is a substance that dissociates in water to produce hydroxide ions, OH-. Two examples of strong, or almost completely dissociated bases are potassium hydroxide, KOH, and sodium hydroxide, NaOH or lye.

Most solutions formed by the reaction of polar molecular compounds with water are observed to have either acidic or basic properties.
Acids
Acidic Properties
Bases
Basic Properties
• are water soluble
• are electrolytes
taste sour
• do not feel slippery
• turn blue litmus red
• neutralize basic solutions
• react with active metals to
produce hydrogen gas
are water soluble
are electrolytes
taste bitter
• feel slippery
• turn red litmus blue
• neutralize acidic solutions


Strong versus Weak
Strong acids are ones that dissolve completely into their ions.  HCl, and HNO3 are strong acids.   Weak acids are usually organic in nature like oxalic acid, citric acid, ascorbic acid and vinegar, etc.

Strong acid        HCl(aq)  +  H2O(l)  ---->  H3O+1(aq)   +  Cl-1(aq)  
100% dissociated - all the HCl  breaks down into ions.

Weak acids     HCH3COO(l) + H2O(l)  ----->  H3O+1(aq) + CH3COO-1(aq) + HCH3COO(aq)
1.8 % dissociated - For every 1000 molecules of vinegar dissolved in water only 18 actually break down into ions.  The rest remain as complete molecules.  This means that the number of  hydrogen ions released into the water is much smaller.


Concentrated versus Dilute
Concentrated and dilute are relative terms about concentration.  Dilute simply  means that there is less solute dissolved per unit volume that a concentrated solution.    We can say things like this solution is more or less concentrated than some other solution. 

Remember from the last unit that concentration is measured as:
       concentration =   amount of solute 
                                  volume of solvent

Usually the concentration in chemistry is mol/L but any units can be used like g/100 mL, ppm etc.


Hydrogen Ion and Hydroxide Ion Concentrations
The ion concentrations depend upon two things:
1)  does the dissolved substance ionize completely and
2)  how much substance actually dissolved?

We will consider several examples to explain the process:
Determine the concentration of hydrogen or hydroxide ions in each of the following solutions of strong acids or bases.

a)    0.5 mol/L HNO
3(aq)
Step_1
Write a balanced ionization equation
HNO3(aq) --->  H+1(aq) + NO3-1(aq)

Step_2
Determine Molar Ratios of Reactants and Products
From the balanced equation, you know  that 1 mol of HNO3(aq)  ionizes to produce 1 mol of H+1(aq) ions and 1 mole of NO3-1(aq) ions.

Step_3
1 HNO3(aq) --->  1 H+1(aq) + 1 NO3-1(aq)
   0.5 mol/L           0.5 mol/L     0.5 mol/L

Therefore 0.5 mol/L nitric acid dissociates into 0.5 mol/L H+1 ions and 0.5 mol/L NO3-1 ions.

b)  0.25 mol/L HCl(aq)
Step_1
Write a Balanced Ionization Equation
HCl(aq)  --->  H+1(aq)  +  Cl-1(aq)

Step_2
Determine Molar Ratios of Reactants and Products
HCl(aq) is a strong acid .  1 mole will ionize to produce one mole of H+1 and 1 mole of Cl-1, therefore

Step_3
Use the Molar Ratios to determine the concentration of Hydrogen Ions or Hydroxide ions.
   HCl(aq)  ----->  H+1(aq)   +    Cl-1(aq)
0.25 mol/L         0.25 mol/L      0.25 mol/L
Homework Practice Exercise Page 300 Questions 1 through 4
Extension Exercise 4.7

pH:   Power of Hydrogen
The pH scale is actually based on pure water which is considered to be perfectly neutral.  As it turns out water self ionizes. The equation for this self-ionization of water is below.

H2O(l)  +  H2O(l)  ---->   H3O+1(aq)  +  OH-1(aq)

The concentration of H3O+1 and OH-1 are both 1.0 X 10-7 mol/L.  
This can be converted into a pH of 7.

The pH scale is out of 14 and 1.0 X 10-7 X 1.0 X 10-7 = 1.0 X 10-14.
This is not a coincidence.  

The pH scale is a logarithmic scale. The p" factor" is defined as the log of the molar concentration of whatever follows the letter p and then multiplied by a negative

So the pH = -log[H+]

For strong acid molar concentrations equal to or less than 1, the pH value would have a value from 0-14.

