Tutorial on Visual Minteq 2.30 Intro Acid-base

Tutorial on Visual Minteq 2.30 operation and input/output for simple problems related to acid/base pH and titrations. To install Visual Minteq click on the following and follow the instructions: http://www.lwr.kth.se/English/OurSoftware/vminteq/#download

Generally, it is recommended that you work problems that you are already familiar with, to see how the program works –open all the buttons and windows with a known problem, for example the pH of 0.001 M NaH2PO4 in water at 25 C is 5.13, neglecting activity coefficients. The layout of the program is based around the “component” concept, e.g., TotPO4 plus the notion of the TotH (proton balance condition) for the solution, but you may alternatively use the mass balance plus the charge balance approach to solve all problems; understanding the difference is one the main sources of confusion using Visual Minteq and related software. You can calculate the pH via mass balances (combine TotPO4 plus TotH) or via mass balances (TotPO4) plus the Charge balance condition {Sum(cations) - Sum(anions) = 0.00}.

Option 1. Combine TotPO4 plus TotH. Option 2. Combine TotPO4 plus Charge balance condition.

At a theoretical level it can be shown that the two options are equivalent, but at a practical level they appear to be quite different. Although Option 2 is often more intuitive for students, it should be noted that the Option 1 is often tens to hundreds of times faster to converge and it is clearly the preferred method of professional programmers. If you enter all the components of a solution such that the solution is electrically neutral (e.g. a solution contains only 0.001 M Na3PO4 or 0.001 M Na2HPO4), i.e., actually how it is prepared so that all anions are counterbalanced by cations, then the two methods of calculation will always yield the same result, although the practical input data is often different, as will be illustrated.

First, we will solve a set of problems based around a solution that is 0.001 M TotPO4. Later we will add PCO2, after we learn how to enter data and interpret the output and then do titrations. This problem is similar to Prob. No. 4.46 in Sawyer, McCarty, and Parkin, Chemistry for Environmental Engineers, 5th” McGraw Hill, 2003. Initially, we will work this problem several ways to better understand how the program options works and the meaning of the various Output boxes.

One way to look at solving problems in Visual Minteq is to use the “Calc. from mass

& charge balance” option, Option 2, all the time. This is likely more intuitive and in keeping with the overall theme in many text books. The key to remember in using this option is that in Visual Minteq using the pH “Calc. from mass & charge balance” option, it is always assumed that the TotPO4, etc., can be viewed as having been added as the fully protonated acids, H3PO4, H2CO3, etc., to which strong bases (e.g., Na+, Mg2+, or Ca2+ as NaOH, Mg(OH)2, or Ca(OH)2, etc.) and strong acids (such as Cl-, NO3

-, or SO42-

as HCl, HNO3, or H2SO4, etc.) are added to make a solution to a given pH. In this approach you do not use the TotH concept, although the program will still calculate TotH

Mason Tomson Visual Minteq 11/2/2004 Acid/base Intro. Page 1

value and report it in the “Equilibrated Mass distribution screen”. Examples of different problem types include:

1. Given a set of neutral compounds added to water, calculate the pH (for example,

0.001 M Na2CO3 plus 0.001 M NaOH or 0.001 M HAc);

2. Given total component concentrations and measured or assumed pH, calculate the species (such as, at 8.1 pH and TotPO4 = 0.001 M);

3. Given the solution composition and pHinitial, calculate the acid or base to change the solution to a second pH, pHfinal (such as, initially TotPO4 = 0.001 M and pHinitial = 8.10, calculate the amount of strong acid needed to lower the solution to pHfinal = 6.00);

4. Given a solution composition, calculate the titration curve (e.g., calculate the titration curve of 0.001 M K2HPO4 with 1 M HCl acid); and

5. Given an pHinitial and the solution composition, calculate the titration curve of the solution (given pHinitial = 8.1 with TotPO4 = 0.001 M, calculate the titration curve upon addition of acid or base.)

First, an example of each problem type will be solved step-by-step and then a bit of background discussion and comparison with the “mass balance” only approach will be presented. Input files with corresponding names are attached, but they must be saved in the C:\vminteq\ folder, such as: “C:\vminteq\ Example 1a calc pH.VDA” for the first example. To run the corresponding input file: On the main menu screen click File, Open Input file, scroll down to the file name, click on the file name, select open to open the file and return you to the main menu, click on Run MINTEQ.

