Tutorial on Visual Minteq 2.50 operation and input/output for simple problems related to acid/base ph and titrations.

Transcription

1 Tutorial on Visual Minteq 2.50 operation and input/output for simple problems related to acid/base ph and titrations. 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 M NaH 2 PO in water at 25 C is 5.13, neglecting activity coefficients. The layout of the program is based around the component concept, e.g., TotPO 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 TotPO plus TotH) or via mass balances (TotPO) plus the Charge balance condition {Sum(cations) - Sum(anions) = 0.00}. Option 1. Combine TotPO plus TotH, i.e., all as mass balances. Option 2. Combine TotPO 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 M Na 3 PO or M Na 2 HPO ), 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. In the mass balance approach, used in Visual Minteq all species are formed from a minimum set of components. By common assumption in Visual Minteq, water, H 2 O, and hydrogen ion, H +, are always considered to be components in solutions and then hydroxide is made by subtracting a hydrogen ion from a water molecule, (OH - = H 2 O - H + ) obviously, you could have used as components: H + and OH - or OH - and H 2 O, but custom is to use H + plus H 2 O. Water is always assumed to be constant at [H 2 O] = 1.00 and is commonly dropped from the table of components and species. In principle for a given solution, components can be just about any set of the species of a compound (e.g., for a solution containing phosphate, either H 3 PO,aq, or H 2 PO 1-, or HPO 2-, or PO 3- could be used to build all the other phosphate species by adding or subtracting H + ions.) In Visual Minteq the specific species used as the component is specified and listed on the opening page under the menu called components. Generally, components are the fully dissociated species. For example, PO 3- is used to build all phosphate species; Fe 3+ is used to make all ferric iron species, Ca 2+ is used for all calcium species (e.g., Ca 2+ = 1Ca 2+, CaOH 1+ = 1Ca H 2 O - 1H +, or CaPO 1- = 1Ca PO 3- ) Exceptions of using the fully dissociated species as the component species are: HS 1- is used for all sulfide species, H 3 AsO 3 is used to build all arsenous (arsenic in the +3 oxidation state) species, and a few others. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 1

2 All species are made from a combination of components, as illustrated in the following table for a solution made by adding NaH 2 PO to pure water. The components are H 2 O, H +, Na +, and PO 3-. All species, X, are made by adding these components together; the integer coefficients can be positive, negative, or zero in the reaction to make the species: X = n 1 H 2 O + n 2 H + + n 3 Na n PO n i = 0, ±1, ±2,... with equilibrium constant, K eq [X] = K eq [H 2 O] n 1 [H+ ] n 2 [Na+ ] n 3 [PO 3- ] n where [X] represents the concentration of X [X] = K eq [H + ] n 2 [Na+ ] n 3 [PO 3- ] n, assuming [H 2O] = First, we will solve a set of problems based around a solution that is M TotPO. Later we will add P CO2, after we learn how to enter data and interpret the output and then do titrations. This problem is similar to Prob. No..6 in Sawyer, McCarty, and Parkin, Chemistry for Environmental Engineers, 5 th McGraw Hill, Initially, we will work this problem several ways to better understand how the program options works and the meaning of the various Output boxes. Components from which all species are made. Equilibrium constants from Visual Minteq along with equation for each species conc. No. Species H 2 O H + Na + 3- PO logk eq Equation for each species: N X n 1 n 2 n 3 n logk eq [X]=10 logkeq [H 2 O] n1 [H + ] n2 [Na + ] n3 [PO 3- ] n 1 H 2 O [H 2 O]= [H 2 O] 1 [H + ] 0 [Na + ] 0 [PO 3- ] 0, or [H 2 O]=1 [H 2 O], which is taken to be constant. 2 H [H + ]= [H 2 O] 0 [H + ] 1 [Na + ] 0 [PO 3- ] 0 =1 [H + ] 3 1- H 2 PO [H 2 PO 1- ]= [H 2 O] 0 [H + ] 2 [Na + ] 0 [PO 3- ] 1, or [H 2 PO 1- ]= [H + ] 2 [PO 3- ] 1 * H 3 PO,aq [H 3 PO,aq ]= [H 2 O] 0 [H + ] 3 [Na + ] 0 [PO 3- ] HPO [HPO 2- ]= [H 2 O] 0 [H + ] 1 [Na + ] 0 [PO 3- ] 1 6 Na [Na + ]= [H 2 O] 0 [H + ] 0 [Na + ] 1 [PO 3- ] 0 7 Na 2 HPO,aq [Na 2 HPO,aq ]= [H 2 O] 0 [H + ] 1 [Na + ] 2 3- [PO 8 1- Na 2 PO [Na 2 PO 1- ]= [H 2 O] 0 [H + ] 0 [Na + ] 2 [PO 3- ] 1 9 NaH 2 PO,aq [NaH 2 PO,aq ]= [H 2 O] 0 [H + ] 2 [Na + ] 1 3- [PO ] 1 ] 1 Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 2

