WATER & WASTE WATER QUALITY
WATER & WASTE WATER QUALITY ENVE 408
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9yAkainityyAciditySO9 Alkalinity and Acidity Alkalinity 1 Definition A measure of the capacity of water to neutralize strong acid the buffer capacity the capacity to resist to drop in pH resulting from acid addition 2 Alkalinity Species a Major species OH C0321 HCOg OH caustic alkalinity b Minor inorganic species silicates ammonia phosphates borates c Organic species hurnic acid acetic acid propionic acid hydrosulfuric acid Acidity 1 Definition A measure of the capacity of water to neutralize strong base 2 Acid Species a Major species H2C03 HCOg H Ht mineral acidity Alkalinity Primarily due to the salts of weak acids e g NaHC03 NaHzPO4 weak or strong bases may also contribute In the Carbonate H2C03 HCOg Cng System H2CO3 ltgt HJr HCO37 carbonic acid H HCOs 31 10 63 PK31263 H2C03l Hco ltgt H co bicarbonate carbonate H C0321 Kaz 210403 PKa22103 9AlkalinityAcidity7809 Sources of Alkalinity Bicarbonates represent the major form of alkalinity from the action of C02 upon basic materials in the soil ave trees CHzO 02 lt gt C02 H20 Organic soil Organics f l J C02 H20 H HZCO3 Carbonic acid CaC03 H2C03 lt gt CaHC032 lt gt Ca2 2HC032 Lime stone Natural alkalinity r Borates silicates and phosphates may be present in small amo nts r A few organic acids that are quite resistant to biological oxidation form salts that add to the alkalinity eg humic acid 7 In anaerobic waters salts of weak acids may be produced and would also contribute to alkalinity eg acetic acid propionic acid hydrosulfuric acid 7 Others may make a contribution to the total alkalinity of a water eg ammonia hydroxides 3 Environment of high alkalinity conditions 7 Natural waters may contain appreciable amounts of alkalinity under certain conditions suc as a The algae remove C02 free and combined from the water to such an extent that pH Values of 9 to 10 are often obtained b Boiler waters always contain carbonate and hydroxide alkalinity c Chemically treated waters eg those produced in limersoda ash softening of water contain carbonates and excess hydroxide 97AlkalinityrAcidity7309 Method of Determining Alkalinity a Alkalinity is measured volumetrically by titration with 002 N 150 N H2804 b Reported in terms of equivalent CaC03 ie mgL alkalinity as CaC03 When the pH of a sample is gt83 initially Phenolphthalein indicator shows pink color Major species are HCOg C0327 andor OH 1 A known volume of sample is titrated with a standard solution of a strong acid 002 N H2804 until the pH is lowered to 83 the point at which phenolphthalein indicator turns from pink to This value corresponds to the equivalence point for the conversion of carbonate ion to bicarbonate ion C0 H lt gt HCOg phenolphthalein pink 9 at pH 83 Volume of titrant acid used 2 VI The alkalinity is carbonate alkalinity often referred to as jzhenolphthalein alkalinity 2 The titration is continued until the pH is lowered to about 45 The end point is indicated by color change of the methyl orange indicator from yellow orange to red This titration corresponds approximately to the equivalence point for the conversion of bicarbonate ion to carbonic