Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Module 4:Acid neutralization
reactorLecture 3:
Acid - base chemistry, pH and the equations for a CSTR for a
reacting system
Mark J. McCreadyChemical Engineering
Acid Neutralization
reactor
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Outline for today
Review: Feedback control of a chemical reactor
Proportional and integral control pH and pH measurement
We will monitor progress of reaction this way pH is a design criterion for outlet stream
Weak acid chemistry and buffer solutions Acetic acid dissociation Neutralization reaction
Design equations for the chemical reactor
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Control of a chemical reactor
The next few slides are a photo gallery showing the expected behavior of a chemical reactor with Proportional control
base flowrate is proportional to error in pH Integral control
flow rate of base is related to accumulated error in pH
Proportional and integral control together.
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Control of a chemical reactor
Setpoint pH is supposed to be 5.
Just proportional control, base flowrate is proportional to error in pH
I should have stayed in
engineering
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Control of a chemical reactor
Just proportional control, base flowrate is proportional to error in pH
Setpoint pH is supposed to be 5.We get a better setpoint by using less gain, but the response is slower
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Control of a chemical reactor
(Just) integral control, flow rate of base is related to accumulated error in pH
Setpoint = pH of 5
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Control of a chemical reactor
Proportional and integral control
Setpoint is 5.
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Control of a chemical reactor
Setpoint is 5.
Proportional and integral control
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Dissociation of water
You may recall that any aqueous solution (or “pure” water) contains hydronium and hydroxide ions because of the dissociation of water. This equilibrium relation is:
If the solution is neutral, the concentrations of the two ions are equal. If not, solution is either basic (more OH-) or acidic (more H3O
+)
2H2O H3O++OH −
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Dissociation of water
We keep track of this dissociation using dissociation constants. For example:
At 25oC, the value for Kw is 1.01 X 10-14 mole2/l. At 100oC, the value for Kw is 5 X 10-13 mole2/l. Hydronium ion concentration of an aqueous solution is often
very important and thus it is useful to be able to measure and easily characterize it.
To do so we talk about pH (hydrogen potential) as
Kw = H3O
+[ ]OH −
[ ]
pH =−log10[H 3O+]
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
How can we measure pH ?
Modern electrodes and electronics allow measurement of pH (except at extreme values) quite easily.
You could make mistakes, but these should be pretty small in this experiment
Electrolyte solutionKCl, pH=7Ag/AgCl reference cellspecial membraneglasspH meter6.5
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Working principle of a pH
electrode If you place two solutions with different
chemical activities on opposite sides of a permeable membrane, an electrical potential will form that can be compared to a standard (i.e., known) voltage.
The way this is used for measurement is there will be a change in potential if the outside pH changes.
permeable membrane,special glassH3O+ activity=a1H3O+ activity=a2outsideinsidewire to pH meter
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Workings of an electrode
The key to a pH electrode is a thin membrane of special glass. Glass is sensitive to H3O
+ ions, but not other singly charged ions.
pH is then determined from the potential between the pH electrode and a standard
reference electrode
Electrolyte solutionKCl, pH=7Ag/AgCl reference cell
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Acid-Base Chemistry
In this project we will use a weak acid (acetic, “HAc”) and a strong base. The equilibrium relation is
The dissociation constant for acetic acid at room temperature is Ka = 1.75X 10-5
moles/liter.
HAc +H2O H3O++Ac−
Ka =
[H3O+][Ac−]
[HAc]
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
pH of acid solution
If we want the pH of our acetic acid solution we start with the equilibrium relation
We want [H3O+] and we know Ka. Thus we need two more
equations to get [Ac-] and [HAc] We have available the mass balance for total acetate.