3 10 NaHPO [NaHPO 1- ]= [H 2 O] 0 [H + ] 1 [Na + ] 1 [PO 3-11 NaOH aq [NaOH aq ]= [H 2 O] 1 [H + ] -1 [Na + ] 1 [PO 3- ] NaPO [NaPO 2- ]= [H 2 O] 0 [H + ] 0 [Na + ] 1 [PO 3- ] 1 13 OH [OH 1- ]= [H 2 O] 1 [H + ] -1 [Na + ] 0 [PO 3- ] 0, or [OH 1- ]= [H 2 O] 1 [H + ] -1, or = [H + ][OH 1- ] with [H 2 O] = PO [PO 3- ]= [H 2 O] 0 [H + ] 0 [Na + ] 0 [PO 3- ] 1 * The subscript aq is used to emphasize that the species is in water as the electronically neutral species, e.g., H 3 PO,aq is [H 3 PO ±0 ] and that the species does not refer to a separate phase of phosphoric acid. Similarly, NaOH aq (Reaction No. 11) refers to the electronically neutral complex in solution [NaOH ±0 ] and not solid sodium hydroxide. ] 1 Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 3

4 Hydroxide, OH 1-, always present in a water solution, is made from water minus one proton (Reaction No. 13, in the Table). The reaction is: [OH - ] = 1 H 2 O 1 H Na PO 3-, or [OH - ] = 1 H 2 O -1 H + with K 13 = , or [OH 1- ]= [H 2 O] 1 [H + ] -1 [Na + ] 0 [PO 3- ] 0, or [OH 1- ]= [H2O] 1 [H + ] -1 and after rearrangement and using [H 2 O] = = [H + ] [OH 1- ] This is the normally reported ionization constant of water at 25 C and 1 atm. HPO 2- (Reaction No. 5) is made as follows: HPO 2- = 0 H 2 O + 1 H Na PO 3- [HPO 2- ] = K 5 [H 2 O] 0 [H + ] 1 [Na + ] 0 [PO 3- ] 1 with K 5 = K 5 = [H ][PO ] = 2 [HPO ] This is the "stability" constant for HPO. The inverse is the normally reported third ionization constant for phophoric acid at 25 C and 1atm.: [HPO ] 10 = + 3 [H ][PO ] NaHPO 1- (Reaction 10) can be used to illustrate that all species are made from the same components and that the listed reported equilibrium constants are sometimes different from what are commonly reported: K K 10 PO Na Na 3 -. plus HPO Note : st.const. NaHPO 1- = 0 H 2 O + 1 H Na PO [NaHPO 1- ] = K 10 [H 2 O] 0 [H + ] 1 [Na + ] 1 [PO 3- ] 1 with K 10 = Normally, textbooks 2- + HPO + HPO 2 = 10 2 = NaHPO NaHPO 1 1 This is the "stability" constant for the fomation of NaHPO + = 10 [H ][Na ][PO = [NaHPO ] 3 with K 1 ] 2 [NaHPO = + [Na ][HPO which correpond to the reaction : ] report the stability constant as a combination of Na stability constant K = ] K = with stability constant, K 1 - st.const. from Na = , H +, and + Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page

