CE4501 Environmental Engineering Chemical Processes, Fall 2008 Problem Set 6

Transcription

1 CE4501 Environmental Engineering Chemical Processes, Fall 2008 Problem Set 6 Due: Friday, 11/21 by 5 p.m. Solutions will be posted on the Web. Problem sets will be graded for completeness, and one problem (selected at random) will be graded in detail. Each problem set contributes 2.5% towards your final grade. Remember that for problems done with MINEQL, you must print out and attach the header file so that I can see your input. 1. Problem 19.3 (p. 454) in the text. Realize that the CdCO 3(s) is added to pure water. Some CdCO 3 is added to pure water that is open to the atmosphere; not all of the solid dissolves and the equilibrium ph is 8.1. What is the solubility product for CdCO 3(s)? SOLUTION: The solubility product will be for the reaction: CdCO 3(s) Cd 2+ + CO 3 2 and the solubility product will be defined as: K s0 = {Cd 2+ }{CO 3 2 } Thus, to find the solubility product, we need to know the activities of Cd 2+ and CO 3 2. Because it is an open system, the CO 3 2 activity can be calculated from the ph (8.1) and the pco 2 in the atmosphere (370 μatm). K H = {H 2 CO 3 }/pco 2 = mole/l atm {H 2 CO 3 } = ( mole/l atm) 370x10 6 atm = 1.2x10 7 K a1 = {HCO 3 }{H + }/{H 2 CO 3 } {HCO 3 ] = K a1 {H 2 CO 3 }/{H + } = 6.7x10 6 M K a2 = {CO 3 2 }{H + }/{HCO 3 } {CO 3 2 } = K a2 {HCO 3 }/{H + } = 3.96x10 8 M To obtain the Cd 2+ activity, it is suggested that we use the charge balance equation: [H + ] + 2[Cd 2+ ] = [OH ] + [HCO 3 ] + 2[CO 3 2 ] If activity coefficients are ignored, all quantities except [Cd 2+ ] are known, so we can rearrange to solve for this unknown: [Cd 2+ ] = ½ ([OH ] + [HCO 3 ] + 2[CO 3 2 ] [H + ]) = ½ ( x (3.96x10 8 ) ) [Cd 2+ ] = 4.0x10 6 M Continuing to ignore activity coefficients, the solubility product would be: K s0 = {Cd 2+ }{CO 3 2 } =( 4.0x10 6 )( 3.96x10 8 ) = 1.59x10 13 = This value is close to the solubility product for Otavite ( ). This discrepancy arises in part because we ignored activity coefficients and thereby underestimated the Cd 2+ concentration. 2. Problem 19.9 (p. 455) in the text. Use MINEQL. The mineral name for Mn(OH) 2(s) is pyrochroite, and for MnCO 3(s) is rhodochrosite. Part C will require some thinking. Solubility diagrams can be generated directly in MINEQL or the data can be copied to the clipboard and pasted into Excel for graphing. 19.9A. Draw a solubility diagram for Mn(+2) as a function of ph if Mn(OH) 2 controls the Mn solubility and MnCO 3(s) does not exist. Assume a total dissolved carbonate concentration of 2x10 3 M SOLUTION: This problem is easily solved using MINEQL. We must be sure that H 2 O, H +, Mn2+, and CO32 are checked as components. The header file would contain: MINEQL+ Header file for pyrochroite.mdo

2 ELECTRONEUTRALITY NOT GUARANTEED TEMPERATURE = 25.0 CELSIUS IONIC STRENGTH CORRECTIONS OFF NO SURFACE MODEL USED EPS = 1.0E 04 ID X LOGX T COMPONENTS D E 18 H2O D E 18 H(+) D E 03 CO3(2 ) D E+00 Mn(2+) ID NAME LOGK DELH SPECIES: TYPE III FIXED SOLIDS PYROCHROITE H2O 2.0 H(+) 2.0 Mn(2+) H2O (Solution) H2O ph (+1) H(+) 1.0 ID NAME LOGK DELH SPECIES: TYPE VI SPECIES NOT CONSIDERED CO2 (g) H2O 1.0 H(+) 2.0 CO3(2 ) RHODOCHROSITE CO3(2 ) 1.0 Mn(2+) 1.0 The resultant solubility diagram is shown below: B. Draw the solubility diagram for Mn(2+) as a function of ph if Mn(OH) 2(s) does not exist and MnCO 3(s) (Rhodochrosite) controls the solubility. SOLUTION: Again, this is easily done in MINEQL. The same components must be checked as before. The only difference is that pyrochroite is moved to species not considered (Type VI) and Rhodochrocite is moved to Type III (fixed solids). The header file should look similar to this:

3 MINEQL+ Header file for rhodo.mdo $$$ INPUT DATA $$$ ELECTRONEUTRALITY NOT GUARANTEED TEMPERATURE = 25.0 CELSIUS IONIC STRENGTH CORRECTIONS OFF NO SURFACE MODEL USED EPS = 1.0E 04 ID X LOGX T COMPONENTS D E 18 H2O D E 18 H(+) D E 03 CO3(2 ) D E+00 Mn(2+) ID NAME LOGK DELH SPECIES: TYPE I COMPONENTS 2 H2O H2O H(+) H(+) CO3(2 ) CO3(2 ) Mn(2+) Mn(2+) 1.0 ID NAME LOGK DELH SPECIES: TYPE II COMPLEXES 3800 OH ( 1) H2O 1.0 H(+) MnOH+ (+1) H2O 1.0 H(+) 1.0 Mn(2+) Mn(OH)3 ( 1) H2O 3.0 H(+) 3.0 Mn(2+) Mn(OH)4 2 ( 2) H2O 4.0 H(+) 4.0 Mn(2+) H2CO3 (aq) H(+) 2.0 CO3(2 ) HCO3 ( 1) H(+) 1.0 CO3(2 ) MnHCO3+ (+1) H(+) 1.0 CO3(2 ) 1.0 Mn(2+) 1.0 ID NAME LOGK DELH SPECIES: TYPE III FIXED SOLIDS RHODOCHROSITE CO3(2 ) 1.0 Mn(2+) H2O (Solution) H2O ph (+1) H(+) 1.0 ID NAME LOGK DELH SPECIES: TYPE VI SPECIES NOT CONSIDERED CO2 (g) H2O 1.0 H(+) 2.0 CO3(2 ) PYROCHROITE H2O 2.0 H(+) 2.0 Mn(2+) 1.0 The solubility diagram should look as shown below.

4 C. Draw a solubility diagram for Mn(2+) as a function of ph assuming both solids exist. SOLUTION: Simply combining the two figures above gives us the required figure as shown below. Mn concentrations are controlled by the least soluble solid; below ph 8 the concentration is controlled by Rhodochrocite, and above ph 8, concentrations are regulated by phyrochroite. 3. Problem (p. 456) in the text. A. Manganese can form two common solids in natural waters, MnO 2(s) and Mn(OH) 2(s). Which of these solids controls the Mn solubility at ph 8 and pe 7? SOLUTION:

5 This problem is as easily done by hand as with MINEQL. Each solid dissolves to yield one mole of Mn 2+ per mole of solid dissolved. Hence, if there is no other source of dissolved Mn, the Mn concentration will equal the mineral solubility. For Mn(OH) 2(s), the solubility (S) may be calculated as: S = [Mn 2+ ] = K s0 /{OH } 2 = /10 12 = = M For MnO 2(s) it is easiest to start with the Nernst equation applied to the following reaction: MnO 2(s) + 2e + 4H + Mn H 2 O pe = pe o + 1/n log({h + } 4 /{Mn 2+ }) At pe = 8 we can solve for {Mn 2+ } which equals the solubility: {Mn 2+ } = {H + } 4 / 10 n (pe pe0) = /10 2(8 20.5) = 10 3 Because the solubility for MnO 2 is smaller, it would control the concentration of Mn under the stated pe and ph conditions. B. Sketch a pe ph diagram for the equilibrium between MnO 2(s) and Mn(OH) 2(s). SOLUTION: We are only asked for the equilibrium line between these two species. We must simply combine the two reactions given above: MnO 2(s) + 2e + 4H + Mn H 2 O pe o = 20.5 logk = 40.9 Mn OH Mn(OH) 2(s) logk = 12.8 MnO 2(s) + 2e +2H + Mn(OH) 2(s) pe 0 = 26.9 logk = 53.7 pe = pe o + 1/n log({h + } 2 /1) = /2 2 ph = 26.9 ph This line is shown on the graph below together with the boundaries for water. The diagram suggests that these two solids should not coexist in water but only at higher oxidation potentials. We would therefore expect Mn(OH) 2(s) to be the predominant mineral encountered in natural waters. 4. Problem (p.456) in the text.

6 5. Use MINEQL and calculate the solubility of cerargyrite (AgCl) as a function of ionic strength over the range < I < 0.1 M. It is not possible to titrate with ionic strength, so you will have to do multiple single runs. Make a plot of solubility vs. ionic strength, and explain qualitatively the trend. SOLUTION: We will assume a ph of 7. The graph below shows that solubility increases only 30% between ionic strength of 10 5 and 10 1 M. The increase in solubility is a result of the decrease in activity coefficients with increasing ionic strength; in other words, the effective concentrations of ions decreases the more ions there are in solution to shield one ion from another. 6. Calculate the solubility of malachite (Cu 2 CO 3 (OH) 2(S) ) in pure water and in ground water that has an alkalinity of 2 meq/l. In which situation is the solubility higher? Why? This problem could be done easily with MINEQL. SOLUTION: To calculate the solubility in pure water using MINEQL, one would enter components of Cu2+ and CO3(2 ). MINEQL should calculate ph based on electroneutrality; total carbonate may be entered as a very low value. All solids except malachite should be moved to type VI (not considered); malachite should be moved to type III (fixed). The header file should contain: MINEQL+ Header file for mala.mdo $$$ INPUT DATA $$$ OPTIONS: IADS= 0 IONIT= 1 IONPH= 0 IPHFX= 0 IPHA= 0 IPHB= 0 ITITL= 0 IPCP=0 ICND=0 SOLUTION IS ELECTRICALLY NEUTRAL TEMPERATURE = 25.0 CELSIUS IONIC STRENGTH CALCULATED FROM SPECIES NO SURFACE MODEL USED EPS = 1.0E 04 ID X LOGX T COMPONENTS D E 18 H2O D E+00 H(+) D E 18 CO3(2 )