If the Hydrogen ion concentration is 0.1 moles/liter

Then the [OH-] could be found by the equation above:

[OH-] = 1 X 10-14 / 1 X 10-1 = 1 X 10-13

The pOH = -log[OH-] = -log(1 X 10-13) = -(log 1 + log 10-13) = -(0 + -13) = -(-13) = 13

For a [H+] = 0.1 = 1 X 10-1

Then pH = -log 1 X 10-1 = -(0 + -1) = 1

Therefore the

pH + pOH = 14

Calculate the pH of a solution that has a [OH-] = 1 X 10-5 M

  1. Determine pOH

    pOH = - log [OH- ] = - [log  1 X 10-5 ]  +5

  2. Determine the pH knowing that pH + pOH = 14

    pH = 14 - pOH = 14 - 5 = 9

Typical strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4

Strong Bases: All the Hydroxide compounds of Group 1 and Group 2 metals

LiOH, NaOH, KOH, etc and Be(OH)2, Ca(OH)2, Mg(OH)2, etc

What would be the [OH-] of a 0.2 M NaOH solution?

NaOH + H2O ----> Na+(aq) + OH- (aq)

0.2 M NaOH will produce 0.2 M OH- since the breakdown is 100%

What would be the [OH-] of a 0.3 M Ca(OH)2

Ca(OH)2 + H2O -----> Ca+2 (aq) + 2 OH- (aq)

0.3 M Ca(OH)2 will produce 0.6M OH- because for every one Ca(OH)2 that breaks apart TWO OH- ions are produced (twice as much) (look at the equation and note the coefficients)

Now here is an example for you to work out.

Given 0.02M Ba(OH)2 solution:

  1. Determine the Hydroxide ion molar concentration
  2. Determine the Hydrogen ion concentration
  3. Determine the pH
  4. Determine to pOH
Homework Practice Questions Page 303 Questions 5 & 6
Homework Page 304 Questions 1 through 13
Investigation 4.9 Page 308 Dilution and pH
Investigation 4.10 Page 312 Acid-Base Reactions

Reactions of Acids and Bases

Acids and bases have a number of characteristic reactions.
1.  Acids react with metals to produce hydrogen gas and a salt of the metal and acid.
 
     Fe(s) + 2 HCl(aq) -->  H2(g) + FeCl2(aq)

2.   Acids react with carbonates to produce a salt and hydrogen carbonate.  The hydrogen carbonate immediately decomposes into carbon dioxide and water.

       2 HCl(aq)  +  Na2CO3(aq)  ------>   2 NaCl(aq) +  H2CO3(aq)
                                H2CO3(aq)  ------>  H2O(l) + CO2(g)                           
       2 HCl(aq)  +  Na2CO3(aq)  ------>   2 NaCl(aq)  +  H2O(l) + CO2(aq)

3.    Acids react with bases to create water and a salt.

       HCl(aq)  +    NaOH(aq)  ----->  H2O(l)  +  NaCl(aq)

Homework Practice Page 318 Questions 1 through 3

Acid-Base Titrations
A titration is a common method in quantitative chemical analysis.  A known volume of a sample to be analysed is titrated against a known concentration of a base.  The burette contains an accurately known concentration called the standard solution.  During the titration, the solution is the burette called the titrant, is added drop by drop.  When the solution reaches the endpoint, where equal concentrations of acid and base have been mixed a chemical indicator signals that neutralization is complete.

Sample Quantitative Analysis Using a Titration
We will make use of the following equation:  Ma
VaCa = MbVbCb
Ma is the molarity of the acid, Mb is the molarity of the base
Va is the volume of acid used, Vb is the volume of base used
Ca is the number of acidic hydrogens in the acid, Cb is the number of hydroxides in the base

Example Analysis:
The concentration of hydrochloric acid can be analysed by titration with sodium hydroxide solution.  Three 10.0 mL sample of HCl are titrated with a standardized 0.200 mol/L solution of sodium hydroxide.  The results for the three trial are shown below.  What is the concentration of the hydrochloric acid?

Trial
1
2
3
Average
Final Burette Reading
13.85 mL
26.95 mL
39.85 mL

Initial Burette Reading
0.70 mL
13.90 mL
26.90 mL

Volume of NaOH(aq) added
13.15 mL
13.05 mL
12.95 mL
13.05 mL

Fill in the following table:
Ma = ?                      Mb = 0.200 mol/L
Va = 10.0 mL            Va = 13.05 mL (experimentally determined)
Ca = 1                        Cb = 1

Rearrange the titration equation to find Ma.