5 The equilibrium constant values, normally reported in text books, can be obtained by multiplying and dividing Visual Minteq constants. Deleting the water column, since [H 2 O] is assumed to be constant and equal to one (1.00), and deleting the listing of equilibrium constants reactions, for simplicity: In solution at equilibirum: Compoents: No. Species H + Na + 3- PO LogK eq 2 H H 2 PO * H 3 PO,aq HPO Na Na 2 HPO,aq Na 2 PO NaH 2 PO,aq NaHPO NaOH aq NaPO OH PO Input compounds to make the solution, assume 100 mg/l NaH 2 PO ( M of NaH 2 PO = 0.100g/120g mole -1, Problem.21 in Sawyer McCarty and Parkin, 5 th ed.: H + Na + 3- PO Molarity Compound(s ) NaH 2 PO Tot NaH2PO = At equilibrium there must be a mass balance for each of the three components (H +, Na +, and PO 3- ): Σ(of all species of each component at equilibrium) = Σ(of all input for that component): 1. H + : 1 [H + ]+2 [ H 2 PO 1- ]+3 [ H 3 PO,aq ]+1 [ HPO 2- ]+1 [ Na 2 HPO,aq ]+ 2 [NaH 2 PO,aq ]+1 [ NaHPO 1- ]-1 [ NaOH aq ]-1 [ OH 1- ] = 2 (0.001M), or 1 [H + ] + 2 [ H 2 PO 1- ] + 3 [ H 3 PO,aq ] + 1 [ HPO 2- ] + 1 [ Na 2 HPO,aq ]+ 2 [NaH 2 PO,aq ] + 1 [ NaHPO 1- ] 1 [ NaOH aq ] 1 [ OH 1- ] = 2(Tot NaH2PO ) Use the above species equalities to express all species in terms of the three component concentrations, [H + ], [Na + ], and [PO 3- ]: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 5

6 [H + ] [H + ] 2 [PO 3- ] [H + ] 3 [PO 3- ] [H + ][PO 3- ] [H + ][Na + ] 2 [PO 3- ] [H + ] 2 [Na + ][PO 3- ] [H + ][Na + ][PO 3- ] [H + ] -1 [Na + ] [H + ] -1 = M 2. Na + : 1 [Na + ]+2 [ Na 2 HPO,aq ]+2 [ Na 2 PO 1- ]+1 [ NaH 2 PO,aq ]+1 [ NaHPO 1- ]+ 1 [ NaOH aq ]+1 [ NaPO 2- ] = 1 ( Tot NaH2PO ) After substituting, as above: [Na + ] [H + ][Na + ] 2 [PO 3- ] [Na + ] 2 [PO 3- ] [H + ] 2 [Na + ][PO 3- ] [H + ] 1 [Na + ][PO 3- ] [H + ] -1 [Na + ] [Na + ][PO 3- ] = M 3. PO 3- : 1 [ H 2 PO 1- ]+1 [ H 3 PO,aq ]+1 [ HPO 2- ]+1 [ Na 2 HPO,aq ]+1 [ Na 2 PO 1- ]+ 1 [NaH 2 PO,aq ]+1 [ NaHPO 1- ]+1 [ NaPO 2- ]+1 [ PO 3- ] = 1 ( Tot NaH2PO ) [H + ] 2 [PO 3- ] [H + ] 3 [PO 3- ] [H + ][PO 3- ] [H + ][Na + ] 2 [PO 3- ] [Na + ] 2 [PO 3- ] [H + ] 2 [Na + ][PO 3- ] [H + ][Na + ][PO 3- ] [Na + ][PO 3- ] +[ PO 3- ] = 1 ( M) Each and every species in the solution is made by combining the three components, H +, Na +, and PO 3- note that water concentration, [H 2 O], is assumed to be constant. Therefore, this must always produce three equations and three unknowns that can always be solved by matrix methods. This is the mass balance method on the input sheet in Visual Minteq. These three equation and three unknowns are solved simultaneously in Visual Minteq. The input screen looks like this as the phosphate is being entered. Note that the ionic strength is set to zero; this is so that the calculated values will be the same as those in most text books. Also, Molarity (moles per liter of solutions) is assumed to be the same as Molality (moles per kg of solvent), this is accurate to about ±0.3%, or better, for most solutions of interest. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 6

7 To add the hydrogen component note that 2 moles of hydrogen ions must be added per mole of phosphate, i.e., Tot NaH2PO = 2 ( M). Sodium ion concentration is simply equal to Tot NaH2PO = M. Then clicking on the Run Minteq button, yields the following output screen: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 7