7 D E 18 Cu(2+) ID NAME LOGK DELH SPECIES: TYPE I COMPONENTS 2 H2O H2O H(+) H(+) CO3(2 ) CO3(2 ) Cu(2+) Cu(2+) 1.0 ID NAME LOGK DELH SPECIES: TYPE II COMPLEXES 3800 OH ( 1) H2O 1.0 H(+) Cu2(OH)2+2 (+2) H2O 2.0 H(+) 2.0 Cu(2+) Cu(OH)3 ( 1) H2O 3.0 H(+) 3.0 Cu(2+) Cu(OH)4 2 ( 2) H2O 4.0 H(+) 4.0 Cu(2+) CuOH+ (+1) H2O 1.0 H(+) 1.0 Cu(2+) Cu(OH)2 (aq) H2O 2.0 H(+) 2.0 Cu(2+) H2CO3 (aq) H(+) 2.0 CO3(2 ) HCO3 ( 1) H(+) 1.0 CO3(2 ) CuHCO3+ (+1) H(+) 1.0 CO3(2 ) 1.0 Cu(2+) CuCO3 (aq) CO3(2 ) 1.0 Cu(2+) Cu(CO3)2 2 ( 2) CO3(2 ) 2.0 Cu(2+) 1.0 ID NAME LOGK DELH SPECIES: TYPE III FIXED SOLIDS MALACHITE H2O 2.0 H(+) 2.0 CO3(2 ) 1.0 Cu(2+) H2O (Solution) H2O 1.0 ID NAME LOGK DELH SPECIES: TYPE VI SPECIES NOT CONSIDERED ph (+1) H(+) CO2 (g) H2O 1.0 H(+) 2.0 CO3(2 ) AZURITE H2O 2.0 H(+) 2.0 CO3(2 ) 2.0 Cu(2+) TENORITE H2O 1.0 H(+) 2.0 Cu(2+) Cu(OH) H2O 2.0 H(+) 2.0 Cu(2+) CuCO CO3(2 ) 1.0 Cu(2+) 1.0 The total soluble Cu under these circumstances is 1.76 μm; because there are two Cu atoms per molecule of malachite, the solubility is one half this value or 0.88 μm. The ph is 8.0 and the total alkalinity is 1.9 μeq/l. In a ground water with alkalinity of 2 meq/l, if all of the alkalinity were associated with Ca 2+, the ph would be about 7.3 (see Fig in the text). At this ph and alkalinity, the header file should look like: MINEQL+ Header file for mala2.mdo $$$ INPUT DATA $$$ OPTIONS: IADS= 0 IONIT= 1 IONPH= 0 IPHFX= 0 IPHA= 0 IPHB= 0 ITITL= 0 IPCP=0 ICND=0 ELECTRONEUTRALITY NOT GUARANTEED TEMPERATURE = 25.0 CELSIUS IONIC STRENGTH CALCULATED FROM SPECIES NO SURFACE MODEL USED EPS = 1.0E 04 ID X LOGX T COMPONENTS D E 18 H2O D E 18 H(+) D E 03 Ca(2+)

8 D E 03 CO3(2 ) D E 06 Cu(2+) ID NAME LOGK DELH SPECIES: TYPE I COMPONENTS 2 H2O H2O H(+) H(+) Ca(2+) Ca(2+) CO3(2 ) CO3(2 ) Cu(2+) Cu(2+) 1.0 ID NAME LOGK DELH SPECIES: TYPE II COMPLEXES 3800 OH ( 1) H2O 1.0 H(+) CaOH+ (+1) H2O 1.0 H(+) 1.0 Ca(2+) Cu2(OH)2+2 (+2) H2O 2.0 H(+) 2.0 Cu(2+) Cu(OH)3 ( 1) H2O 3.0 H(+) 3.0 Cu(2+) Cu(OH)4 2 ( 2) H2O 4.0 H(+) 4.0 Cu(2+) CuOH+ (+1) H2O 1.0 H(+) 1.0 Cu(2+) Cu(OH)2 (aq) H2O 2.0 H(+) 2.0 Cu(2+) CaHCO3+ (+1) H(+) 1.0 Ca(2+) 1.0 CO3(2 ) H2CO3 (aq) H(+) 2.0 CO3(2 ) HCO3 ( 1) H(+) 1.0 CO3(2 ) CuHCO3+ (+1) H(+) 1.0 CO3(2 ) 1.0 Cu(2+) CaCO3 (aq) Ca(2+) 1.0 CO3(2 ) CuCO3 (aq) CO3(2 ) 1.0 Cu(2+) Cu(CO3)2 2 ( 2) CO3(2 ) 2.0 Cu(2+) 1.0 ID NAME LOGK DELH SPECIES: TYPE III FIXED SOLIDS MALACHITE H2O 2.0 H(+) 2.0 CO3(2 ) 1.0 Cu(2+) H2O (Solution) H2O ph (+1) H(+) 1.0 ID NAME LOGK DELH SPECIES: TYPE VI SPECIES NOT CONSIDERED CO2 (g) H2O 1.0 H(+) 2.0 CO3(2 ) LIME H2O 1.0 H(+) 2.0 Ca(2+) PORTLANDITE H2O 2.0 H(+) 2.0 Ca(2+) AZURITE H2O 2.0 H(+) 2.0 CO3(2 ) 2.0 Cu(2+) TENORITE H2O 1.0 H(+) 2.0 Cu(2+) Cu(OH) H2O 2.0 H(+) 2.0 Cu(2+) ARAGONITE Ca(2+) 1.0 CO3(2 ) CALCITE Ca(2+) 1.0 CO3(2 ) CuCO CO3(2 ) 1.0 Cu(2+) 1.0 Under these circumstances, the total dissolved Cu is predicted to be 1.05 μm or 40% lower than in the absence of the background bicarbonate. The solubility again is ½ the Cu concentration of 0.52 μm. The presence of carbonate ions from other sources lowers the solubility (common ion effect). 7. The total dissolved concentration of Cu(II) in Torch Lake is 20 μg/l. Given the ion concentrations that you measured (summarized in the table below), is the lake water in equilibrium with any of the copper minerals included in MINEQL? Which mineral do you think may be regulating the concentration of Cu in the lake? Explain your answer. Substance Conc. Units Temperature 18.9 o C ph 8.09 Ionic Strength 1.6 mm Na μm