We can also be pretty sure that all of the [H3O+] comes from
the acetic acid (it is not from the little that would be present
in pure water) so that (each Ac- must have an H3O
+)
Ka =
[H3O+][Ac−]
[HAc] HAc +H2O H3O
+ +Ac−⎛ ⎝
⎞ ⎠
cHAc =[Ac−]+[HAc]
[Ac−]=[H3O+]
(cHAc
is the concentration you made up)
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
pH calculation
We have three equations that we can combine
Ka =
[H3O+][Ac−]
[HAc] cHAc =[Ac−]+[HAc] [Ac−]=[H3O
+]
Ka =[H3O
+][Ac−]
cHAc −[Ac−]
Ka =[H3O
+][H3O+]
cHAc −[H3O+]
Ka =
[H3O+]2
cHA −[H3O+] [H3O
+]2 +Ka[H3O+]−KacHA =0
[H3O
+]=−Ka + Ka
2+4KacHA
2
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
pH calculation
We have three equations that we can combine
Ka =
[H3O+][Ac−]
[HAc] cHAc =[Ac−]+[HAc] [Ac−]=[H3O
+]
Ka =[H3O
+][Ac−]
cHAc −[Ac−]
Ka =[H3O
+][H3O+]
cHAc −[H3O+]
Ka =
[H3O+]2
cHA −[H3O+] [H3O
+]2 +Ka[H3O+]−KacHA =0
[H3O
+]=−Ka + Ka
2+4KacHA
2
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
pH calculation
We have three equations that we can combine
Ka =
[H3O+][Ac−]
[HAc] cHAc =[Ac−]+[HAc] [Ac−]=[H3O
+]
Ka =[H3O
+][Ac−]
cHAc −[Ac−]
Ka =[H3O
+][H3O+]
cHAc −[H3O+]
Ka =
[H3O+]2
cHA −[H3O+] [H3O
+]2 +Ka[H3O+]−KacHA =0
[H3O
+]=−Ka + Ka
2+4KacHA
2
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
pH numbers
We can put in numbers Ka = 1.75X 10-5 moles/liter, cHAc = 1
mole/liter
[H3O
+]=−Ka + Ka
2+4KacHAc
2
[H3O
+]=−1.75X10−5 + (1.75X10−5)2 +4(1.75X10−5)(1)
2
[H3O+]=0.0042moles / liter
pH =−
log(0.0042)log(10)
pH =2.38
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
An Aside
So, you don’t like lectures !!! If you have come to all of them, so far you have been to
about 26 lectures, there are 6 remaining.
You would have been to 81% of the total classes so far..
19%
81%
remaining
done
0 5 10 15 20 25 30
remaining
done
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Buffer solutions
Once you begin to titrate the acetic acid with sodium hydroxide, you begin producing sodium acetate and water.
The presence of both a weak acid and its “conjugate base” (i.e., sodium acetate) constitutes a “buffer solution”. It is called this because it resists changes in pH caused by the addition of a strong base or pure water.
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
pH of buffer solutions
Calculation of pH for buffer solutions is even easier than for just a weak acid.
We have the same relation,
Suppose we know the acid concentration: 1 moles/liter and the acetate concentration: 0.5 mole/liter
We would expect that [Ac-] = 0.5 and [HAc]=1,
Ka =
[H3O+][Ac−]
[HAc]
Ka =
[H3O+][.5]
[1] [H3O
+]=1.75X10−5 [1][.5]
pH =−log10(3.5X10−5)=4.5
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Batch titration
How can we model a simple batch titration? We will take fixed volume of acid and add
base until the solution is neutralized.
Mass Balance equations -- of course!
2
Base in
HAc
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
RecallRecall Mass Balances
General mass balance equation for a fixed control volume Rate of Accumulation = Rate In - Rate Out + Production by reaction- Consumption by reaction
Overall
Component mass (mole) balance
dρVdt
=∑jqj ρj
masstime
⎛
⎝ ⎜
⎞
⎠ ⎟
dciVdt
=∑jqjc j i −riV
molestime
⎛
⎝ ⎜
⎞
⎠ ⎟
j - density of stream j, (mass/length3 )qj -- volumetric flow rate of stream j, (length3 /time)V -- active volume of reactor,(length3)
cji-- molar concentration of species i in stream j, (moles/ length3 )ri -- molar reaction rate per volume (moles/ (length3 -time))
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Model equations for batch titration
We will consider a tank of acetic acid to which we will add NaOH. Let’s look at our mass balance equations
dV
dt=q2
dVcHAc
dt=−rHAcV
dVcAc−
dt=r
Ac−V
Total volume
Acetic acid
Ka =
[H3O+][Ac−]
[HAc]
Acetate ions
None flows in, just reacts with each OH- that shows up
There is no flow in, all of the acetate comes from neutralized
HAc
[H3O
+] =Ka[HAc]
[Ac−]
In addition to the mass balance equations, we have the equilibrium relation.