8 From the Output screen the concentrations of the three species that were used as components are: 1. [H + ] = 6.85E-6 M = M (ph = 5.165); 2. [Na + ] = 8.32E- M; and 3. [PO 3- ] =.70E-13 M. From the equilibrium concentration of these three component species, the concentrations of all species can be calculated using the equilibrium constants in the table, e.g., [ H 2 PO 1- ] = [H + ] 2 [PO 3- ] = (6.85E-6) 2 (.70E-13) = 8.23E- M. The concentration of the sodium complexes, such as [NaHPO - ], are small and are often not included in textbook calculations, for simplicity. It is important to note that the term, hydrogen ion, is used in two contexts, one as the name of the component hydrogen with a plus one oxidation state (vs. hydrogen gas), and two as the actual concentration of the hydrogen ion in solution, [H + ], related to Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 8

9 ph = -log 10 [H + ]. This dual use of the term, hydrogen ion, is the source of much of the confusion in how to input data into Visual Minteq. For example, to input M phosphoric acid, H 3 PO, into water would require adding the component, hydrogen ion, or H +, as M and the component phosphate, or PO 3- as M. The second method to solve for species in solution is based on the concept of charge balance (Ch. Bal.) of ions in solution coupled with the mass balance equations which are essentially the components other than hydrogen ions, i.e., Na + and PO 3- : 1. Charge Balance: Σ(concentration of all positive ions) = Σ(concentration of all negative ions) [H + ] + [Na + ] = [ H 2 PO 1- ] + 2 [ HPO 2- ]+ 3 [PO 3- ] + [ NaHPO 1- ] + [ Na 2 PO 1- ] + 2 [ NaPO 2- ] + [ OH 1- ] Use the same above species equalities to express all species in terms of the three component concentrations, [H + ], [Na + ], and [PO 3- ]: [H + ] + [Na + ] = [H + ] 2 [PO 3- ] [H + ][PO 3- ] + 3 [PO 3- ] [H + ][Na + ][PO 3- ] [Na + ] 2 [PO 3- ] [Na + ][PO 3- ] [H + ] Na + mass balance: [Na + ] + 2 [ Na 2 HPO,aq ] + 2 [ Na 2 PO 1- ] + [ NaH 2 PO,aq ] + [ NaHPO 1- ] + [ NaOH aq ] + [ NaPO 2- ] = 1 Tot NaH2PO = M After substituting, as above: [Na + ] [H + ][Na + ] 2 [PO 3- ] [Na + ] 2 [PO 3- ] [H + ] 2 [Na + ][PO 3- ] [H + ] 1 [Na + ][PO 3- ] [H + ] -1 [Na + ] [Na + ][PO 3- ] = M 3. PO 3- mass balance: [ H 2 PO 1- ] + [ H 3 PO,aq ] + [ HPO 2- ] + [ Na 2 HPO,aq ] + [ Na 2 PO 1- ] + [NaH 2 PO,aq ] + [ NaHPO 1- ] + [ NaPO 2- ] + [ PO 3- ] = 1 Tot NaH2PO = M or after substitution: [H + ] 2 [PO 3- ] [H + ] 3 [PO 3- ] [H + ][PO 3- ] [H + ][Na + ] 2 [PO 3- ] [Na + ] 2 [PO 3- ] [H + ] 2 [Na + ][PO 3- ] [H + ][Na + ][PO 3- ] [Na + ][PO 3- ] +[ PO 3- ] = 1 ( M) Again, these are three equations (charge balance plus two mass balance equations) in terms of three unknowns ([H + ], [Na + ], and [PO 3- ]) and they can be solved by matrix methods. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 9