9 K + 25 μm Ca μm Mg μm Cl 365 μm NO 3 50 μm 2 SO 4 5 μm Alkalinity 988 μeq/l SOLUTION: If all components are entered into MINEQL, the header file resembles this: MINEQL+ Header file for torch.mdo $$$ INPUT DATA $$$ OPTIONS: IADS= 0 IONIT= 1 IONPH= 0 IPHFX= 0 IPHA= 0 IPHB= 0 ITITL= 0 IPCP=0 ICND=0 ELECTRONEUTRALITY NOT GUARANTEED TEMPERATURE = 19.0 CELSIUS IONIC STRENGTH CALCULATED FROM SPECIES NO SURFACE MODEL USED EPS = 1.0E 04 ID X LOGX T COMPONENTS D E 18 H2O D E 18 H(+) D E 04 Ca(2+) D E 04 Cl( ) D E 04 CO3(2 ) D E 07 Cu(2+) D E 05 K(+) D E 04 Mg(2+) D E 04 Na(+) D E 05 NO3( ) D E 06 SO4(2 ) ID NAME LOGK DELH SPECIES: TYPE I COMPONENTS 2 H2O H2O H(+) H(+) Ca(2+) Ca(2+) Cl( ) Cl( ) CO3(2 ) CO3(2 ) Cu(2+) Cu(2+) K(+) K(+) Mg(2+) Mg(2+) Na(+) Na(+) NO3( ) NO3( ) SO4(2 ) SO4(2 ) 1.0 ID NAME LOGK DELH SPECIES: TYPE II COMPLEXES 3800 OH ( 1) H2O 1.0 H(+) CaOH+ (+1) H2O 1.0 H(+) 1.0 Ca(2+) 1.0