2
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
batch titration (cont.)
The problem is, what are the reaction terms? Let’s do a mass balance for OH-. We can see that it comes in
by our inlet stream and it will all react away until all of the acid is used up. Further, the concentration of OH- in the reactor will be almost 0 (solution is acidic) until the acid is used up.
For every mole of OH- used up we must react one mole of H
3O+. Thus we use up one mole of HAc.
We likewise make a mole of Acetate,
dVcOH−
dt=0=q2c2OH − −rOH −V q2c2OH − =rOH −V
rHAV =rOH −V =q2c2OH −
rA−V =rOH −V =q2c2OH −
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
batch titration (cont.)
The problem is, what are the reaction terms? Let’s do a mass balance for OH-. We can see that it comes in
by our inlet stream and it will all react away until all of the acid is used up. Further, the concentration of OH- in the reactor will be almost 0 (solution is acidic) until the acid is used up.
For every mole of OH- used up we must react one mole of H
3O+. Thus we use up one mole of HAc.
We likewise make a mole of Acetate,
dVcOH−
dt=0=q2c2OH − −rOH −V q2c2OH − =rOH −V
rHAV =rOH −V =q2c2OH −
rA−V =rOH −V =q2c2OH −
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Model equations for batch titration
Thus our equations become:
dVdt
=q2
dVcHAc
dt=−q2c2OH −
dVcAc−
dt=q2c2OH −
Total volume
Acetic acid
Acetate ions
We have substituted for the reaction term !!!
[H3O
+] =Ka[HAc]
[Ac−]
We then obtain the concentration of H3O+ from the equilibrium relation.
We have substituted for reaction term !!!
2
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Titration curve
If we take 50 ml of 0.1M HAc and add 0.1 M NaOH. Here is the result
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Yikes, they won’t go Yikes, they won’t go away!!away!!
General mass balance equation for a fixed control volume Rate of Accumulation = Rate In - Rate Out + Production by reaction- Consumption by reaction
Overall
Component mass (mole) balance
dρVdt
=∑jqj ρj
masstime
⎛
⎝ ⎜
⎞
⎠ ⎟
dciVdt
=∑jqjc j i −riV
molestime
⎛
⎝ ⎜
⎞
⎠ ⎟
j - density of stream j, (mass/length3 )qj -- volumetric flow rate of stream j, (length3 /time)V -- active volume of reactor,(length3)
cji-- molar concentration of species i in stream j, (moles/ length3 )ri -- molar reaction rate per volume (moles/ (length3 -time))
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Model equations for flowing
system Now consider our stirred tank, neutralization reactor:
dVdt
=q1 +q2 −q3
dVc3HAc
dt=q1c1HAc −q3c3HAc −rHAcV
dVc3Ac−
dt=−q3c3Ac− +r
Ac−V
Total volume
Acetic acid
Ka =
[H3O+][Ac−]
[HAc]
Acetate ions
In addition to the mass balance equations, we again have the equilibrium relation.
3
3
21 NaOH
HAc
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Flowing system model (cont.)
Again, we don’t yet know the reaction terms. What are they?
Again we do a mass balance for OH-. We can see that it comes in by our inlet stream and it will all react away until all of the acid is used up. Further, the concentration of OH- in the reactor will be almost 0 (solution is acidic) until the acid is used up.
For every mole of OH- used up we must react one mole of H
3O+. Thus we use up one mole of HAc. So we again get:
We likewise make a mole of Acetate,
dVcOH−
dt=0=q2c2OH − −rOH −V q2c2OH − =rOH −V
rHAV =rOH −V =q2c2OH −
rA−V =rOH −V =q2c2OH −
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Model equations for flowing
system Now consider our stirred tank, neutralization reactor:
dVdt
=q1 +q2 −q3
dVc3HAc
dt=q1c1HAc −q3c3HAc −q2c2OH −
dVc3Ac−
dt=−q3c3Ac− +q2c2OH −
Total volume
Acetic acid
Ka =
[H3O+][Ac−]
[HAc]
Acetate ions
In addition to the mass balance equations, we again have the equilibrium relation.