10 It might appear that the component balance for [H + ] and charge balance equations are actually two independent constraints on the system, but it can be proven in general that the charge balance can always be expressed as a linear combination of the component balance equations and therefore you can use either, but not both, because they are linear combinations of each other. For the example of Tot NaH2PO = M in water, the charge balance equation can be obtained by adding: {1 (Hydrogen component equation) + 1 (Sodium component equation, i.e., Mass Balance for TotNa + ) - 3 (Phosphate component equation, i.e., Mass Balance for TotPO ) } = Charge Balance equation {1 {[H + ] + 2 [ H 2 PO 1- ] + 3 [ H 3 PO,aq ] + [ HPO 2- ] + [ Na 2 HPO,aq ]+ 2 [NaH 2 PO,aq ] + [ NaHPO 1- ] [ NaOH aq ] [ OH 1- ] = 2 Tot NaH2PO + 1 {[Na + ]+2 [ Na 2 HPO,aq ]+2 [ Na 2 PO 1- ]+[ NaH 2 PO,aq ]+[ NaHPO 1- ]+ [ NaOH aq ]+[ NaPO 2- ] = 1 Tot NaH2PO } 3 {[ H 2 PO 1- ]+[ H 3 PO,aq ]+[ HPO 2- ]+[ Na 2 HPO,aq ]+[ Na 2 PO 1- ]+ [NaH 2 PO,aq ]+[ NaHPO 1- ]+[ NaPO 2- ]+[ PO 3- ] = 1 Tot NaH2PO } = {[H + ] + [Na + ] = [ H 2 PO 1- ] + 2 [ HPO 2- ]+ 3 [PO 3- ] + [ NaHPO 1- ] + [ Na 2 PO 1- ] + 2 [ NaPO 2- ] + [ OH 1- ]}, which is the Charge Balance equation. Thus illustrating that, in this particular case, that the charge balance equation can be obtained as a linear combination of the component balance equations and is therefore not an independent constraint; you can use either but not both equations. The general proof is simply a tedious extension of these methods. It is clear that the sodium complexes are all negligible fractions of the total sodium, hydrogen, or phosphate and are therefore commonly not included in calculations. Excluding the sodium complexes, for clarity of presentation and discussion: 1. H + : 1 [H + ]+3 [ H 3 PO,aq ]+2 [ H 2 PO 1- ]+1 [ HPO 2- ]-1 [ OH 1- ] = 2 (Tot NaH2PO ), or Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 10

11 [H + ] [H + ] 3 [PO 3- ] [H + ] 2 [PO 3- ] [H + ][PO 3- ] [H + ] -1 = TotH + = 2 (Tot NaH2PO ) 2. Na + : 1 [Na + ] = 1 ( TotNa + ) = 1 (Tot NaH2PO ) 3. PO 3- : 1 [ H 3 PO,aq ] + 1 [ H 2 PO 1- ] + 1 [ HPO 2- ] + 1 [ PO 3- ] = 1 ( Tot NaH2PO ) [H + ] 3 [PO 3- ] [H + ] 2 [PO 3- ] [H + ][PO 3- ] +[ PO 3- ] = 1 (TotPO ) = TotPO = 1 (Tot NaH2PO ) This is still three equations and three unknowns, [H + ], [Na + ], and [PO 3- ]. First, we will solve a set of problems based around a solution that is M TotPO. Later we will add P CO2, after we learn how to enter data and interpret the output and then do titrations. This problem is similar to Prob. No..6 in Sawyer, McCarty, and Parkin, Chemistry for Environmental Engineers, 5 th McGraw Hill, 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 TotPO, etc., can be viewed as having been added as the fully protonated acids, H 3 PO, H 2 CO 3, etc., to which strong bases (e.g., Na +, Mg 2+, or Ca 2+ as NaOH, Mg(OH) 2, or Ca(OH) 2, etc.) and strong acids (such as Cl -, NO 3 -, or SO 2- as HCl, HNO 3, or H 2 SO, 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 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, M Na 2 CO 3 plus M NaOH or M HAc); 2. Given total component concentrations and measured or assumed ph, calculate the species (such as, at 8.1 ph and TotPO = M); 3. Given the solution composition and ph initial, calculate the acid or base to change the solution to a second ph, ph final (such as, initially TotPO = M and ph initial = 8.10, calculate the amount of strong acid needed to lower the solution to ph final = 6.00);. Given a solution composition, calculate the titration curve (e.g., calculate the titration curve of M K 2 HPO with 1 M HCl acid); and 5. Given an ph initial and the solution composition, calculate the titration curve of the solution (given ph initial = 8.1 with TotPO = M, calculate the titration curve upon addition of acid or base.) Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 11

12 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; or follow these steps listed: 1 a. Example 1a calc ph.vda. Given a set of neutral compounds added to water, calculate the ph (0.001 M Na 2 CO 3 plus 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;. Component window select CO3; and concentration to 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 (i.e., 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: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 12

13 The calculated ph = , it took 86 iterations to converge, if you click on Gases the equil. P CO2 = 8.7E-8 atm, the (sum of cations) = (sum of anions) = 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: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 13

