1. Equilibrium Chemistry
CE 541
Equilibrium Chemistry

2. Limitations of Equilibrium Calculations
Important to:
Determine the relationship between constituents in water
Understand the effect of alterations of water on the different chemical species present
Limitations of Equilibrium Calculations
Dynamic changes (in wastewater and surface water)
Due to exposure to sun
Due to exposure to pollution (organic and inorganic)
Rapid reactions (reaction between acids and bases)
Very slow reactions (oxidation-reduction in natural waters)
Precipitation reactions
Lack of information on accurate equilibrium constants for many of the reactions taking place in natural waters

3. Ion Activity Coefficients
“The activity of ion or molecule can be found by multiplying its molar concentration by an activity coefficient, ”
{A} =  [A]
{A} = activity
[A] = concentration
 = activity coefficient
For practical reasons and rough calculations [A] is used in place of {A}

4. Some investigators found that “the activity coefficients for ions in an electrolyte were related to the concentration of charged particles in the solution”. They introduced the ionic strength as an empirical measure of the interactions among all the ions in a solution.
 = ionic strength
Ci = molar concentration of the ith ion
Zi = charge of the ith ion
Langelier estimated  as (TDS  2.5  10-5)

5. Other investigators found that, for dilute solution, there is a relationship between  and  as follows:
This relationship is used for dilute solutions with ionic strength < 0.1 (in Environmental Engineering, most waters of interest are more dilute than this, except seawater)
This relationship is used for solutions with  up to 0.5 M.

6. Conclusion
There is no good relationship that provides a satisfactory estimate of  for  > 0.5 M. In this course,  will be assumed to be equal to 1 unless otherwise mentioned.
Study Example 1 page 108.

7. Problem 4.2
Calculate the activity coefficient and activity of each ion in a solution containing 75 mg/l Na+, 25 mg/l Ca2+, 10 mg/l Mg2+, 125 mg/l Cl-, 50 mg/l HCO3-, and 48 mg/l SO42-.

8. Solutions to Equilibrium Problems
Le Chatelier’s Principle states that “A chemical system will respond to change with processes which tend to reduce the effect of the change”
For any chemical reaction
Principle of conservation of mass must be obeyed
Electroneutrality must be maintained “All positively charged species in solution must be balanced by equivalent numbers of negatively charged species”
Proton condition “Species with an excess of protons must be balanced by the species with a deficiency in protons”
All reactions involved must proceed towards a state of equilibrium.

9. Steps to Solve Equilibrium Problem Involving Aqueous Phase only
Define the equilibrium problem
what chemical reactions are taking place
what is reacting with what
List all constituents of the system
all systems involving water include
H2O H+
OH-
Include all ions, elements and neutral species present initially

10. Steps to Solve Equilibrium Problem Involving Aqueous Phase only
For each element present initially
list all forms or species which are likely to contain the element and which are likely to present after equilibrium is attained
Identify concentrations of all species for each element so that mass and charge balances can be made
List all appropriate equilibrium relationships between species of concern

11. Steps to Solve Equilibrium Problem Involving Aqueous Phase only
List associated equilibrium constants
List all mass and charge balance relationships for the system
List proton conditions
Steps 5 to 8 will produce a number of equations equal to the unknown species.
Solve the equations simultaneously

12. Steps to Solve Equilibrium Problem Involving Aqueous Phase only
If gaseous or solid phase are involved, then equations expressing mass and charge balances between and within each phase must be included.
Study Example 2

13. Problem 4.4
A solution is prepared by diluting 10-2 mol of ammonia to 1 liter with distilled water. Calculate the equilibrium concentration for each chemical species in the water.

15. Acids and Bases
strong acids and bases ionize completely in dilute solutions (water)
weak acids and bases ionize partially in dilute solutions (water)
acids increase H+ concentration
Bases increase OH- concentration

16. [H+][OH-] = Kw [ ] = activity or approximately molar concentration
[H+] is expressed in terms of pH. pH has an effect on:
Equilibrium between most of the chemical species
Effectiveness of coagulation
Potential of water to be corrosive
Suitability of water to microorganisms
Other quality characteristics of the water
Thus it is very important to understand the factors that have an effect on pH of water.

17. The pH and p(x) Concept In pure water (no other materials):
Activity = Molar concentration
[H+] = [OH-]
Kw = 25 C
[H+][OH-] = 10-14
[H+] = [OH-] = 10-7
pH = 7 (is the neutral pH)

18. pH Meters
scale 0 to 14
pH less than 7 indicates acidic condition; [H+] > [OH-]
pH more than 7 indicates basic condition; [H+] < [OH-]
electrodes measure hydrogen-ion activity not molar concentration

19. The concept of expressing [H+] activity can be used with other ions. So
x = concentration of a given chemical species or equilibrium constant. Then

20. Since, [H+][OH-] = Kw = 10-14 @ 25 C
Then, -log[H+] – log[OH-] = -log Kw
pH + pOH = pKw
pKw = 25 C
for weak acids and bases:
pKA is the negative log of the ionization constant for weak acids
pKB is the negative log of the ionization constant for weak bases
Tables 4-1 and 4-2 show KA, pKA, KB, and pKB for weak acids and bases.