acid HCO3T HJr lt gt H2CO3 methyl orange yellow orange 9 red at pH 45 Volume of titrant acid used an the bicarbonate alkalinity is often referred to as quotmethylorange alkalinityquot Note sometimes total alkalinity is called quotmethylrorange alkalinityquot 3 When the pH of a sample is less than 83 a single titration is made to a pH of 45 At pH lt 83 C0327 0 phenolphthalein alkalinity 0 9AlkalinityAcidity7809 Alkalinity Calculation Equation N1 N1 V1 N2 V2 V1 unit V1 N1 HIL eqL eq N2 V2 mL L 00 V2 Note alkalinity is reported as mgL CaC03 eq 103 meq mg CaC03 mg CaC03 N L eq meq L eq wt of CaC03 50 mgIneq V1 mLN1Lq1000m meq 5 L V2 mL V1 mLN1 10mm 503 CaCO3 mg L eq meq alkallmty as CaCO3 L V2 mL or V N 50000 Balkalinity as CaC03 L L V2 mL or V1 kamk N1 50000 alkalinity as CaCO3 L V2 mL 97AlkalinityrAcidity7309 Example 434 SJ 174 A sample of water has a pH of 90 200 mL of this water require 11 mL of 002N HZSO4 to titrate it to the phenolphthalein end point and require 229 mL of 002N HZSO4 to titrate it further to the methyl orange end point 1 What are the total and carbonate alkalinities of the sample in meqL and in mgL CaC03 2 What is the major alkalinity species and their concentrations Solution N1 002NH2304 v Given Sample water volume V2 2 200 ml pH 9 Titrant conc N1 002 N HZSO4 002 eqL O Vp phenolphthalein end point V1 2 11 mL 0 Vmo methyl orange end point V1 2 229 mL N2 v2 200 mL Solution Note At pH 90 major alkalinity species are C0327 and HCOg because Vp lt Vmo 1 Calculating carbonate alkalinity or phenolphthalein alkalinity C0 H lt gt Hco 9AlkalinityAcidity7809 Phenolphthalein alkalinity Carbonate alkalinity eq 1000 meq VP mL 002 Carbonate Alkalinity meqL eq J 011 meqL 200 mL Carbonate Alkalinity mg L as CaC03 11mL 002 M 50 L 6 meq J55 mgLas CaCO3 200 mL or Carbonate alkalinity or phenolphthalein alkalinity meq of acid required to reach the phenolphthalein end point L of sample Vp002 eqL103 meqeq 11 m1002 eqL103 meqeq F i quot quot39 011meqL 200ml 011 meq 50 mg CaC03 quot 77777777777777777 w 55 rugL as CaC03 an002 eqL103 Ineqeq Methyl orange alkalinity Methyl orange alkalinity 229 Inqu 229 meq 50 mg CaC03 1145 mgL as CaC03 L meq 97AlkalinityrAcidity7309 Vp Vmo 002 eqL103 rneqeq Total alkalinity 200 mL 11 mL 229 rnL002 eqL103 rneqeq Total alkalinity 24 meqL 200 mL 24 meq 50 mg CaC03 120 mgL as CaCO3 L meq 2 Vp 11 mL Vmo 229 mL Since Vmo gt VP the major species present are HCOg and C032 Vp 2 mL acid required to titrate C032 according to the reaction H co3 239 lt gt Hco3 Vp002 eqL 11 mL002 eqL C03 2 200 mL 200 mL 11X10 4eqL NzM 21 H C032 lt gt Hcoy 003239 11x10394M Vrno Vp 002 eqL 229 7 11 mL 002 eqL 218 x103 eqL HC03 200 mL 200 mL NzM 21 H Hcog lt gt Hzcog Hco3 39 218 x 10393 M 9AlkalinityAcidity7809 Note a When pH is very high pH 14 if OH is predominant species HJr 9 H20 VP m0 0 b At lower pH if OH and C032 are predominant species OH H 9 H20 C0 H 9 HCO Vp HCOg H 9 H2C03 Vmo c At lower pH if C032 is predominant species C0 H 9 HCO Vp HJr 9 H2CO3 Vmo d As pH is lower if C032 and HCOg are predominant species C0 H 9 HCO Vp HCOg H 9 H2C03 HJr 9 H2CO3 Vmo Vp lt Vmo d At lower pH if HCOg is predominant species HCOg H a H2C03 Vmo szo 97AlkalinityrAcidity7309 Acidity SJ 177 