CE 420/520 Environmental Engineering Chemistry
Dr. S.K. Ong
ACID-BASE CHEMISTRY
Examples of acid-base reactions in environmental engineering and science
•
•
•
•
•
pH adjustment and neutralization of water and wastewater are common acid-base reactions
interactions of acid rain with natural bodies of water (rivers, lakes, etc.)
biological activities, in general, are dependent on the acidity of the water. Most organisms prefer a certain
range of hydrogen ion activities under which they grow and flourish.
Hydrolysis reactions of certain organic compounds are usually a function of H +, e.g., triazine
(pesticide) hydrolysis.
Adsorption reactions are usually regulated by nature of the surface that changes with H + ions,
e.g., the charge on oxide surfaces changes with the pH of the solution.
ACID AND BASES
Three different definitions of acids and bases
•
Arrhenius acid-base concept (water system only)
acid - a substance which upon addition to water forms hydrated protons (i.e., H 3O+ or H+)
base - a substance which produces OH- ions upon addition to H2O
(doesn't describe any other solvents beside water)
•
Bronsted-Lowry concept
An acid is a proton donor e.g.,
A base is a proton acceptor e.g.,
•
HCl + H2O ---> H3O+ + Clacid
base
H2O + NH3 ---> NH4+ + OH-
Lewis Concept
- a more comprehensive definition that defines acids and bases in terms of formation of coordinate covalent
bonds
A Lewis acid can accept and share a pair of electrons donated by a Lewis base
Using this definition, metal cations may be defined as acids
e.g.,
H+ +
H 2O
====> H3O+
A base is an electron pair donor.
The more commonly used definition is that of Bronsted-Lowry.
If HA is an acid which donates a proton to form a base A-, then HA and A- is termed as an
acid-conjugate base pair.
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ACID-BASE REACTIONS
A compound that acts as an acid or a base (i.e., accepts or donates protons as in Bronsted-Lowry concept) is termed
as____________. Water is an________________.
Based on the Bronsted-Lowry concept, we can write the following general equation
…………………………………..(1)
where Ka is the equilibrium constant for a reaction in which an acid donates a proton to a water molecule.
Similarly, for a base reaction, we can write:
A- + H2O <===> OH- + HA
……………………………………(2)
where Kb is the equilibrium constant for a reaction in which a base accepts a proton from a water molecule.
Adding reactions (1) and (2), we have
2H2O <==> H3O+ + OH-
or
K = K a • Kb
The Kb value of a conjugate base may be computed if Ka of the acid is known.
•
•
strong acids or bases dissociate completely in water,
weak acid or bases only dissociate partially in water
______ or _______ means strong acid
______ or ______ means strong base
•
•
•
conversely, small Ka values or high values of pKa indicate weak acids.
dividing line between strong and weak acids is at approximately pKa = ______ while for strong bases and
weak bases is at approximately pKb = _______
Examples of strong acids :
HCl, H2SO4 and HNO3
Examples of strong bases:
NaOH, H2S
Examples of weak acids:
________________________________
Examples of weak bases:
________________________________,
Acid-base calculations
What are the final pH of a solution and the concentrations of the various chemical species when one or more acids or
bases is added to water? Various approaches may be employed in finding the pH and the concentrations of various
species in the mixture, i.e.,
Numerical method
Graphical method
(Tableau Method)
Note that the graphical method employs most of the equations used in the numerical method. It is therefore critical
to understand the equations derived for the numerical method. The steps in applying the numerical method or
graphical method are as follows:
2
1.
2.
3.
4.
List all the unknowns
List all equations that inter-relate the unknowns
List mass balances of all species
Do a charge balance of cations or/and anions
Acid-base equations for a weak acid HA.
Assume HA added to distilled water. Determine the pH of the solution.
1.
List all unknowns, i.e., H2O, H+, OH-, HA, AThere are four unknowns excluding H2O.
e.g., for H2CO3, we have water (H2O, H+, OH-) and carbonic acid (H2CO3, HCO3-, CO32-)
2.
List all equilibrium equations interrelating the species.
i.
Dissociation of water
H2O <===> H+ + OH-14
Kw = 10 = [H+] [OH-]
...........................(1)
ii.
Dissociation or equilibrium constant of the acid (or base)
[H  ][ A  ]
Ka 
……………………………………(2)
[HA ]
If we add a salt MA, the basicity constant will suffice.