 MaVaCa = MbVbCb    rearranges to give

 MaVaCa = MbVbCb
    Va•Ca          Va•Ca

 Ma = MbVbCb
             Va•Ca


  Ma = 0.200 mol/L •  13.05 mL • 1
                  10.0 mL
• 1
     
         = 0.261 mol/L for the HCl

Homework Practice Page 322 Questions 4 through 8
Extension Exercise 4.10 Standardization of a Stock Acid
Homework Page 323 Questions 1 through 8
Extension Exercise 4.11  Balancing Equations and Solving Titration Problems
Extension Exercise 4.11 Concentration of a Base
Extension Exercise 4.11 Concentration of an Acid
Self Quiz 4.7 - 4.13

Acid Rain
Read Section 4.13 Pages 325 to 328
Homework Page 329 Questions 1 through 9

 

Gases: The Empirical Gas Laws: Pressure

Quantitative measurements on gases were first made by the English chemist Robert Boyle (1627 - 1691). Boyle used two instruments to measure pressure: the manometer, which measures differences in pressure, and the barometer, which measures the total pressure of the atmosphere.

Units of Pressure

Units of pressure were originally all based on the length of the column of liquid, usually mercury, supported in a manometer or barometer. By far the most common of these units was the mmHg, however, the modern SI unit of pressure is derived from the fundamental units of the SI.  Pressure is force per unit area, and force is the product of mass times acceleration, so the SI unit of pressure is the kg m s-2/m2 or newton/m2, which is called the pascal (Pa).



All of the older units of pressure have now been redefined in terms of the pascal. One standard atmosphere, the pressure of the atmosphere at sea level, is by definition exactly 101,325 Pa. The torr, named in honour of Torricelli, is defined as 1/760 of a standard atmosphere or as 101,325/760 Pa. The mmHg can be considered identical to the torr. The term bar is used for 100000 Pa, which is slightly below one standard atmosphere.
 
Robert Boyle and his Law
Boyle used the manometer and barometer to study the pressures and volumes of different samples of different gases. The results of his studies can be summarized in a simple statement which has come to be known as Boyle's law:
 
At any constant temperature, the product of the pressure and the volume of any size sample of any gas is a constant.
 
For a particular sample of any gas, Boyle's law can be shown graphically as is done in the Figure below. It is more common to express it mathematically as P1V1 = P2V2.
 
The pressure and the volume vary inversely; as the pressure increases the volume of the sample of gas must decrease.
 
Example:  A sample of gas occupies a volume of 47.3 cm3 at 20oC when the pressure is 30 cm of mercury.  If the pressure is increased to 75 cm of mercury, the sample will occupy a volume of 47.3 cm3 (30 cmHg/75 cmHg) = 18.9 cm3.
 
Homework Practice Page 335 questions 1 through 5
Extension Exercise 4.14
Go to the Boyle's Law Worksheet

Gases: The Empirical Gas Laws - Celcius and Kelvin Temperatures

In SI Metric the temperature scale is defined as Kelvin temperature scale. The degree unit is the kelvin (K).  The symbol for the unit is K, not oK.  Kelvin temperatures must be used in many gas law equations in which temperature enters directly into the calculations.
 
The Celcius and kelvin scale are related unit for unit. One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale.  The only difference between these two scales is the zero point.  The zero point on the Celcius scale was defined as the freezing point of water, which means that there are higher and lower temperatures around it.  The zero point on the kelvin scale - called absolute zero - corresponds to the lowest temperature that is possible.  It is 273.15 units lower than the zero point on the Celcius scale.  So this means that 0 K equals -273.15oC and 0oC equals 273.15 K. Thermometers are never marked in the kelvin scale. If we need degrees in kelvin the following relationships are to be used.
 
 TK =  tc  +  273.15             or       tc  =  TK  -  273.15

   

Gases: The Empirical Gas Laws - Temperature

The conventional liquid-in-glass thermometer was invented in the seventeenth century. This bulb-and-tube device is still in use; it is shown in the Figure below. In these thermometers the diameter of the bulb is much greater than the diameter of the tube so that a small change in the volume of liquid in the bulb will produce a large change in the height of the liquid in the tube. Two things were not clear about the thermometer at this time. The first question was what it was that the thermometer measured. As the temperature or "degree of hotness" apparent to one's fingers increased, the height of the liquid obviously did also, and this was useful in medicine for checking fevers, but there was no quantitative measurement made, merely the relative degree of hotness between this and that. The second question was whether the degree of hotness of any particular thing was a constant everywhere so that the temperatures of other things could be measured relative to it. Suggested fixed temperatures included that of boiling water, that of melting butter, and the apparently uniform temperature of deep cellars.
 