Major acidity species H H2C03 HCOg H OH lt gt H20 H2C03 on lt gt cogz ZHZO H2C03 OH39 H HC0339 H20 HC0339 OH39 H C032 H20 HCOg OH lt gt C032 H20 Methyl orange Phenolphthalein Acidity Acidity orange Mineral acidity or Methyl orange acidity Adding base OH to achieve a pH of 45 will complete the reaction H OH lt gt H20 C02 acidity or Phenolphthalein acidity Adding base OH to achieve a pH of 83 from pH 45 will complete the reaction H2CO3 0H7 lt gt 7 H20 9AlkalinityAcidity7809 Example 4 35 SJ 179 i 250 mL of a freshly sampled well water approximately pH 67 requires the addition of 58 mL 01N NaOH to raise its pH to 83 and ii the same volume of the water requires the addition of 122 mL Vmo of 01 N HCl to lower its pH to 43 1 What are the total alkalinity and total acidity in meqL 2 What is the carbon dioxide C02 acidity 3 What are the approximate concentrations of the major carbonate species N1 01NHCI N1 01NHaOH v 122 mL V158mL Hco H gt ch03 N2 N2 v2 250 mL v2 250 mL pH 67 gt 43 pH 67 gt 83 1 1 Determine total alkalinity Note At pH 67 predominant species HCOg and H2C03 major alkalinity species is HC03 Alkalinity is determined by Vmo Vmo 122 mL because pH lt 84 HCO37 HJr lt gt H2CO3 97AlkalinityrAcidity7309 122 mL0leq L103 meq eq Total alkalinity 250 L l m samp e 488 meqL Bicarbonate alkalinity methyl orange alkalinity HCO339 488 meqL 488 mM Note N ZM z 1 as HCOg H lt gt H2C03 2 Determine total acidity At pH 67 major acidity species HC03 H2C03 H is very low at pH 67 NaOH H OH lt gt H70 0 mL H2C03 OH lt gt HCOg H2O 58 mL pH 84 HCO 7 is acidity species created HCOg OH lt gt COg H2O 58mL theoretically HCOg OH lt gt COg H2O 122mL theoretically HC03 is acidity species present originally Total volume of base to be added 238 mL Note HCOg OH lt gt C037 H2O 122mLofOH HCOg H 9 H2C03 122mLof H Require same volume of OH or H 238mL0leqL103 meq eq 250 mL sample 952 meq L Total acidity Note Total acidity C02 acidity 1 eq fraction C02 acidity C1 coz C02 acidity total alkalinity Total alkalinity r carbonate alkalinity CT CO3 9AlkalinityAcidity7SO9 2 Carbon dioxide C02 acidity 2 amount of base required to raise pH to 83 phenolphthalein acidity HZCO3 OH H H005 H20 58 mL pHe 84 3 C02 acidity W 250 mL sample 232 meq L 3 At pH 67 Vp 0 mL Vmo 123 mL The major alkalinity species is HC03 see step 1 HC03 H lt gt H2C03 122 mL because pH lt 84 3 HCO3 122mL0leqL10 meg61 488 megL 250 mL sample HCOg 488 mM z l HCOg OH lt gt C0 H20 At pH 67 mineral acidity 0 mL Volume of OH 39 titrant to pH 45 is 0 mL The major acidity species is H2C03 H2C03 OH lt gt HCOg H20 58an 3 HZCOB 58mL0leqL10 meqeq 232 quotmgL 250 mL sample H2C03232 mM zzl H2C03 OH lt gt HCOg H20 97A1kahnityrAcidity7309 Note a When pH is very low if H is predominant H 0H lt gt H20 Mineral Acidity Methyl orange indicator Red to yellow orange Vmo Vp 0 b At higher pH if H and H2C03 are predominant HJr 0H7 lt gt H20 H2CO3 OH7 H HCO37 H20 Vmo HC03 0H lt gt C03 H20 Vp vmo gt vp c At higher pH if H2C03 and HC03 are predominant H2C03 0H lt gt HC03 H20 Vmo HC03 0H lt gt C0 H20 HC03 0H lt gt C03 H20 VP vmo lt vp d At higher pH if HC03 is predominant HC03 0H lt gt C03 H20 Vm00 Vp 411 Mixtures of Acids and Bases 7 pH Calculation SJ 136 Examples of Applications 