Acid Neutralization reactor

1. Acid Neutralization reactor Module 4: Acid neutralization reactor Lecture 3: Acid - base chemistry, pH and the equations for a CSTR for a reacting system Mark J. McCready Chemical Engineering

2. 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

3. 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.

4. Control of a chemical reactor Just proportional control, base flowrate is proportional to error in pH I should have stayed in engineering Setpoint pH is supposed to be 5.

5. 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

6. Control of a chemical reactor (Just) integral control, flow rate of base is related to accumulated error in pH Setpoint = pH of 5

7. Control of a chemical reactor Proportional and integral control Setpoint is 5.

8. Control of a chemical reactor Proportional and integral control Setpoint is 5.

9. 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+)

10. 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

11. 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

12. 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.

13. 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

14. 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.

15. 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+) (cHAc is the concentration you made up)

16. pH calculation • We have three equations that we can combine

17. pH calculation • We have three equations that we can combine

18. pH calculation • We have three equations that we can combine

19. pH numbers • We can put in numbers • Ka = 1.75X 10-5 moles/liter, cHAc = 1 mole/liter

20. 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..

21. 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.

22. 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,

23. 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! Base in 2 HAc

24. Recall 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 • r 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))

25. Model equations for batch titration 2 • We will consider a tank of acetic acid to which we will add NaOH. Let’s look at our mass balance equations Total volume None flows in, just reacts with each OH- that shows up Acetic acid There is no flow in, all of the acetate comes from neutralized HAc Acetate ions In addition to the mass balance equations, we have the equilibrium relation.

26. 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 H3O+. Thus we use up one mole of HAc. • We likewise make a mole of Acetate,

27. 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 H3O+. Thus we use up one mole of HAc. • We likewise make a mole of Acetate,

28. Model equations for batch titration 2 Total volume • Thus our equations become: Acetic acid We have substituted for the reaction term !!! Acetate ions We have substituted for reaction term !!! We then obtain the concentration of H3O+ from the equilibrium relation.

29. Titration curve • If we take 50 ml of 0.1M HAc and add 0.1 M NaOH. Here is the result

30. Yikes, they won’t go 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 • r 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))

31. Model equations for flowing system HAc 1 NaOH 2 • Now consider our stirred tank, neutralization reactor: 3 Total volume Acetic acid 3 Acetate ions In addition to the mass balance equations, we again have the equilibrium relation.

32. 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 H3O+. Thus we use up one mole of HAc. So we again get: • We likewise make a mole of Acetate,

33. Model equations for flowing system 1 2 3 • Now consider our stirred tank, neutralization reactor: Total volume 3 Acetic acid Substituted reaction terms Acetate ions In addition to the mass balance equations, we again have the equilibrium relation.

34. 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; %

35. 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

36. Model equations for flowing system 1 2 3 • Some results • For tank initially filled with HAc • q1 = 10 ml/s, c1=0.85 mole/l • q2= 3 ml/s, c2=0.55 mole/l 3 4.4 V=.3, 1, 5 liters

37. Model equations for flowing system 1 2 3 • Some results • For tank initially filled with HAc • q1 = 10 ml/s, c1=0.85 mole/l • q2= 10 ml/s, c2=0.55 mole/l 3 5.0 V=.3, 1, 5 liters

38. 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

39. 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.

40. 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!

41. Recap:model equations for flowing system 1 2 3 • Now consider our stirred tank, neutralization reactor: Total volume 3 Acetic acid Substituted reaction terms Acetate ions In addition to the mass balance equations, we again have the equilibrium relation.