1 a. Example 1a calc pH.VDA. Given a set of neutral compounds added to water, calculate the pH (0.001 M Na2CO3 plus 0.001 M NaOH, calculate the pH);

1. Open Visual Minteq software, everything is initialized; 2. pH box click “Calc. from mass & charge balance”; 3. Ionic strength equal 0.00; units Molality; 4. Component window select CO3; and concentration to 0.001 M; 5. Units Molality (m) ≅ Molarity (M) to within about ± 0.3 % for most solutions

and the difference is never of concern to us in any practical problems; 6. Click “add to list”; 7. Component window select Na+; and concentration to 0.003 (i.e., 0.001 +

0.002 M); 8. Click “add to list”; 9. Click Run Minteq and then OK and you are taken to the Output screen:


and the Output screen looks like this:

The calculated pH = 11.056, it took 86 iterations to converge, if you click on Gases the equil. PCO2 = 8.7E-8 atm, the (sum of cations) = (sum of anions) = 2.951 millimoles of charge/L. Since ionic strength is 0.00 the activity coefficients are all 1.00 and therefore the Concentrations = Activities.


10. Click on View species distribution:

Click on Display saturation indexes and note that natron, Na2CO3⋅10H2O is greatly undersaturated. Click on Equilibrated mass distributions:

and observe that the TotCO3 = 0.001 M, TotNa = 0.003 M, and that TotH = -0.001 M. The negative value of TotH means that there is an excess of OH- added equal to the concentration of NaOH. As a component the way to add an excess of OH- is to add a negative value for TotH. This can be done by selecting H+ and typing –0.001 in the Total concentration bar on the main menu:

or in the View/edit list type in –0.001:


At this point, you could calculate the same pH using either pH option. Try it. Notice the dual use of the same symbol, “H+1”, to refer to the concentration of free hydrogen ions in solution, [H+1], as related to pH in the Output screen and to refer to the total concentration of the component hydrogen added to solution, TotH, in the Component screen and on the main menu input. This dual use of the same symbol is a common source of confusion. 1 b. Example 1b pH of HAc.VDA. Data for numerous organic acids is included, but

not normally listed in the Component window on the main menu. To list the available organic acids and calculate the pH of 0.001 M TotAc:

1. Click on Parameters then Various default settings in the main menu and select

the yes button next to Show organic components and then Save and quit:

2. Select Component name box and the following will appear. Set Acetate-1 to 0.001 M.

3. Select Calculate pH from mass and charge balance. 4. Set ionic strength = 0.00 and click Run Minteq:


The pH = 3.907. View the various Output windows.

2. Example 2 Fixed pH.VDA. Given total component concentrations and measured

or assumed pH, calculate the species concentrations in solution (at 8.1 pH and TotPO4 = 0.001 M, calculate the concentrations);

1. Open Visual Minteq software, everything is initialized; 2. pH box click “Fixed at...” and enter 8.10; 3. Ionic strength equal 0.00; units Molality; 4. Component window select PO4; and concentration to 0.001; 5. Click “add to list”; 6. Click Run Minteq then OK and you are taken to:

For example: [PO4-3] = 4.7174E-08 M from the Output menu and note that the Concentrations and activities are the same since we set ionic strength to 0.00. It was not necessary to choose the method to calculate pH, because pH was specified. Since the sum of anions is 1.89 mM greater than the sum of cations, to prepare 0.001 M TotPO4 with a pH = 8.10, one could add 1.89 mM strong base


cations, probably as NaOH, to 1.00 mM H3PO4 –of course there would be numerous other combinations of chemicals that would produce the same result, such as, 1.00 mM of KH2PO4 plus 0.89 mM NaOH, etc.