10 13100 Cu2(OH)2+2 (+2) H2O 2.0 H(+) 2.0 Cu(2+) Cu(OH)3 ( 1) H2O 3.0 H(+) 3.0 Cu(2+) Cu(OH)4 2 ( 2) H2O 4.0 H(+) 4.0 Cu(2+) CuOH+ (+1) H2O 1.0 H(+) 1.0 Cu(2+) Cu(OH)2 (aq) H2O 2.0 H(+) 2.0 Cu(2+) MgOH+ (+1) H2O 1.0 H(+) 1.0 Mg(2+) CaHCO3+ (+1) H(+) 1.0 Ca(2+) 1.0 CO3(2 ) H2CO3 (aq) H(+) 2.0 CO3(2 ) HCO3 ( 1) H(+) 1.0 CO3(2 ) CuHCO3+ (+1) H(+) 1.0 CO3(2 ) 1.0 Cu(2+) MgHCO3+ (+1) H(+) 1.0 CO3(2 ) 1.0 Mg(2+) NaHCO3 (aq) H(+) 1.0 CO3(2 ) 1.0 Na(+) HSO4 ( 1) H(+) 1.0 SO4(2 ) CaCO3 (aq) Ca(2+) 1.0 CO3(2 ) CaNO3+ (+1) Ca(2+) 1.0 NO3( ) CaSO4 (aq) Ca(2+) 1.0 SO4(2 ) CuCl3 ( 1) Cl( ) 3.0 Cu(2+) CuCl2 (aq) Cl( ) 2.0 Cu(2+) CuCl4 2 1 ( 2) Cl( ) 4.0 Cu(2+) CuCl+ (+1) Cl( ) 1.0 Cu(2+) CuCO3 (aq) CO3(2 ) 1.0 Cu(2+) Cu(CO3)2 2 ( 2) CO3(2 ) 2.0 Cu(2+) MgCO3 (aq) CO3(2 ) 1.0 Mg(2+) NaCO3 ( 1) CO3(2 ) 1.0 Na(+) Cu(NO3)2 (aq) Cu(2+) 1.0 NO3( ) CuNO3+ (+1) Cu(2+) 1.0 NO3( ) CuSO4 (aq) Cu(2+) 1.0 SO4(2 ) KSO4 ( 1) K(+) 1.0 SO4(2 ) MgSO4 (aq) Mg(2+) 1.0 SO4(2 ) NaSO4 ( 1) Na(+) 1.0 SO4(2 ) 1.0 ID NAME LOGK DELH SPECIES: TYPE III FIXED SOLIDS 3801 H2O (Solution) H2O ph (+1) H(+) 1.0 ID NAME LOGK DELH SPECIES: TYPE VI SPECIES NOT CONSIDERED CO2 (g) H2O 1.0 H(+) 2.0 CO3(2 ) LIME H2O 1.0 H(+) 2.0 Ca(2+) PORTLANDITE H2O 2.0 H(+) 2.0 Ca(2+) ATACAMITE H2O 3.0 H(+) 3.0 Cl( ) 1.0 Cu(2+) AZURITE H2O 2.0 H(+) 2.0 CO3(2 ) 2.0 Cu(2+) MALACHITE H2O 2.0 H(+) 2.0 CO3(2 ) 1.0 Cu(2+) ARTINITE H2O 5.0 H(+) 2.0 CO3(2 ) 1.0 Mg(2+) HYDROMAGNESITE H2O 6.0 H(+) 2.0 CO3(2 ) 4.0 Mg(2+) TENORITE H2O 1.0 H(+) 2.0 Cu(2+) Cu(OH) H2O 2.0 H(+) 2.0 Cu(2+) Cu2(OH)3NO H2O 3.0 H(+) 3.0 Cu(2+) 2.0 NO3( ) ANTLERITE H2O 4.0 H(+) 4.0 Cu(2+) 3.0 SO4(2 ) CuOCuSO H2O 1.0 H(+) 2.0 Cu(2+) 2.0 SO4(2 ) BROCHANTITE H2O 6.0 H(+) 6.0 Cu(2+) 4.0 SO4(2 ) LANGITE H2O 7.0 H(+) 6.0 Cu(2+) 4.0 SO4(2 ) PERICLASE H2O 1.0 H(+) 2.0 Mg(2+) BRUCITE H2O 2.0 H(+) 2.0 Mg(2+) 1.0

11 Mg(OH)2 (active) H2O 2.0 H(+) 2.0 Mg(2+) GYPSUM H2O 2.0 Ca(2+) 1.0 SO4(2 ) NESQUEHONITE H2O 3.0 CO3(2 ) 1.0 Mg(2+) THERMONATRITE H2O 1.0 CO3(2 ) 1.0 Na(+) NATRON H2O 10.0 CO3(2 ) 1.0 Na(+) CHALCANTHITE H2O 5.0 Cu(2+) 1.0 SO4(2 ) EPSOMITE H2O 7.0 Mg(2+) 1.0 SO4(2 ) MIRABILITE H2O 10.0 Na(+) 2.0 SO4(2 ) ARAGONITE Ca(2+) 1.0 CO3(2 ) CALCITE Ca(2+) 1.0 CO3(2 ) HUNTITE Ca(2+) 1.0 CO3(2 ) 4.0 Mg(2+) DOLOMITE (ordered) Ca(2+) 1.0 CO3(2 ) 2.0 Mg(2+) ~ Ca(2+) 1.0 CO3(2 ) 2.0 Mg(2+) ANHYDRITE Ca(2+) 1.0 SO4(2 ) MELANOTHALLITE Cl( ) 2.0 Cu(2+) HALITE Cl( ) 1.0 Na(+) CuCO CO3(2 ) 1.0 Cu(2+) MAGNESITE CO3(2 ) 1.0 Mg(2+) CuSO Cu(2+) 1.0 SO4(2 ) THENARDITE Na(+) 2.0 SO4(2 ) 1.0 From the special reports, the solids saturation index may be used to see how close different minerals are to equilibrium. The report shows to following for minerals containing Cu: SOLIDS SATURATION INDEX SUMMARY SI Sample ph AZURITE MALACHIT ~ Sample ARTINITE HYDROMAG TENORITE Cu(OH)2 Cu2(OH) Sample ANTLERIT CuOCuSO4 BROCHANT LANGITE PERICLAS The only two minerals with SI close to zero are malachite (SI = 0.368) and tenorite (SI = 0.271). Both are slightly supersaturated. Because we did not account for the DOC in the lake, MINEQL likely overestimated the free Cu 2+ concentration and thereby overestimated the SI for all of the minerals. All that we can say with certainty is that malachite and tenorite are the two minerals that might potentially be regulating Cu concentrations in Torch Lake as evidenced by the fact that they are the closest to equilibrium concentrations estimated by MINEQL. Because all other minerals are predicted to be highly undersaturated, it is highly unlikely that these other minerals are present. If any of them are present, they are not regulating the Cu concentration because they are far from equilibrium.