MA + H2O <===> M+ + HA + OHor
A- + H2O <===> HA + OH-
3.
Mass balances of all species containing a given element
.........................(3)
e.g., for H2CO3, then CT,CO32-
4.
= _____________________________
Charge balance, i.e , the solution must be neutral at all times.
Sum of negative charges = Sum of positive charges
...........................(4)
e.g.,
with H2CO3, we have :
________________________________________
with CaCO3, we have:
________________________________________
We have 4 unknowns and four equations, concentrations of individual species can be found.
Substitute [OH-] in equation (4) for equation (1)
.................................(5)
Substitute [HA] in equation (2) into equation (3)
[A  ] 
C T,A
[H  ]
1
Ka
………………………………………(6)
3
................(7)
Equation 7 is a quadratic equation and [H+] i.e. pH of a weak acid can be determined from the equation.
Acid-base equations for a strong acid.
For a strong acid CT,A
giving
= [A-] (complete dissociation) and [H+] = [OH-] + [A-]
[H+] = CT,A + Kw / [H+]
or
or
[H+]2 - [H+] CT,A
- Kw
= 0
Note that CT,A >> 4 Kw if CT,A > 10-6 M, then pH = CT,A added. Lower concentrations (CT,A < 10-6 M) must
take into consideration the dissociation of water in computing the proton concentrations.
Concentration of the other species can be computed from the following equations.
The concentrations of [A-] is given by equation (6).
[A  ] 
C T,A
................................(6)
[H  ]
1
Ka
[HA] can be obtained by substituting equation (6) into equation (3), i.e.,
C T,A
[HA ] 
................................(8)
Ka
1 
[H ]
Example: Find the pH of a solution containing 1 x 10 -3 M HAc, if the pKa = 4.75. We will solve the problem two
ways: numerically and by approximation.
Numerically. Let us use Newton's method to solve the quadratic equation. Newton's equation is defined as:
where f '(y1) is the differential of f(y1), y1 is the (n-1) th solution for the equation, yn is the nth solution for the
equation. By iteration, y2 is the solution of the quadratic equation if the difference between y2 and y1 is within a
specified value.
We have:
substitute CT,A = 1x 10-3 and Ka = 10 -4.75 = 1.78 x 10-5
then:
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Multiply by 1021 and let y = 1x 107 [H+], therefore for the above two equations, we have:
Use Newton's Method
y2 = y1 - f (y1) / f '(y1)
Make a guess of the initial value of y1 (assume y1 = 1,000), then tabulating the results as shown:
y1
1,000
f(y1)
f '(y1)
y2
_________________________________________________________________
Therefore y = 1x 107 [H+] = 1,248. i.e. [H+] = 1.25 x 10-4
i.e.
pH = 3.9
By Approximation
[H+]3 + Ka [H+]2 - [H+] (CT,A • Ka + Kw ) - Ka • Kw = 0
or
[H+]2 + Ka [H+] - (CT,A • Ka + Kw ) - Ka • Kw / [H+] = 0
Since we are adding an acid, pH can be assumed to be low, i.e. [OH -] = Kw / [H+] ≈ 0
also let us assume that CT,A • Ka >> Kw . Note Kw = 10-14 (a small number in comparison to
1.78 x 10-8). Therefore, we have
solve for
[H+] = 1.248 x 10-4, i.e. pH = 3.9
Check that [H+] >> [OH-]
[OH-] = 8.01 x 10-11 therefore [H+] >> [OH-]
ΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩΩ
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Example: If instead of HAc, NaAc was added, what will be the final pH. Use the same approach:
1.
Unknowns are H2O, H+, OH-, HAc, Ac- and Na+
2.
3.
4.
Equilibrium equations interrelating the species would be the same.
Kw = 10-14 = [H+] {OH-]
...........................(1)


[H ][ Ac ]
Ka 
………………..(2)
[HAc]
Mass Balances of all species containing a given element
CT,Ac
= [Ac-] + [HAc]
.........................(3)
and
CT,Na
= [Na+]
.........................(9)
Charge balance would become:
Sum of negative charges = Sum of positive charges
[Na+] + [H+] = [OH-] + [Ac-]
...........................(10)
Develop quadratic equation based on above equations
GRAPHICAL METHOD
The graphical method uses a plot of pC versus pH.
The y axis is plotted as pC = -log (concentration of a given species) including [H+] and [OH-].