3. Example 3a(b) amount of acid or base to change pH.VDA. Given total component composition and pHinitial, calculate the acid or base to change the solution to a second pH, pHfinal (initially 8.1 pH and TotPO4 = 0.001 M, calculate the acid to lower the solution pH to 6.00 pH);

1. Open Visual Minteq software, everything is initialized; 2. pH box click “Fixed at...” and enter 8.10 (this is 3 a); 3. Ionic strength equal 0.00; units Molality; 4. Component window select PO4; and concentration to 0.001; 5. Click “add to list”; 6. Click Run Minteq and OK and the Output screen:

7. CBase(1) = 1.8900E-03 - 7.9433E-09 ≈ 1.89 mM 8. pH box click “Fixed at...” and enter 6.00 (this is 3 b); 9. Click Run Minteq, for the second time for the solution;

10. CBase(2) = 1.0595E-03 - 1.0000E-06 = 1.0585E-3 M ≈ 1.06 mM 11. CAcid(Added) = CBase(1) - CBase(2) = 1.89 mM - 1.06 mM = 0.83 mM.

That is, to lower 1 L of 1 mM TotPO4 at 8.10 pH to 6.00 pH would require that you add 0.83 millimoles of strong acid (e.g., 0.83 ml of 1.00 M HCl or 0.415 ml of 1.00 M H2SO4). This type of calculation, in two parts, can be used to solve many practical water chemistry problems of how to change the solution pH from one value to another.

4. Example 4 titration curve.VDA. Given a solution composition, calculate the titration curve (calculate the titration curve of 0.001 M K2HPO4 with 1 M HCl acid);

1. Open Visual Minteq software, everything is initialized; set ionic strength to 0.00 2. Component window select PO4; and concentration to 0.001; 3. click “add to list”; 4. Select K+ and set concentration to 0.002 M and click add, This is the amount of

net K+ to make the solution charge balanced at the starting point. 5. Select pH box “Calc. from mass & charge balance”;


6. Select Cl- and set concentration to 1E-8 and click add –this will make the Cl- ion available to you to output the value of Cl- in the titration (quirk of program);

7. Select tool bar Multiproblem/sweep;

8. Select titration box and click on the “go to titration manager” bar; 9. Click the Titration box, put 40 in no. of titration points, make volume = 1.00 L

and volume of titrant = 0.0001 L (0.1 ml and 0.1 meq./L); 10. Click the box “Start addition on the 2nd step”, this gives you the pH, etc. at the

start of the titration; 11. Type 1.00 in Concentration box and select H+1 and click on “Save and Next” and

select Cl-1 and “Save and Next.” This will add 1/10th of a milli-equivalent of HCl per titration point. Since most solutions are in the range of about 1 to 3 milli-equivalents of weak acid, this will normally give a nice looking titration curve. Adjustments can be made to give finer looking titrations.


12. Click “Save and back to multisweep menu.”; 13. Add “Total dissolved” Cl-1 and H+1 (TotCl and TotH, the total dissolved option

includes the concentrations of titratnt that form complexes, etc. in solution and is generally what you want to use for a titration plot). If we had selected H+1 in the Add comp./species window and Concentration in the Which type? window, the value of free hydrogen ion in solution related to pH would have been output, instead of TotH;

14. Click “Save and Back”; 15. Click Run Minteq. 16. The output menu will appear. Note that the speciation for the 1st titration point

will be shown in the output screen; you can scroll down in the “Select problem No.” box at the top middle and any of the 40 iteration points will be displayed.


17. Select Sweep output and print to Excel, delete Row 2, and drag the pH column to the right of H+1:

18. Plot the data against Cl-1, amount (M) of HCl added. Again, notice that the line labeled “H+1” is actually the TotH value that starts at TotH = 0.001 M. If in the titration sweep window we had selected “H+1 concentration,” instead of “H+1 Total dissolved” we would have gotten [H+]=10-pH. Since this is the titration curve of 0.001 M K2HPO4, there is an inflection point at 0.001 M HCl added at pH ≈ 5. There is not a second inflection point at 0.002 M HCl because the first ionization constant of phosphoric acid is K1 ≈ 10-2 which is moderately strong acid compared with the concentration of TotPO4 = 10-3 M.

0

2

4

6

8

10

0 0.001 0.002 0.003 0.004 0.0050

0.001

0.002

0.003

0.004

0.005

0.006

pHH+1

5. Example 5 titration curve pH.VDA. Given an initial pH and total composition, calculate the titration curve of the solution (given pHinital = 8.1 with TotPO4 = 0.001 M, calculate the titration curve.)