12 8. It is generally desirable to have a Langelier Index of slightly greater than 0.5 in drinking water to prevent dissolution of metals (Cu, Pb) from the pipes into the water. What is the Langelier Index of the Houghton tap water at room temperature (21 o C)? What would be the Langelier Index of the same water when it is heated to 50 o C? Is scale formation likely to occur at either temperature? The ph sat is easily calculated using MINEQL. The measured ph of the water was 8.03 and the alkalinity was 2.66 meq/l. The ionic strength was 4.8 mm. SOLUTION: I did not do a very good job of giving you the correct information that you needed to do this problem correctly. To calculate a Langelier Index, you need measured values for Ca, HCO 3, and ph. I misread the table when I gave the ph value above; the correct value is 7.8. I neglected to give the Ca concentration which was 1355 μm. The problem can be done either manually or with MINEQL. The equation for LI is: LI = ph act ph sat where ph sat is given by: ph sat = log(γγ) +pca + phco 3 + pk a2 pk sp At an ionic strength of , the activity coefficients may be calculated to be 0.93 and 0.74 (log(γ) = Az 2 I 0.5 /(1+I 0.5 )). The equilibrium constants must be corrected for temperature using the van t Hoff equation: ln(k 2 /K 1 ) = ΔH/R (1/T 1 1/T 2 ) Values of ΔH rxn may be calculated from values of ΔH formation that are tabulated in MINEQL as well as in most Aquatic Chemistry texts except the one that we are using this year. Temperature ΔH (kj/mole) log(k a2 ) ΔH (kj/mole) log(k sp ) 25 o C o C o C Having the values above, one can calculate the ph sat to be 7.52 at 21 o and 7.08 at 50 o C. These, in turn yield LI values of 0.28 and The LI values indicate that calcite is supersaturated at both temperatures, but the degree of supersaturation increases with increasing temperature. Calcite is more likely to precipitate at high temperature than at low temperature because of the lower solubility at high temperature (seen in the negative enthalpy value). To solve this with MINEQL one enters the total Ca given above, allows MINEQL to calculate CO3(2 ) from alkalinity and measured ph, but also allows MINEQL to calculate ph based on the total H concentration (TotH). The total hydrogen concentration is the sum of all of the forms in which H is bound: TotH = [H + ] + [HCO 3 ] + 2[H 2 CO 3 ] [OH ] In this case, it is essentially equal to the [HCO 3 ] or 2.66x10 3 M. All solids except calcite are moved to type VI (not considered), and calcite is moved to a fixed solid (Type III). Ionic strength is set at the stated value, and temperature is set to 21 o C. The ph sat calculated by MINEQL is which gives a LI of 0.18, slightly below the value of 0.28 calculated by hand. The discrepancy is likely due to MINEQL s ability to consider all other simultaneous equilibria. Similarly, at 50 o C, the calculated ph sat is which gives an LI value of 0.46, again lower than the hand calculated value. 9. The Chassell well water is slightly corrosive (i.e., LI < 0) due to the oxidation of reduced Fe. One option to prevent pipe corrosion would be to add CaCO 3 to the water. Make a plot of the Langelier Index vs. the concentration of CaCO 3 added to the water. What dose would be required to reach a LI of 0.6? The ionic composition of this water is summarized in the table below.

13 Substance Conc. Units Temperature 25 o C ph 8.1 Ionic Strength 2.3 mm Na μm K + 95 μm Ca μm Mg μm Cl 290 μm NO μm 2 SO 4 25 μm Alkalinity 960 μeq/l DIC 975 μm SOLUTION: This problem is similar to the previous one. To solve it manually is challenging because we must calculate the actual ph after each addition of CaCO 3 ; we know the new total Ca and carbonate concentrations as well as the new alkalinity. The nice thing about the carbonate system is that knowing any two parameters allows one to calculate the others. To calculate ph from C T CO3 and alkalinity, one must solve a quartic equation that expresses {H + } as a function of DIC and alkalinity. [H + ] 4 + (Alk + K a1 )[H + ] 3 + (Alk K a1 K w + K a1 K a2 DIC K a1 )[H + ] 2 + (Alk K a1 K a2 K w K a1 2DIC K a1 K a2 )[H + ] K w K a1 K a2 = 0 The calculated initial ph (6.93) is lower than the estimate given above, but may be just as accurate of an estimate. Calculation of ph sat is straightforward because one knows the Ca 2+ and alkalinity after each addition of CaCO 3. ph sat = log(γγ) +pca + phco 3 + pk a2 pk sp The data for these hand calculations are shown below. CaCO 3 added (mmole/l) ph actual ph sat LI (6.93)

14 To solve the problem using MINEQL, is much more challenging than I realized. 10. A local entrepreneur has proposed to use the Houghton area stamp sands on roofing shingles. His rationale is that some copper will dissolve from the stamp sands into the rain, and the Cu concentrations will prevent the growth of algae on the roof; it is the decomposition of algae and bacteria that provides the nutrients for mosses to grow on roofs. Free Cu 2+ concentrations of 0.01 μm are enough to inhibit the growth of algae. The two primary copper minerals present in the aged stamp sands are tenorite (CuO) and malachite (CuCO 3 (OH) 2 ). Given the composition of rain water measured in class (see table below), what concentration of free Cu 2+ would you predict to be in rain water in contact with tenorite and malachite? Which mineral will control the concentration of Cu? Use MINEQL to answer this problem; think carefully about how you specify the carbonate system. Is the idea to use the stamp sands in roofing materials a good idea based on your calculations? Substance Conc. Units Temperature 25 o C ph 7.1 Ionic Strength 0.1 mm Na + 5 μm K + 0 μm Ca μm Mg 2+ 7 μm Cl 0 μm NO 3 0 μm 2 SO 4 2 μm Alkalinity 100 μeq/l SOLUTION: One thing that MINEQL reveals is that not all of our measurements in the lab are internally self consistent. Rain is likely to be in equilibrium with atmospheric CO 2 and so we