14 Click on Display saturation indexes and note that natron, Na 2 CO 3 10H 2 O is greatly undersaturated. Click on Equilibrated mass distributions: and observe that the TotCO3 = M, TotNa = M, and that TotH = 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 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 M TotAc: 1. Show organic components button is displayed on the Main menu, just below Component name pull down menu. Click and organic acids will be shown: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 1

15 2. Select Component name box and the following will appear. Set Acetate-1 to M. 3. Select Calculate ph from mass and charge balance.. Set ionic strength = 0.00 and click Run Minteq: The ph = 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 TotPO = 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;. Component window select PO; and concentration to 0.001; 5. Click add to list ; 6. Click Run Minteq then OK and you are taken to: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 15

16 For example: [PO-3] =.717E-08 M from the Output menu and note that the Concentrations and activities are the same since we set ionic strength to 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 M TotPO with a ph = 8.10, one could add 1.89 mm strong base cations, probably as NaOH, to 1.00 mm H 3 PO of course there would be numerous other combinations of chemicals that would produce the same result, such as, 1.00 mm of KH 2 PO plus 0.89 mm NaOH, etc. 3. Example 3a(b) amount of acid or base to change ph.vda. Given total component composition and ph initial, calculate the acid or base to change the solution to a second ph, ph final (initially 8.1 ph and TotPO = 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;. Component window select PO; and concentration to 0.001; 5. Click add to list ; 6. Click Run Minteq and OK and the Output screen: 7. C Base (1) = E E 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; Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 16

17 10. C Base (2) = E E-06 = E-3 M 1.06 mm 11. C Acid (Added) = C Base (1) - C Base (2) = 1.89 mm mm = 0.83 mm. That is, to lower 1 L of 1 mm TotPO 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 0.15 ml of 1.00 M H 2 SO ). 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.. Example titration curve.vda. Given a solution composition, calculate the titration curve (calculate the titration curve of M K 2 HPO with 1 M HCl acid); 1. Open Visual Minteq software, everything is initialized; set ionic strength to Component window select PO; and concentration to 0.001; 3. click add to list ;. Select K + and set concentration to 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; Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 17

18 9. Click the Titration box, put 0 in no. of titration points, make volume = 1.00 L and volume of titrant = L (0.1 ml and 0.1 meq./l); 10. Click the box Start addition on the 2 nd 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 1/10 th of a milli-equivalent of HCl per titration point. Since most solutions are in the range of about 1 to 3 milliequivalents 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; 1. Click Save and Back ; 15. Click Run Minteq. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 18

19 16. The output menu will appear. Note that the speciation for the 1 st 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 0 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 = 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 M K 2 HPO, there is an inflection point at M HCl added at ph 5. There is not a second inflection point at M HCl because the first ionization constant of phosphoric acid is K which is moderately strong acid compared with the concentration of TotPO = 10-3 M ph H+1 Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 19

20 5. Example 5 titration curve ph.vda. Given an initial ph and total composition, calculate the titration curve of the solution (given ph inital = 8.1 with TotPO = M, calculate the titration curve.) 1. Open Visual Minteq software, everything is initialized; 2. ph box click Fixed at ; 3. Ionic strength equal 0.00; units Molality;. Component window select PO; and concentration to 0.001; click add; 5. Select Run Minteq, which will take you to the Output page with Sum of cations = 7.933E-09 (simply the [H+] concentration) and Sum of anions = E-03 M; 6. Select Na + and set concentration to 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 0 in no. of titration points, make volume = 1.00 L and volume of titrant = L; 12. Click the box Start addition on the 2 nd 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. 1. 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 1 st 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 0 iteration points will be displayed. 19. Select Sweep output and print to Excel and rearrange as before ph H+1 This plot can also be used to calculate the amount of acid or base to go between any two ph values from and below. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 20

21 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 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);. In Component name window select PO (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: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 21

22 The ph = , 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: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 22