1. Open Visual Minteq software, everything is initialized; 2. pH box click “Fixed at...” 8.100; 3. Ionic strength equal 0.00; units Molality; 4. Component window select PO4; and concentration to 0.001; click add; 5. Select Run Minteq, which will take you to the Output page with Sum of cations =

7.9433E-09 (simply the [H+] concentration) and Sum of anions = 1.8900E-03 M; 6. Select Na+ and set concentration to 0.00189 M and click add, This is the amount

of net Na+ to make the solution charge balanced at the starting point, at 8.10 pH. 7. Select pH box “Calc. from mass & charge balance”; 8. Select Cl- and set concentration to 1E-8 and click add –this will make the Cl- ion

available to you to output the value of Cl- in the titration (quirk of program); 9. Select tool bar Multiproblem/sweep; 10. Select titration box and click on the “go to titration manager” bar; 11. Click the Titration box, put 40 in no. of titration points, make volume = 1.00 L

and volume of titrant = 0.0001 L; 12. Click the box “Start addition on the 2nd step”, this gives you the pH, etc. at the

start of the titration; 13. Type 1.00 in Concentration box and select H+1 and click on “Save and Next” and

select Cl-1 and “Save and Next.” 14. Click “Save and back to multisweep menu.”; 15. Add “Total dissolved” Cl-1 and H+1, as shown above; 16. Click “Save and Back”; 17. Click Run Minteq. 18. The output menu will appear. Note that the speciation for the 1st titration point


19. Select Sweep output and print to Excel and rearrange as before.

0

2

4

6

8

10

0 0.001 0.002 0.003 0.004 0.0050

0.001

0.002

0.003

0.004

0.005

0.006

pHH+1

This plot can also be used to calculate the amount of acid or base to go between any two pH values from 8.100 and below.


The following examples will compare and contrast the use of the mass balance and the charge balance methods of pH calculation. Some discussion will also be included. 6. Example 6 pH mass balances only.VDA. What is the pH of 0.001 M total

phosphoric acid in water?

1. Open Visual Minteq software; everything is initialized; 2. pH box click “mass balances”; 3. Ionic strength equal 0.00 (sets all activity coefficients equal to 1.00); 4. In Component name window select PO4 (in the blank window you can type the

letters “po” and you will be automatically taken to that component; and enter concentration to 0.001;

5. Click “add to list” and the screen will look something like this:

6. Click Run Minteq and then click on OK and the output window will automatically open:


The pH = 10.980, clearly this is not phosphoric acid into water, as will be explained.

7. Click on View species distribution:

8. If you click on Display saturation indices, the screen will be blank, because no minerals are possible. Click on Equilibrated mass distribution:


In the Output Window and there are several things to note in the top boxes in the Output window: the pH = 10.980; Ionic Strength = 0.00; number of iterations = 6; sum of cations = 1.0473E-11 moles of charge/kg water (≈M of charge); sum of anions is = 3.0000E-03 moles of charge/kg water; % dif. in cations vs. anions is 99.999999 %. What the computer solved was the mass balance equations for total phosphate, TotPO4, and the total proton balance for the solution relative to the components, PO4

3-, H2O, and H+. The TotH2O is always assumed to be constant at 55.5 M and is not solved for by the software. That leaves two equations, TotPO4 and TotH (the default option to calculate pH) and two unknowns, [H+] and [PO4

3-] when all the equilibrium constants are substituted into the following equations and ignoring activity coefficients for simplicity:

TotPO4 = [PO4

3-] + [HPO42-] + [H2PO4

1-] + [H3PO4] = 0.001 M and TotH = 1⋅[HPO4

2-] + 2⋅[H2PO41-] + 3⋅[H3PO4] + 1⋅[H+] - 1⋅[OH-] = (-4.1E-12) ≈ 0.00

with

]H[K][OH and ,

K]PO[]H[]HPO[ ,

KK]PO[]H[]POH[ ,

KKK]PO[]H[]POH[ w-

3

342

432

34

2

42321

34

343 +

−−

−−

−=

⋅=

⋅⋅

=⋅⋅

⋅=

The species are given below, as copied to Excel from the output file is:

"Name" "Calc mol" # of H's vs. H2O or PO4-3 "H3PO4" 2.3383688598170800E-16 7.0151065794512400E-16 =+3*H3PO4 "OH-" 9.6145765999262200E-04 -9.6145765999262200E-04 =-1*OH- "HPO4-2" 9.6114004614617200E-04 9.6114004614617200E-04 =+1*HPO4-2"H2PO4-" 1.5879961785328100E-07 3.1759923570656200E-07 =+2*H2PO4-“H+” 1.0472800548262400E-11 1.0472800548262400E-11 =+1*H+

-4.1372413223813100E-12 =TotH

The value of TotH (Sum of protons gained minus the sum of protons lost, i.e., proton balance, or TotH) is zero to within convergence error, either from the above Excel addition or from Visual Minteq “Equilibrated Mass Distribution” window TotH = -4.1372E-12 –when I transferred data from the Output screen to Excel and re-added them the sum was -2.38E-09, which is round off difference in the copy and transfer,


but when the default output file in the Visual Minteq folder (C:\vminteq\vmint.out) is opened and all digits are saved the TotH value is exactly what is listed in the Equilibrated mass distribution window, above. The value of TotPO4 component was 1.000E-03. That is, both mass balance conditions were satisfied! as required when we instructed the program to use the “Mass Balance” equations to solve the problem, because TotPO4 = 0.001 M and TotH ≈ 0.00 ± convergence tolerance of ±10-4 of TotComponent. As can be seen from the top of the “Output” window, the solution is not electrically neutral. That is to say, this solution can not be made up as simply phosphate, PO4

3-, into water. If you added Na+ = 0.003 M as another component, you will get the same answer, except that the solution will be electrically neutral, and the % Charge difference box would be 0.0000 %. Try it. Also, if you select H+ on the Main menu in the components box and make the concentration 0.003 and click on Add to list:

and click on Run the pH = 3.051, the pH of 0.001 M phosphoric acid in water since TotPO4 = 0.001 and TotH = 0.003.

TotPO4 = [PO4

3-] + [HPO42-] + [H2PO4

1-] + [H3PO4] = 0.001 M and TotH = 1⋅[HPO4

2-] + 2⋅[H2PO41-] + 3⋅[H3PO4] + 1⋅[H+] - 1⋅[OH-] = 0.003 M

Try it. 7. Example 7 mass and charge balance.VDA. Next, solve the same problem with the

same input data with pH “Calc. from mass & charge balance” selected:

a. Open Visual Minteq software, everything is initialized; b. pH box click “Calc. from mass & charge balance”; c. Ionic strength equal 0.00; d. Component window select PO4; and concentration to 0.001; e. Click “add to list”; f. Click Run Minteq and OK:


The program will automatically open the Output Window and there are several things to note in the top boxes in the Output window: the pH = 3.051; Ionic Strength = 0.00; number of iterations = 103; sum of cations = 8.8896E-04 moles of charge/kg water (1 L); sum of anions is = 8.8896E-04 moles of charge/kg water; % dif. in cations vs. anions is 0.000001 % (this is because we required charge balance equation to be solved). What the computer solved was the mass balance equation for total phosphate, TotPO4, plus the Charge balance equation for the solution. When all the equilibrium constants are substituted in there are two equations, TotPO4 and Charge balance and two unknowns, [H+] and [PO4

3-]. TotPO4 = [PO4

3-] + [HPO42-] + [H2PO4

1-] + [H3PO4] = 0.001 M and ChargeBalance = Sum Cations – Sum Anions =0.00 ={1⋅[H+]} – {3⋅[PO4

3-] + 2⋅[HPO42-] + 1⋅[H2PO4

1-] + 1⋅[OH-]} = 0.00

As you can see, since the charge balance equation has no strong acid anions, e.g. NO3

-or Cl-, and no strong base cations, e.g., Na+ or 2⋅Ca2+, it is the same as having added 0.001 M H3PO4 to the solution and then calculate the pH, pH = 3.051. Note also that the TotH component calculation, reported in the Equilibrated Mass Distribution window is TotH = 1⋅[HPO4

2-] + 2⋅[H2PO41-] + 3⋅[H3PO4] + 1⋅[H+] -

1⋅[OH-] = 0.003000 M, although this equation was not used in the calculation, because the charge balance option was selected and no strong base cations, Na+, were added. If you add 0.003 M Na+ to the component list on the Main menu, and use Charge balance option to calculate the pH you will get pH = 10.978, as before. Try it.