1 Acidrbase titrations eg for the determination of acidity and alkalinity Coagulation addition of metal ion coagulan s Water softening 7 addition of lime and soda ash for softening Water stabilization in water treatment Acid rain Dissolution of minerals eg CaC03 by groundwater bearing C02 338338 107AcidrBase Titration7309 Neutralization of acid wastes by inrplant base addition or by discharge and dilution in surface waters Note When a strong acid is titrated with a strong base an indicator called bromthymol blue is used because bromthymol blu changes color when the solution reaches a pH of about 7 4111 Strong AcidStrong Base Mixtures SJ 136 Example 424 SJ 136 A 01 M NaOH solution is added to 1 L of 10 3 M HCl solution What is the pH after addition of a 5 mL h 10 mL and c 20 mL of 01 M NaOH to 1 L of 10 3 M HCl solution Conditions the solution temperature 25 C neglect ionic strength effects Solution HCl is a strong acid and NaOH is a strong base thus their dissociation in aqueous solution is assumed to be complete Equilibria HC1 a H or 10 3 10 3 NaOH 9 Na OH 10 1 10 1 H20 lt gt H OH KWH OH 10 14 Mass Balance number of moles CT total volume NaOH 01 M V 5 1o 20 mL C HCI 1x10393M V1 L 1 10AcidyBase Titration7309 10 3mol L 1L C C 2 TH l 1L Vaddzd L 10 1moz L Valid d in L C N quot 3 TYNE a 1LVadd2d L Charge balance Nal Hl C1 OH 4 a When 0005 L 5 mL of NaOH is added 10 3mol L 1L 4 C Cl 9950 10 IL 2 m 1L0005L x quotw 0 71 CT NR Nf W 4975x10 4molL 3 39 1 L 0005 L Substitute 2 3 into Charge balance 4 Nal W C1 OH 4 4975 x104 H 9950 x104 OH We assume H gtgt OH because acidic solution as 10393 molL 1L HC1 gtgt 10391 molL 0005 L NaOH H 9950 x 10394 4975 x 10394 4975 x104 We obtain pH 33 OH KWH 103914 4975 x104 2010 x1011 We obtain OH 107 p Check with charge balance Na PF C1 OH 4 4975 x104 4975 x104 9950x10394 2010x 103911 998687 x104 99500 x104 error lt 5 OK 10AcidyBase Titration7309 b When 001 L 10 mL of NaOH is added 10 3moz IL 1L C Cl 9901 101 IL 2 m 1L001L x quotm U 71 CT Na Nal W 9901x10 4molL 3 39 1L 001 L From charge balance Nal H Cl OH 4 9901x10 4 H 9901x10 4 OH39 Since H OH pH7 Note NaOH HCl 9 NaCl H20 We therefore haveasalt solution WithapH of7l c When 002 L 20 mL of NaOH is added 73 CT CI Cl W 9804x10 4moz L 2 v 1 L 002L 10 1mozL 002 L C N W a 1L002L 1961x10 3mozL 3 From charge balance Na H C1 OH 4 1961 x 103 H 9804x10 4 OH39 We assume OH gtgt H basic solution because 10 3 molL 1L HCl gtgt 10 1 molL 002 L NaOH OH 1961 x 103 9804 x 10394 9806 x 10394 We obtain pOH 301 H KW OH 10149806 x 10 4 10198 x 10 11 We obtain pH 1099 Check with charge balance Na H C1 OH 1961 x 103 10198x1039 9804x10394 9806x10394 1961x10393 1961x10393 errorlt5 OK IOVAcidrBase Titration7309 H250A 1 M Example 4 25 SJ 138 v mL What volume of a 1 M H2804 must be added per 1 L of 10 1 M NaOH solution to pH 10 Note that 01 M of NaOH yields 1 01 0 Initially lons M z Ci Z 2 Na 01 1 0112 OH 01 1 0112 quotquotquotquot 39E 255275mmquot39 HZCizi2 01 c NaOH01 M v 1 L Conditions 1 T 25 C 2 The ionic strength 1 01 after titration to pH 10 Equilibria NaOH 9 Na OH 10391 10391 H2804 9 2H 04 2 HSO4 is negligible at high pH 1 1 H20 lt gt H OH Kw H OH Kw YH Hl 39 Y OH7 OH 10 14 Mass Balance 1 CT Na Na 10 mol L 1 L 01moles 