23 In the Output Window and there are several things to note in the top boxes in the Output window: the ph = ; Ionic Strength = 0.00; number of iterations = 6; sum of cations = 1.073E-11 moles of charge/kg water ( M of charge); sum of anions is = E-03 moles of charge/kg water; % dif. in cations vs. anions is %. What the computer solved was the mass balance equations for total phosphate, TotPO, and the total proton balance for the solution relative to the components, PO 3-, H 2 O, and H +. The TotH 2 O is always assumed to be constant at 55.5 M and is not solved for by the software. That leaves two equations, TotPO and TotH (the default option to calculate ph) and two unknowns, [H + ] and [PO 3- ] when all the equilibrium constants are substituted into the following equations and ignoring activity coefficients for simplicity: TotPO = [PO 3- ] + [HPO 2- ] + [H 2 PO 1- ] + [H 3 PO ] = M and TotH = 1 [HPO 2- ] + 2 [H 2 PO 1- ] + 3 [H 3 PO ] + 1 [H + ] - 1 [OH - ] = (-.1E-12) 0.00 with [H] [PO ] [H] [PO ] [H] [PO ] K [H PO ], [H PO ], [HPO 2 - ], and [OH ] w 3 = 2 = = = K1 K2 K3 K2 K3 K + 3 [H ] The species are given below, as copied to Excel from the output file is: "Name" "Calc mol" # of H's vs. H2O or PO-3 "H3PO" E E-16 =+3*H3PO "OH-" E E-0 =-1*OH- "HPO-2" E E-0 =+1*HPO-2 "H2PO-" E E-07 =+2*H2PO- H E E-11 =+1*H E-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 = E-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, Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 23

24 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 TotPO 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 TotPO = M and TotH 0.00 ± convergence tolerance of ±10 - 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, PO 3-, into water. If you added Na + = 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 %. Try it. Also, if you select H+ on the Main menu in the components box and make the concentration and click on Add to list: and click on Run the ph = 3.051, the ph of M phosphoric acid in water since TotPO = and TotH = TotPO = [PO 3- ] + [HPO 2- ] + [H 2 PO 1- ] + [H 3 PO ] = M and TotH = 1 [HPO 2- ] + 2 [H 2 PO 1- ] + 3 [H 3 PO ] + 1 [H + ] - 1 [OH - ] = 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 PO; and concentration to 0.001; e. Click add to list ; f. Click Run Minteq and OK: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 2

25 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 = E-0 moles of charge/kg water (1 L); sum of anions is = E-0 moles of charge/kg water; % dif. in cations vs. anions is % (this is because we required charge balance equation to be solved). What the computer solved was the mass balance equation for total phosphate, TotPO, plus the Charge balance equation for the solution. When all the equilibrium constants are substituted in there are two equations, TotPO and Charge balance and two unknowns, [H + ] and [PO 3- ]. TotPO = [PO 3- ] + [HPO 2- ] + [H 2 PO 1- ] + [H 3 PO ] = M and ChargeBalance = Sum Cations Sum Anions =0.00 ={1 [H + ]} {3 [PO 3- ] + 2 [HPO 2- ] + 1 [H 2 PO 1- ] + 1 [OH - ]} = 0.00 As you can see, since the charge balance equation has no strong acid anions, e.g. NO 3 - or Cl -, and no strong base cations, e.g., Na + or 2 Ca 2+, it is the same as having added M H 3 PO to the solution and then calculate the ph, ph = Note also that the TotH component calculation, reported in the Equilibrated Mass Distribution window is TotH = 1 [HPO 2- ] + 2 [H 2 PO 1- ] + 3 [H 3 PO ] + 1 [H + ] - 1 [OH - ] = 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 M Na + to the component list on the Main menu, and use Charge balance option to calculate the ph you will get ph = , 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 ph. Then Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 25

26 only the mass balance equations, excluding TotH, are solved with TotPO 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 M TotPO 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 PO; 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 = 7.933E-09 moles of charge/kg water (1 L); sum of anions is = E-03 moles of charge/kg water; % dif. in cations vs. anions is %. What the computer solved was the mass balance equations for total phosphate, TotPO, using the ph. After substituting in the equilibrium constants, the free [PO 3- ] concentration is calculated see equations listed above: TotPO = [PO 3- ] + [HPO 2- ] + [H 2 PO 1- ] + [H 3 PO ] = M, or [PO 3- ] = H H H = function of only ph K1K2K3 K2K3 K3 Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 26