8. Example 8 Fixed pH. Commonly, as with a water analysis, you don’t know all the

other strong base cation and anion concentrations very well, but you have a measured pH value. At this point you enter a fixed pH in the pH box of e.g. 8.10 pH. Then


only the mass balance equations, excluding TotH, are solved with TotPO4 for the species distributions. At this point the anion/cation imbalance is calculated and reported from which you can calculate much strong base cations or strong acid anions must be added for the solution to be neutral. Example for 0.001 M TotPO4 at 8.10 pH:

a. Open Visual Minteq software, everything is initialized; b. pH box click “Fixed at...” and enter 8.10; c. Ionic strength equal 0.00; units Molality; d. Component window select PO4; and concentration to 0.001; e. click “add to list”; f. click Run Minteq and OK:

The program will automatically open the Output Window and there are several things to note in the top boxes in the Output window: the pH = 8.100; Ionic Strength = 0.00; number of iterations = 0; sum of cations = 7.9433E-09 moles of charge/kg water (1 L); sum of anions is = 1.8900E-03 moles of charge/kg water; % dif. in cations vs. anions is 99.999159 %. What the computer solved was the mass balance equations for total phosphate, TotPO4, using the 8.100 pH. After substituting in the equilibrium constants, the free [PO4

3-] concentration is calculated –see equations listed above: TotPO4 = [PO4

3-] + [HPO42-] + [H2PO4

1-] + [H3PO4] = 0.001 M, or

[PO43-] =

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+++ 1KH

KKH

KKKH001.0

332

2

321

3 = function of only pH


32134

343

3234

2142

334

24

100.8ww

KKK/]PO[]H[]POH[

KK/]PO[]H[]POH[

K/]PO][H[]HPO[

10/K]H/[K]OH[

⋅⋅=

⋅=

=

==

−+

−+−

−+−

−+−

The charge balance is not used, but it is computed from the above and reported at the top of the Output menu and this is often valuable in various calculations.

ChargeBalance = Sum Cations – Sum Anions = ={1⋅[H+]} – {3⋅[PO4

3-] + 2⋅[HPO42-] + 1⋅[H2PO4

1-] + 1⋅[OH-]} = 7.9433E-09 - 1.8900E-03 = -1.8900E-03

This means that to prepare a solution of TotPO4 = 0.001 M at pH = 8.100, you would have to add 0.00189 M Na+ or K+, etc., positively charged strong base cations to the solution, e.g., as 0.00189 M NaOH or KOH to a solution of 0.001 M H3PO4. You could write the formula as follows: 0.001 moles of Na1.89H1.11PO4 per liter. Most likely, what you would have done was to add 0.001 moles of H3PO4 plus 0.00189 moles of NaOH per liter, at this point the net TotH = 0.00111 moles of H+ plus 0.00189 moles of Na+. See next section for verification.

9. Example 9 QC check.VDA. Check on the consistency of the calculations with the

“Calc. from mass & charge balance” option:

a. Open Visual Minteq software, everything is initialized; b. pH box click “Calc. from mass & charge balance”; c. Ionic strength equal 0.00; units Molality; d. Component window select PO4; and concentration to 0.001; click “add to list”; e. Select Na+ and set concentration to 0.00189; f. Click Run Minteq.


The program will automatically open the Output Window and there are several things to note in the top boxes in the Output window: the pH = 8.092 (approx. to 8.100); Ionic Strength = 0.00; number of iterations = 26; sum of cations = 1.8704E-03 moles of charge/kg water (1 L); sum of anions is = 1.8704E-03 moles of charge/kg water; % dif. in cations vs. anions is 0.000008 %. The calculated pH would be exactly 8.100, if you delete all Na-complexes in solution: on the Main menu you click on the drop down menu “Solid phases and excluded species” and “Specify excluded species,” then click on “Aqueous species” and select all species that are Na-complexes (such as NaHPO4

-, etc.). Try it. What the computer solves is the mass balance equations for total phosphate, TotPO4 = 0.001 M, and the charge balance for the solution and since TotNa = 0.00189 M, this mass balance is also included. This approach can be thought of as if TotPO4 is added as 0.001 M H3PO4 and TotNa is added as NaOH.