The x axis is pH = - log[H+]
In our earlier example, 1x 10-3 M of a weak monoprotic acid, Hac was added to distilled water. Ka is 1.78 x 10 -5.
As before:
1.
Species:
HAc, Ac-, H+, OH-, H2O
2.
Equilibrium equations.
Kw = 10-14 = [H+] [OH-]
...................
(1)


[H ][ Ac ]
Ka 
........................ (2)
[HAc]
3.
Mass Balances of all species containing a given element
CT,Ac
= [Ac-] + [HAc]
........
............ (3)
4.
Charge balance would become:
[H+] = [OH-] + [Ac-]
........................... (4)
5.
In order to develop the pC-pH diagram, we need to use the equations relating [HAc] and [Ac -]
and the various equilibrium constants. i.e.,
C T, Ac
................................(6)
[Ac  ] 
[H  ]
1
Ka
[ HAc] 
and
C T , Ac
1
................................(8)
Ka
[H  ]
A. The first step in plotting the pC - pH diagram is to plot the [H+] and [OH-] lines.
i.e.,
pH = -log [H+] = pC
at pH = 0, pC = 0
at pH = 14, pC = 14, which is a straight line with a slope of +1
in the case of -log [OH-] or pOH we have
or
i.e.,
Kw = 10-14 = [H+] [OH-]
14 = pH + pOH
pH = 0, pOH = 14
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to pH
and pH = 14, pOH = 0, which is a straight line with a slope of -1
B. The second step is to determine the system points. The system points are equal to the pK values and the pCT,Ac
values for the acid or base or salt that was added to the solution. In this example the system point is (pK a, pCT,Ac) =
(4.75, 3). Locate the system point on your graph.
Using the system point, the diagram may be divided into two regions.
Region 1. pH < pKa
Region 2. pH > pKa
They are:
Let us consider Region 1 for species [HAc] (see equation (8)). In this region, [H +] >> Ka or
1 > (Ka / [H+])
For equation (8)
[HAc] =
CT,Ac
= 10-3
or
pCT,Ac =
p[HAc]
= 3
This is plotted as a straight line (line 1) on the graph from pH = 0 to pK a. This line is terminate at approximately
one pH unit before system point.
In Region 2, pH > pKa, Ka >> [H+]
or
1 < (Ka / [H+])
Equation 8 becomes:
[HAc] 
C T,Ac
Ka

C T,Ac [H  ]

Ka
[H ]
Taking the -log for the above equation.
p[HAc] = pCT,Ac + pH
-
p Ka
For region 2, p[HAc] varies with pH according to the above equation. With regards to pH the slope of the line is 1.
Plot line as shown but terminating the line about one pH unit before the system point.
At the system point the variation of p[HAc] with pH can be plotted as follows. The equilibrium constant gives:
[H  ][ Ac  ]
Ka 
 10  4.75
[HAc]
When pH = pKa then
At this point
Therefore
or
CT,Ac
[Ac-] = [HAc],
= [Ac-] + [HAc]
CT,Ac
= 2 [HAc]
log (CT,Ac) = log[HAc] + log (2)
p[HAc]
= pCT,Ac + 0.3
This equation implies that at the system point, p[HAc] is 0.3 units below the system point.
For species [Ac-], equation (6) is used. The same reasoning is used:
In this region 1, [H+] >> Ka
or
1 > (Ka / [H+])
For equation (6)
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[Ac  ] 
[Ac  ] 
C T, Ac
................................(6)
[H  ]
1
Ka
C T,Ac K a
[H  ]
Taking - log throughout the equation
p[Ac-] = pCT,Ac -
pH
+
pKa
For region 1, p[Ac-] varies with pH according to the above equation. The slope of the line is - 1. Plot line as shown
but terminating the line about one pH unit before the system point.
In Region 2 [H+] << Ka or
For equation (6)
[Ac-]
or
pCT,Ac
1 > ( [H+] / Ka )
=
=
CT,Ac
p[Ac-]
=
=
10-3
3
This plotted as a straight line on the graph from pH = pKa to pH = 14.
unit before system point.
Terminate this line at approximately one pH
At the system point the variation of p[Ac-] with pH is similar to the above discussion on [HAc]. i.e.,
p[Ac-] = pCT,Ac + 0.3
This equation implies that at the system point, p[Ac-] is 0.3 units below the system point.