21. For weak acid and its conjugate base or weak base and its conjugate acid:
pKA + pKB = 25 C or
KAKB = = Kw
Example
Boric acid has pKA = 9.24
Borate has pKB = 4.76
pKA + pKB = 14

22. Solving Acid – Base Equilibrium Problems
Assumptions
equilibrium occurs very fast (thus neglect kinetic considerations)
strong acids and bases are completely ionized in water (except when the added concentration is ≈ 10-7)

23. Study Examples 3, 4, 5, and 6 Tools to be used in Solving Problems
equilibrium relationships
mass balance
charge balance
proton condition
How to Solve Problems
identify unknowns
generate equations = unknowns
solve equations simultaneously
use graphical solutions
use computers for complex problems
Study Examples 3, 4, 5, and 6

24. Problem 4.5
Calculate the equilibrium pH of a solution containing (a) 10-3 M H2SO4; (b) 10-8 M H2SO4

25. Logarithmic Concentration Diagram
Log C – pH diagram represents mass balance of each constituent at every pH value. To construct the diagram, develop equations of C as a function of pH, Kw, KA, CT

26. These curves show the log of the concentration as a function of pH
How does a logarithmic concentration diagram change when the concentration is changed?
How does a logarithmic concentration diagram change when the Ka is changed?
How does a logarithmic concentration diagram change when the Ka2 is changed?

27. Logarithmic Concentration Diagram for Monoprotic Acids and Bases
Diagram of Monoprotic Acid
0.02 M acetic acid solution. Use:
the above acetic acid equations can be used for all monoprotic acids. Monoprotic acid is an acid which yields one proton when ionized.

29. Line of [H+] is obtained from:
log [H+] = -pH
Line of [OH-] is obtained from:
log [OH-] = pH – pKw
Concentration of Acetic Acid and Acetate can be obtained from:

30. Intersection of [HAc] and [Ac-] lines is called the System Point and is located at:
pH = pKA
To left of the system point, [H+] > KA, so
Both lines pass through the system point

31. To the right of the system point, [H+] < KA, so
Both lines pass through the system point

32. log [(1/2)CT] = log CT + log (1/2) = log CT – 0.3
Just below the system point and from [HAc] + [Ac-] = CT
[HAc] = [Ac-] = (1/2)CT
log [(1/2)CT] = log CT + log (1/2) = log CT – 0.3
So, curves of horizontal and diagonal lines intersect at a point 0.3 below the system point.

33. General Procedure for Construction of logC-pH Diagrams for Monoprotic Acids and Bases
Draw horizontal line representing log CT
Locate the system point at pH = pKA
Draw 45 lines sloping to left and right of the system point
Locate a point of 0.3 logarithmic units below the system point
Connect horizontal and 45 lines with short curves passing through the points
[H+] and [OH-] lines are drawn as 45 lines which intersect at pH = 7 and log C = -7
Change in concentration will only shift the [HAc] and [Ac-] curves up or down

34. Diagram of Monoprotic Base
Similar procedure
System point at pH = pKw – pKB
0.01 M NH3 Solution
For ammonia,
pKB = 4.74
so, system point is located at:
14 – 4.74 = 9.26

35. Logarithmic Concentration Diagram for a Weak Acid and a Weak Base
0.1 M acetic acid and 0.1 M ammonia
The logC-pH diagram is made by superimposing the curves for each material on a single diagram.
Study Examples 11 to 15

36. Logarithmic Concentration Diagram for Polyprotic Acids and Bases
Take a solution containing 0.01 M carbonic acid (H2CO3) as an example.

37. Solving the equation for individual carbonic species gives

38. The diagram was constructed in the same manner as in the case of monoprotic acids and bases except that:
Slope of the line for [CO32-] changes from +1 to +2 when pH drops below pKA1 ([H+] >> KA1)
Slope of the line for [H2CO3*] changes from -1 to -2 when pH becomes greater than pKA2 ([H+] << KA2)

39. General Procedure for Construction of logC-pH Diagrams for Diprotic Acids and Bases
Draw horizontal line representing log CT
Locate the system point at pH values equal to pKA1 and pKA2 (pKw-pKB1 and pKw-pKB2 for a base)
Draw 45 lines sloping to left and right of each system point to the adjacent system point
The slope of lines changes from -1 to -2 and from +1 to +2
The procedure for construction of logC-pH diagram for polyprotic acids and bases are similar except that the slope of the diagonal changes from +1 to +2 and then from +2 to +3 as it reaches the adjacent system points.

41. Problem 4.25
Draw a log C-pH diagram for a 10-2 M solution of hydrogen sulfide. Assume a closed system. From the diagram, determine the pH for solutions that contain the following:
10-2 M H2S
10-2 M Na2S
0.5  10-2 M HS- and 0.5  10-2 M S2-
0.5  10-2 M H2S and 0.5  10-2 M HS-