10. Example 10 QC check.VDA. Check on the consistency of the calculations with the “Calculated from mass balance” option:

a. Open Visual Minteq software, everything is initialized; b. pH box click “Calculated from mass balance”; c. Ionic strength equal 0.00; units Molality; d. Component window select PO4; and concentration to 0.001; click “add to list”; e. Select Na+ and set concentration to 0.00189 and click on Add to list; f. Select H+ and set concentration to 0.00111 and click on Add to list; g. Click Run Minteq and OK:


In the top boxes in the Output window: the pH = 8.092; Ionic Strength = 0.00; sum of cations = 1.8704E-03 moles of charge/kg water (1 L); sum of anions is = 1.8704E-03 moles of charge/kg water, TotPO4 = 0.00100 M, and the mass balance for TotH = 0.00111 M and mass balance for TotNa = 0.00189 M. Essentially all the TotNa is in solution as simply [Na+] ions, although the trace concentrations of complexes are calculated and cause the pH to drop from 8.100 to 8.092.

11. Example 11 titration.VDA. Use the calculate pH from “Calculated from mass

balance” option, first: a. Open Visual Minteq software, everything is initialized; b. pH box click “Calculated from mass balance”; c. Ionic strength equal 0.00; units Molality; d. Component window select PO4; and concentration to 0.001; click add; e. Select H+ and set concentration to 0.00111 and click add; f. Select Cl- and set concentration to 1E-8, or any “small” value, and click add –this

will make the Cl- ion available to you to output the value of Cl- in the titration (quirk of program);

g. Select Na+ and set concentration to 0.00189 and click add; h. Select tool bar Multiproblem/sweep; i. Select titration box and click on the “go to titration manager” bar; j. Click the Titration box, put 40 in State the number of titration steps, make volume

= 1.00 L and volume of titrant = 0.0001 L (0.1 ml or 0.1 meq./L per step); k. Click the box “Start addition on the 2nd step”, this gives you the pH, etc. at the

start of the titration; l. Type 1.00 in Concentration box and select H+1 and click on “Save and Next” and

select Cl-1 and “Save and Next.” Adjustments to number of titration points and volume of titrant can be made to give finer looking titration.


m. Click “Save and back to multisweep menu.”;

n. Add “Total dissolved” Cl-1 and H+1, as below (the total dissolved option includes the concentrations of titratnt that form complexes, in solution and is generally what you want to use):

o. Click “Save and Back”; p. Click Run Minteq and OK; q. The output menu will appear. Note that the speciation for the 1st titration point


r. Click on Select Sweep output and Print to Excel:

Plot the titration against Cl-1. In the Excel Sheet delete the second row and select and move the pH data to the right of the H+ column.


In the graph below, to put TotH on a second axis, click on the diamonds along the bottom of the plot, click on Format, Selected data series, Axis, Secondary –this scales the second axis for the y-axis. The titration curve looks like the following after a little more formatting of the axes.

Titration of 0.001 M TotPO4 starting at 8.1 pH

02468

10

0 0.0005 0.001 0.0015 0.002

HCl M added

pH

0

0.002

0.004

0.006

TotH

, M,

calc

ulat

ed pHH+1

The pH “Calculated from mass balance” and the “Calc. from mass & charge balance” options will both work the same since the initial solution was made to be the actual solution by initially adding TotH = 0.00111 M and TotNa = 0.00189 M. The “Calculated from mass balance” used the TotH equation and added 0.0001 M H+ to each point and calculated the pH. The “Calc. from mass & charge balance” option added 0.0001 M Cl- to the charge balance equation each time and calculated the result, this method is much slower to converge and a couple of the points in latter part of the calculation may not converge completely and might be marked, but they will be very close. Try it.

12. Example 12a(b) titration with PCO2.VDA. Repeat the titration of 0.001 M

TotPO4 starting at 8.100 pH, but include PCO2 = 0.0005 atm.