15 should model the system as open. If we also fix the ph at our measured value of 7.1, we will find that MINEQL predicts a carbonate concentration of only 70 mm; thus, the measured alkalinity is too high for the measured ph. You must also realize that you can only specify one mineral as being in equilibrium at a time; one can put the other mineral in the dissolved solids category and see if it precipitates. So if we enter the measured cation concentrations as input, specify an open system that is in equilibrium with tenorite, MINEQL predicts that the ph will be 7.3, total carbonate will be 126 μm, total dissolved copper will be 0.2 μm, and free Cu 2+ will be μm. Malachite is predicted to be undersaturated (SI = 1.05), and thus tenorite is controlling the solubility at this ph. Free Cu 2+ is predicted to be high enough (> 0.01 μm) to inhibit algal growth. If we tried to specify that Malachite was the fixed species and allowed tenorite to be a dissolved solid (Type V changing to Type IV if precipitation occurs), the program would give an error message because equilibrium with malachite could not be maintained. As shown in the lecture slides on the web, at circum neutral ph, tenorite is much less soluble than malachite. Malachite could continue to dissolve and tenorite to precipitate until the computer processor wore out; for that reason, an error message is generated. The header file and Cu summary output file are shown below. MINEQL+ Header file for rain2.mdo $$$ INPUT DATA $$$ OPTIONS: IADS= 0 IONIT= 1 IONPH= 0 IPHFX= 0 IPHA= 0 IPHB= 0 ITITL= 0 IPCP=0 ICND=0 SOLUTION IS ELECTRICALLY NEUTRAL TEMPERATURE = 25.0 CELSIUS IONIC STRENGTH FIXED CONCENTRATION NO SURFACE MODEL USED EPS = 1.0E 04 ID X LOGX T COMPONENTS D E 18 H2O D E 04 H(+) D E 05 Ca(2+) D E 18 CO3(2 ) D E 18 Cu(2+) D E 06 Mg(2+) D E 06 Na(+) D E 06 SO4(2 ) ID NAME LOGK DELH SPECIES: TYPE I COMPONENTS 2 H2O H2O H(+) H(+) Ca(2+) Ca(2+) CO3(2 ) CO3(2 ) Cu(2+) Cu(2+) Mg(2+) Mg(2+) Na(+) Na(+) SO4(2 ) SO4(2 ) 1.0