3
3
21
Substituted reaction terms
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Matlab code to solve the equations
%titration_cstr.m %this m file solves the "titration" problem for a stirred tank with %two input streams %The volume can remain constant or vary depending upon the initial conditions. % q1 = 10/1000; %acid flow rate in l/s c1 = .85; %acid concentration in moles/liter q2 = 10/1000; %base flow rate in l/s c2 = .55;%base concentration in moles/liter
q3(1) = q1 + q2;% initial exit flow rate
v(1) =1; %initial volume Vmax = 1; %maximum volume ka = 1.75 *10^-5;%dissociation constant for acetic acid ch(1) = (-ka+sqrt(ka^2+ 4*c1*ka))/2;%initial H3O+ concentration ph(1) = -log(ch(1))/log(10); %initial pH
t(1) = 0; %set the initial time to 0 cv1(1) = .85; %initial concentration*volume for the acetic acid cv2(1) = 0; %initial concentration * volume for the acetate. index = 3000;% number of time steps to take dt = .1; %
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Matlab code to solve these
equations (pg2) % here is the loop for i=1:index t(i + 1) = t(i) + dt; %advance the time counter v(i + 1) = v(i) + (q1 + q2 - q3(i))*dt; %get a new volume with Euler int. if v(i + 1) < Vmax %check to see if overflowing q3(i + 1) = 0; %if not, don't change flow rate else q3(i + 1) = q1 + q2 ; %if so, set outlet flow equal to inflow end %find the new V*c1 using Euler integration cv1(i + 1) = cv1(i) + (q1*c1 - q2*c2 - q3(i)*cv1(i)/v(i))*dt; %find the new V*c2 using Euler integration cv2(i + 1) = cv2(i) + ( q2*c2 - q3(i)*cv2(i)/v(i))*dt; ch(i + 1) = ka*cv1(i + 1) /cv2(i + 1) ; %calculate the new H3O+
concentration ph(i + 1) = -log(ch(i + 1))/log(10); %calculate the pH
end
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Model equations for flowing
system
Some results For tank initially filled with HAc q
1 = 10 ml/s, c
1=0.85 mole/l
q2= 3 ml/s, c
2=0.55 mole/l
3
3
21
V=.3, 1, 5 liters
4.4
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Model equations for flowing
system
Some results For tank initially filled with HAc q
1 = 10 ml/s, c
1=0.85 mole/l
q2= 10 ml/s, c
2=0.55 mole/l
3
3
21
V=.3, 1, 5 liters
5.0
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Recap pH
We keep track of this dissociation using dissociation constants. For example:
At 25oC, the value for Kw is 1.01 X 10-14 mole2/l. At 100oC, the value for Kw is 5 X 10-13 mole2/l. Hydronium ion concentration of an aqueous solution is often
very important and thus it is useful to be able to measure and easily characterize it.
To do so we talk about pH (hydrogen potential) as
Kw = H3O
+[ ]OH −
[ ]
pH =−log10[H 3O+]
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
recap pH electrode
If you place two solutions with different chemical activities on opposite sides of a permeable membrane, an electrical potential will form that can be compared to a standard (i.e., known) voltage.
The way this is used for measurement is there will be a change in potential if the outside pH changes.
permeable membrane,special glassH3O+ activity=a1H3O+ activity=a2outsideinsidewire to pH meter
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Recap:Reaction terms
To get the reaction terms for Ac- and Hac, we do a mass balance for OH- and determine that there will never be a significant amount of OH- present thus:
Rate of OH- reaction is the rate at which it flows in This is also the rate at
which HAc reacts away and the rate at which Ac- is produced!
rHAV =rOH −V =q2c2OH −
rA−V =rOH −V =q2c2OH −
Introduction to Engineering Systems
Copyright ©2001, University of Notre Dame
Module 4- Acid Neutralization Reactor
Recap:model equations
for flowing system
Now consider our stirred tank, neutralization reactor:
dVdt
=q1 +q2 −q3
dVc3HAc
dt=q1c1HAc −q3c3HAc −q2c2OH −
dVc3Ac−
dt=−q3c3Ac− +q2c2OH −
Total volume
Acetic acid
Ka =
[H3O+][Ac−]
[HAc]
Acetate ions
In addition to the mass balance equations, we again have the equilibrium relation.
3
3
21
Substituted reaction terms