The pC - pH diagram for HAc is now completed. Note that similar diagrams are drawn even when NaAc is added to
the solution. In order, to determine the pH of the solution, the proton condition or a charge balance is needed.
C. Proton Condition
- this concept can be visualized as comparing the proton level of the solution relative to a reference level or
zero level.
1. Assign zero level to the species being added
2. The sum of the species that have gained H+ must equal the sum of the
concentration of species that have lost an H+ relative to the zero level
i.e.,
∑ excess = ∑ deficit
Example 1
addition of NaAc to distilled water
List all possible species: Na+, Ac-, H+, OH-, HAc, H2O
Zero level: H2O and NaAc (or Na+ and Ac- since NaAc completely dissociates)
Excess
(H+ gain)
[H+ ] (or H3O+)
[HAc]
Therefore:
Zero Level
Deficit
(H+ loss)
[OH-]
[H2O]
[Ac-]
[H+] +
[HAc]
=
[OH-]
Example 2:
NaH2PO4 added to H2O
1.
Species: H3PO4, H2PO4-, HPO42-, PO43-, H+, OH-, H2O, Na+
2.
Zero level: H2O and NaH2PO4 (or Na+ and H2PO4 - since NaH2PO4 is
assumed to completely dissociate)
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Excess
(H+ gain)
[H+]
[H3PO4]
Zero Level
Deficit
(H+ loss)
[OH-]
[HPO42-]
[H2O]
[H2PO4-]
Therefore:
(2H+ loss)
2[PO43-]
[H3PO4 ] + [ H+] = [OH-] + [HPO42- ] + 2[PO43-]
Note that obtaining the proton condition by using a reference level becomes more difficult for a system with a large
number of acids/bases/salts added to the system. Another approach is to obtain the proton condition from the mass
balance and charge balance equation. For example 2:
Charge balance: [ H+] + [Na+] = [OH-] + [H2PO4-] + 2[HPO42- ] + 3[PO43-]
Mass Balance: CT = [Na+] = [H3PO4] + [H2PO4-] + [HPO42- ] + [PO43-]
...........(a)
.............(b)
Substitute [Na+] in (b) into (a),
[ H+] + [H3PO4] + [H2PO4-] + [HPO42- ] + [PO43-]
= [OH-] + [H2PO4-] + 2[HPO42- ] + 3[PO43-]
Therefore:
[H3PO4 ] + [ H+] = [OH-] + [HPO42- ] + 2[PO43-]..........................(c)
Why use the proton condition?
The proton condition is a simpler expression than the charge balance, i.e. 5 variables
(c)) vs. 6 variables (equation (a)) meaning some species need not be plotted on the diagram.
(eqn.
Returning to the problem, the proton condition for HAc solution is:
Zero level: H2O and HAc
Excess
(H+ gain)
[H+ ]
Zero Level
Deficit
(H+ loss)
[OH-]
[Ac-]
[H2O]
[HAc]
Therefore, proton condition is:
[H+ ] = [OH-] + [Ac-]
To determine the pH, the above equation must be satisfied.
Start with the left-hand side of the proton condition that has one term i.e. [H +]. Solution to the problem must lie on
this line.
Point A: [H+] = [HAc], note that [HAc] is not in the proton condition therefore this point is not a solution.
Point B: [H+] = [Ac-], check to see the contribution of the other terms, i.e., [OH -] is much smaller than [Ac-].
Point C: [H+] = [OH-], this point is not the solution since the sum of the concentrations of the species on the right
hand side of the proton condition is much higher than the species on the left hand side.
Therefore the solution is point B. The pH at this point is 3.8.
Comments: The advantage of using the graphical method is that the same diagram may be used to solve similar
problems of different concentrations and different conjugate compounds of the same acids.
Example: Instead of HAc, 10-3 M of NaAc was added, the same diagram may be used.
i.e., The proton condition is now (see example 1 above):
[H+] +
[HAc]
=
[OH-]
This is satisfied at pH _________.
9
CE 520 Environmental Engineering Chemistry
Dr. S.K. Ong
ACID- BASE EQUILIBRIA (Cont'd)
Example:
Add 10-3 M NaAc pKa = 4.75
Add 10-4 M HF
pKa = 3.25
Find the equilibrium pH
Plot the system independently
Proton condition:
Excess
H+
HAc
i.e.
Zero Level
H2O
Na Ac
HF
Deficit
OHF-
[H+] + [HAc] = [OH-] + F(See Figure)
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