27 [OH [H [H 2 3 [HPO PO ] = K 2 ] 1 PO w = [H ] = [H ] = [H /[H + + ][PO + ] + ] 3 ] = K 2 [PO w 3 ]/ 3 3 [PO /10 K 3 ]/ K K ]/ K K 2 3 K 3 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 [PO 3- ] + 2 [HPO 2- ] + 1 [H 2 PO 1- ] + 1 [OH - ]} = 7.933E E-03 = E-03 This means that to prepare a solution of TotPO = M at ph = 8.100, you would have to add M Na + or K +, etc., positively charged strong base cations to the solution, e.g., as M NaOH or KOH to a solution of M H 3 PO. You could write the formula as follows: moles of Na 1.89 H 1.11 PO per liter. Most likely, what you would have done was to add moles of H 3 PO plus moles of NaOH per liter, at this point the net TotH = moles of H + plus 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 PO; and concentration to 0.001; click add to list ; e. Select Na + and set concentration to ; f. Click Run Minteq. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 27

28 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 = (approx. to 8.100); Ionic Strength = 0.00; number of iterations = 26; sum of cations = 1.870E-03 moles of charge/kg water (1 L); sum of anions is = 1.870E-03 moles of charge/kg water; % dif. in cations vs. anions is %. 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 NaHPO -, etc.). Try it. What the computer solves is the mass balance equations for total phosphate, TotPO = M, and the charge balance for the solution and since TotNa = M, this mass balance is also included. This approach can be thought of as if TotPO is added as M H 3 PO 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 PO; and concentration to 0.001; click add to list ; e. Select Na+ and set concentration to and click on Add to list; f. Select H+ and set concentration to and click on Add to list; g. Click Run Minteq and OK: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 28

29 In the top boxes in the Output window: the ph = 8.092; Ionic Strength = 0.00; sum of cations = 1.870E-03 moles of charge/kg water (1 L); sum of anions is = 1.870E-03 moles of charge/kg water, TotPO = M, and the mass balance for TotH = M and mass balance for TotNa = 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 to 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 PO; and concentration to 0.001; click add; e. Select H + and set concentration to 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 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 0 in State the number of titration steps, make volume = 1.00 L and volume of titrant = L (0.1 ml or 0.1 meq./l per step); k. Click the box Start addition on the 2 nd 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. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 29

30 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 1 st 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 0 iteration points will be displayed. 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. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 30

31 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 M TotPO starting at 8.1 ph ph HCl M added TotH, M, calculated ph H+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 = M and TotNa = M. The Calculated from mass balance used the TotH equation and added M H + to each point and calculated the ph. The Calc. from mass & charge balance option added 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 M TotPO starting at ph, but include P CO2 = atm. 1. Open Visual Minteq software, everything is initialized; 2. ph box click Fixed at ; 3. Ionic strength equal 0.00; units Molality;. Component window select PO; and concentration to 0.001; click add; Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 31

32 5. Select Gases menu bar and set 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 = 7.933E-09 0 M (simply the [H+] concentration) and Sum of anions = E-03 M, i.e., there is too little cations, such as Na + = E-03 M, in solution. This is the end of Example 12a; 7. This is Example 12b. Select Na + and set concentration to E-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 1 st 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 0 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. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 32

33 HCl, M, added ph TotCO3 SecAxis TotH SecAxis Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 33

34 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 H SiO (96.1 g/mol)/sio 2 (60.1 g/mol) since input is as H SiO, not SiO 2 ; TotCO3 input is as CO 3 = 73 mg/l {CO 3 (60 g/mol)/hco 3 (61 g/mol)} + 6 mg/l = { mg/l} = 77.8 mg/l as CO 3. Iron was assumed to be TotFe(III) since the solution is probably aerated, but the results would not change Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 3

35 much if it were assumed to be TotFe(II) or equally divided, but the percentage to precipitate would be quite different. Temp = 62 F = C was used. Click on Run Minteq and OK to Output screen: The ionic strength is.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. Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 35

36 The following is a listing to TotComponent for each component in the solution: Note that TotH = E-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 (CaCO 3 ), 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: Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 36

37 Now the ionic strength is 6.2 mm and Charge difference would require (8.76 mm 1.7 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-02 1.E-03 1.E-0 1.E-05 1.E-06 1.E E-07 1.E-08 1.E-09 1.E-10 1.E-11 TotCa+2 TotCO3-2 TotMg+2 TotFe+3SecAxis Mason Tomson Visual Minteq /18/2007 Acid/base Intro. Page 37