16 ID NAME LOGK DELH SPECIES: TYPE II COMPLEXES 3800 OH ( 1) H2O 1.0 H(+) CaOH+ (+1) H2O 1.0 H(+) 1.0 Ca(2+) Cu2(OH)2+2 (+2) H2O 2.0 H(+) 2.0 Cu(2+) Cu(OH)3 ( 1) H2O 3.0 H(+) 3.0 Cu(2+) Cu(OH)4 2 ( 2) H2O 4.0 H(+) 4.0 Cu(2+) CuOH+ (+1) H2O 1.0 H(+) 1.0 Cu(2+) Cu(OH)2 (aq) H2O 2.0 H(+) 2.0 Cu(2+) MgOH+ (+1) H2O 1.0 H(+) 1.0 Mg(2+) CaHCO3+ (+1) H(+) 1.0 Ca(2+) 1.0 CO3(2 ) H2CO3 (aq) H(+) 2.0 CO3(2 ) HCO3 ( 1) H(+) 1.0 CO3(2 ) CuHCO3+ (+1) H(+) 1.0 CO3(2 ) 1.0 Cu(2+) MgHCO3+ (+1) H(+) 1.0 CO3(2 ) 1.0 Mg(2+) NaHCO3 (aq) H(+) 1.0 CO3(2 ) 1.0 Na(+) HSO4 ( 1) H(+) 1.0 SO4(2 ) CaCO3 (aq) Ca(2+) 1.0 CO3(2 ) CaSO4 (aq) Ca(2+) 1.0 SO4(2 ) CuCO3 (aq) CO3(2 ) 1.0 Cu(2+) Cu(CO3)2 2 ( 2) CO3(2 ) 2.0 Cu(2+) MgCO3 (aq) CO3(2 ) 1.0 Mg(2+) NaCO3 ( 1) CO3(2 ) 1.0 Na(+) CuSO4 (aq) Cu(2+) 1.0 SO4(2 ) MgSO4 (aq) Mg(2+) 1.0 SO4(2 ) NaSO4 ( 1) Na(+) 1.0 SO4(2 ) 1.0 ID NAME LOGK DELH SPECIES: TYPE III FIXED SOLIDS TENORITE H2O 1.0 H(+) 2.0 Cu(2+) CO2 (g) H2O 1.0 H(+) 2.0 CO3(2 ) H2O (Solution) H2O 1.0 ID NAME LOGK DELH SPECIES: TYPE V DISSOLVED SOLIDS MALACHITE H2O 2.0 H(+) 2.0 CO3(2 ) 1.0 Cu(2+) 2.0 ID NAME LOGK DELH SPECIES: TYPE VI SPECIES NOT CONSIDERED ph (+1) H(+) LIME H2O 1.0 H(+) 2.0 Ca(2+) PORTLANDITE H2O 2.0 H(+) 2.0 Ca(2+) AZURITE H2O 2.0 H(+) 2.0 CO3(2 ) 2.0 Cu(2+) ARTINITE H2O 5.0 H(+) 2.0 CO3(2 ) 1.0 Mg(2+) HYDROMAGNESITE H2O 6.0 H(+) 2.0 CO3(2 ) 4.0 Mg(2+) Cu(OH) H2O 2.0 H(+) 2.0 Cu(2+) ANTLERITE H2O 4.0 H(+) 4.0 Cu(2+) 3.0 SO4(2 ) CuOCuSO H2O 1.0 H(+) 2.0 Cu(2+) 2.0 SO4(2 ) BROCHANTITE H2O 6.0 H(+) 6.0 Cu(2+) 4.0 SO4(2 ) LANGITE H2O 7.0 H(+) 6.0 Cu(2+) 4.0 SO4(2 ) PERICLASE H2O 1.0 H(+) 2.0 Mg(2+) BRUCITE H2O 2.0 H(+) 2.0 Mg(2+) Mg(OH)2 (active) H2O 2.0 H(+) 2.0 Mg(2+) GYPSUM H2O 2.0 Ca(2+) 1.0 SO4(2 ) NESQUEHONITE H2O 3.0 CO3(2 ) 1.0 Mg(2+) THERMONATRITE H2O 1.0 CO3(2 ) 1.0 Na(+) NATRON H2O 10.0 CO3(2 ) 1.0 Na(+) CHALCANTHITE H2O 5.0 Cu(2+) 1.0 SO4(2 ) EPSOMITE H2O 7.0 Mg(2+) 1.0 SO4(2 ) MIRABILITE H2O 10.0 Na(+) 2.0 SO4(2 ) ARAGONITE Ca(2+) 1.0 CO3(2 ) CALCITE Ca(2+) 1.0 CO3(2 ) 1.0

17 HUNTITE Ca(2+) 1.0 CO3(2 ) 4.0 Mg(2+) DOLOMITE (ordered) Ca(2+) 1.0 CO3(2 ) 2.0 Mg(2+) ~ Ca(2+) 1.0 CO3(2 ) 2.0 Mg(2+) ANHYDRITE Ca(2+) 1.0 SO4(2 ) CuCO CO3(2 ) 1.0 Cu(2+) MAGNESITE CO3(2 ) 1.0 Mg(2+) CuSO Cu(2+) 1.0 SO4(2 ) THENARDITE Na(+) 2.0 SO4(2 ) 1.0 Cu Component Species summary Obs.,Species ID, Name, Type, Conc., LogC, LogK, %Total, Stoch., 1, 30, Cu(2+), 1, 8.31e 8, 7.08, 0, 40.5, 1, 1, 13100, Cu2(OH)2(+2), 2, 9.33e 11, 10, 10.6, 0, 2, 1, 13200, Cu(OH)3( 1), 2, 1.39e 12, 11.9, 26.9, 0, 1, 1, 13300, Cu(OH)4( 2), 2, 2.68e 18, 17.6, 40, 0, 1, 1, 13400, CuOH(+1), 2, 6.02e 8, 7.22, 7.51, 29.3, 1, 1, 13500, Cu(OH)2 (aq) 2, 2.82e 9, 8.55, 16.2, 1.4, 1, 1, 31900, CuHCO 3 (+1), 2, 5.74e 10, 9.24, 12.1, 0, 1, 1, 95100, CuCO3 (aq 2, 5.85e 8, 7.23, 6.73, 28.5, 1, 1, 95200, Cu(CO3)2 ( 2) 2, 2.06e 11, 10.7, 10.2, 0, 1, 1, , CuSO4 (aq 2, 3.43e 11, 10.5, 2.32, 0, 1, 1, , TENORITE 3,,, 7.65,, 1, 1, , MALACHITE , 5.26,, 2, 1, , AZURITE 6, , 3.46, 16.8,, 3, 1, , Cu(OH)2 6, , 1.03, 8.68,, 1, 1, , ANTLERITE 6, 4.73e 7, 6.33, 8.85,, 3, 1, , CuOCuSO4 6, 3.28e 16, 15.5, 10.3,, 2, 1, , BROCHANTITE 6, 7.68e 6, 5.12, 15.3,, 4, 1, , LANGITE 6, 4.15e 8, 7.38, 17.6,, 4, 1, , CHALCANTHITE 6, 6.54e 11, 10.2, 2.6,, 1, 1, , CuCO3 6, , 2.5, 11.5,, 1, 1, , CuSO4 6, 1.72e 16, 15.8, 2.98,, 1, 1, , TOTAL Cu(2+) 7, 2.05e 7, 6.69,, 100,,