Robert Boyle knew of the thermometer, and also was aware that a gas expands when heated, but since no quantitative temperature scale existed he could not, and did not, determine the relationship between degree of hotness (temperature) and volume of a gas quantitatively. Boyle did propose a scale of temperature, suggesting that use of a specific fluid in a standardized thermometer bulb with a capacity of 10,000 units filled at the boiling point of water would give a proper scale if changes were at the one-unit level; that is, one degree would have a volume of 10,001 units. His scale was not adopted.
 
Guillaume Amontons (d. 1705) developed the air thermometer, which uses the increase in the volume of a gas with temperature rather than the volume of a liquid. The air thermometer is an excellent demonstration of Charles' law because the atmosphere maintains a fixed downward pressure above a small mercury plug of constant mass. The volume of a trapped sample of air increases on heating until the pressure of the trapped air equals the pressure of the atmosphere plus the small pressure due to the plug. Nevertheless, Amontons failed to achieve formulation of Charles' law for the same reason as did Boyle: a quantitative scale of temperature was needed.
 
A quantitative scale of temperature could only be developed after it was realized that at a fixed pressure any pure substance undergoes a phase change at a single fixed temperature which is characteristic of that substance. The melting point of ice to water was taken as 0oC and the boiling point of water was taken as 100oC to give our common Celsius scale of temperature.
 
The study of the effect of temperature upon the properties of gases took considerably longer to achieve a simple quantitative relation than did study of the effect of pressure, primarily because the development of a quantitative scale of temperature was a difficult process. However, once such a scale was developed, the appropriate measurements were made, primarily by the French chemist Jacques Charles (1746 - 1823). On the modern Celsius scale and using modern pressure units, a typical set of Charles' 1787 data would appear as shown in the Figure below.
 
The experimental data were formulated into a general law which became known as the law of Charles or Charles' law:
 
At any constant pressure, the volume of any sample of any gas is directly proportional to the temperature.
 
However, as the graph above shows, the volume extrapolates to zero at a temperature of -273.15oC. If this temperature were taken as the zero of a temperature scale then all negative temperatures could be eliminated.  Such a temperature scale is now the fundamental scale of temperature in the SI.  It is called the absolute scale, the thermodynamic scale, and the Kelvin scale. Temperature on the Kelvin scale, and only on the Kelvin scale, is symbolized by T.
 
A useful formula when the volume of one particular sample of gas changes with temperature, is: 
 
V1T2= V2T1
Example: The volume of a sample of gas is 23.2 cm3 at 20oC. If the gas is ideal and the pressure remains unchanged, its volume at 80oC will be given by 23.2 cm3 (353.15 K/293.15 K) = 27.95 cm3.

Homework Practice Page 338 questions 6 through 9
Go to the Charles' Law Worksheet
  
Gases: The Empirical Gas Laws - Partial Pressures of Gases
When Dalton was conducting his studies, which led him to the atomic-molecular theory of matter, he also included studies of the behaviour of gases.  These led him to propose, in 1803, what is now called Dalton's law of partial pressures:
 
For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone.
 
This law can be expressed in equation form as:   p = p1 + p2 + p3 + ...
 
where p is the total or measured pressure and p1, p2, ... are the partial pressures of the individual gases. For air, an appropriate form of Dalton's law would be:
 
p(air) = p(N2) + p(O2) + p(CO2) + ...
 
Homework Practice Page 339 Questions 10 through 12
Homework Page 341 Questions 1 through 13
Go to the Partial Pressures Worksheet

The Kyoto Protocol
Homework Read Section 4.15 Pages 343 through 348
Extension Exercise 4.15
Homework Page 349 Questions 1 through 6

Air Quality
Homework Read Section 4.16 Pages 350 through 355
Activity 4.16 Interpreting the Air Quality Index
Homework Page 356 Questions 1 through 9
Self Quiz 4.14 - 4.16

Unit 4 Summary
Homework Unit Review Page 363 Questions 1 through 38
Unit 4 Self Quiz