1. Open Visual Minteq software, everything is initialized; 2. pH box click “Fixed at...” 8.100; 3. Ionic strength equal 0.00; units Molality; 4. Component window select PO4; and concentration to 0.001; click add;


5. Select Gases menu bar and set 0.0005 atm. and select Add and go back Main menu;

6. Select Run Minteq, which will take you to the Output page with Sum of cations = 7.9433E-09 ≈ 0 M (simply the [H+] concentration) and Sum of anions = 2.8539E-03 M, i.e., there is too little cations, such as Na+ = 2.8539E-03 M, in solution. This is the end of Example 12a;

7. This is Example 12b. Select Na+ and set concentration to 2.8539E-03 M and click add, This is the amount of net Na+ to make the solution charge balanced at the starting point.

8. Select pH box “Calc. from mass & charge balance”; 9. Select Cl- and set concentration to 1E-8 and click add –this will make the Cl- ion

available to you to output the value of Cl- in the titration (quirk of program), as before;

10. Select tool bar Multiproblem/sweep and set the titration up, as above; 11. The Output menu will appear. Note that the speciation for the 1st titration point

will be shown in the output screen; you can scroll down in the “Select problem No.” box at the top middle and any of the 40 iteration points will be displayed. Select Sweep output and print to Excel. The following is the plot after a little editing and putting TotCO3 and TotH on the secondary axis, as before.


0

2

4

6

8

10

0 0.001 0.002 0.003 0.004

HCl, M, added

0

0.001

0.002

0.003

0.004

0.005

pH TotCO3 SecAxis TotH SecAxis


13. Example 13a(b,c) Houston Tap Water.VDA. The following is an analysis of

Houston tap water, obtained from the City of Houston Water Quality Labs.

This is Example 13a. The following is the corresponding input into Visual Minteq, using the left-most analysis and note that input concentrations have been input as mg/l. The pH of 8.68 is used as a fixed pH and ionic strength has been calculated. SiO2 is increased by the ratio of H4SiO4(96.1 g/mol)/SiO2(60.1 g/mol) since input is as H4SiO4, not SiO2; TotCO3 input is as CO3 = 73 mg/l⋅{CO3(60 g/mol)/HCO3(61 g/mol)} + 6 mg/l = {71.8 + 6 mg/l} = 77.8 mg/l as CO3. Iron was assumed to be TotFe(III) since the solution is probably aerated, but the results would not change


much if it were assumed to be TotFe(II) or equally divided, but the percentage to precipitate would be quite different. Temp = 62 F = 16.67 C was used.

Click on Run Minteq and OK to Output screen:

The ionic strength is 4.3 mM, which would correspond to 251 mg/l as NaCl vs. the observed value of 222 mg/l TDS, which is reasonable. The close agreement between sum of cations and sum of anions is excellent and may suggest that they adjusted the charge balance by adding Na+ or K+ to match or that the analyses are exceptionally good –in either case the analysis is quite self consistent.


The following is a listing to TotComponent for each component in the solution:

Note that TotH = 1.2537E-03 is calculated, which is the proton balance of all species relative to the Component that was entered. This is Example 13b. Next, it is often necessary to raise the pH to some value to precipitate the hardness ions, mostly TotCa and TotMg, and this is illustrated below by allowing possible solid precipitates of Calcite (CaCO3), Brucite (Mg(OH)2), and Ferrihydrate (Fe(OH)3). On the main menu Click on Solid Phases and Excluded Species, Specify possible solid phases, then select calcite, brucite, and ferrihydrate. Then, on the main menu select Run:


Now the ionic strength is 6.2 mM and Charge difference would require (8.76 mM – 1.47 mM) = 7.29 mM of strong base to be added. In the process of increasing the pH most of the Ca, Mg, and Fe precipitate, as shown:

The new 12 pH equilibrium composition would be:

This is Example 13c. By keeping the “Possible solids to precipitate,” as above, and performing a pH sweep from 7 to 12 pH would produce the following equilibrium amounts of solids left in solution vs. pH.

1.E-071.E-061.E-051.E-041.E-031.E-02

7 8 9 10 11 12 131.E-11

1.E-10

1.E-09

1.E-08

1.E-07

TotCa+2 TotMg+2TotCO3-2 TotFe+3SecAxis


Date post:	02-May-2017
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Tutorial on Visual Minteq 2.30 Intro Acid-base

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