1 1LV L 1V L 1 molL V L 1V moles C so 2 2 T5042 4 1LVL 1VL U 10AcidyBase Tiu ation7309 Charge balance Na H 2SO42 OH 3 From given conditions At pH 10 H 1010 and OH 10 4 From Fig 34 p 78 we obtain VH 083 and y OH 075 at u 01 Since H yH H H HyH 103910083120x103910 4 Since OH y OH OH OH OH y OH 104075 133 x 10394 5 Substitute 1 2 4 and 5 into 3 i 120640 27V 133x104 1V 1V Assume Na gtgt H or 12 x1010ltlt 01 1V at pH 10 i27VL33x10 4 1V 1V Solve for V 017 2V 133 x 10394 1V 133 x104 133x10394 V 2V 133 x104 V 017133 x104 20001 V 00999 V 00499 L 499 mL of H2804 10AcidyBase Titration7309 Titration of Weak Acids and Strong Bases SJ 139 Weak Acid estrong Base and Strong Acid Weak Base Mixtures Example 4 26 p140 Consider a solution contains 10 3 M of acetic acid HAc a What is the solution pH after 00025 0005 and 001 L of 01 M NaOH and 003 L of lM NaOH has been added per 1 L b Sketch a titration curve equivalent fraction f added vs pH and mL titrant added vs pH and shows the relationship between the titration curve and the pC pH diagram Conditions T 25 C and neglecting ion strength effect NaOH 0 01M 00025 0005 001 L v HAc lt gt H Ac 10M 003 L The equivalent fraction f is defined as number of equivalents NaOH added f O CTAc number of equivalents NaOH added 0 HAcAc Co HAc 1x10S M f C V V0 1 L C0 V0 Assumptions 1 the addition of NaOH causes no volume change since the volume of base added per liter is very small Reactions HAc lt gt H Ac H20 lt gt H OH NaOH gt Na OH HAc Na OH gt NaAc H20 Equilibria Ka 104 7 pKa 47 HA6 Kw H0H1014 Mass Balance CTVAc HAc Ac 2 10 3 01 V or 1 V 1V 1V CTJVa N011 where V is in liter Charge balance Na H 2 Ac OH Proton condition Reference sp HAc H20 Hl AC l OH 1 1 2 3 4 5 6 IOVAcidrBase Titration7S09 1 We plot a pC pH diagram for the acetic acid HAc Ac system using Eqs 1 2 and 3 pCTVAc 3 and pKa 47 2 Determine the initial pH 1V0mL Assumptions initially Na 0 and H gtgt OH because HAc is weak acid From Charge balance 5 Na H 2 Ac OH Thus from charge balance H 2 Ac From pC pH diagram pH 36 Equivalent fraction f IOVAcidrBase Titration7309 f CV z LZWJM Co Va 10 3 mol L J00 L Based on approximation method Assumption acidic solution H gtgt OH From proton condition 4 Hr 2 Ac 6 From mass balance 3 HAc 10 3 Ac 10 3 H 7 Substitute 6 and 7 into 1 2 FEM 11 Kg 2 210 407 HA6 10 3 11 Let X 2 Hr and solve for X x2 1047 x 4077 0 10 7 two 472 4x10 7 7 2 x 2632 x 10394 H 2632 x 10 4 pH 358 2 V 00025 L 01 M NaOH From 4 01 lL 00025 L Cm Ni W 25 2610 4 molL 10 molL pCTNa 36 IOVAcidrBase Titration7S09 From the charge balance 5 Nal Hl AC l OH 1 5 25 X 1074 H 2 AC OH 7 Assume the solution is still acidic OH 0 The Na and H are the same order of magnitude So plot p Na H vs pH Preliminary calculation for the p Na H vs pH plot og Na H Na H Na H pH p Na H 10393 8 10393 10392 9 3 290 103988 103985 10325 35 325 10393 8 10394 10393 45 4 345 10393 8 103945 1035489 45 355 10393 8 10395 103583 5 358 103988 103955 103595 55 350 From the plot Na H 2 Ac we obtain pH 44 C V 00025 L f 025 Co Va 10 3 mol1L L If you assume H ltlt Na and OH ltlt Ac on the charge balance you will get Ac 2 25 X 10 4 and find pH 43 at pAc 36 Charge balance Nal Hl AC l OH 1 5 Na 2 Ac 2 25 x10 4 pAc 36 10AcidyBase Titration7309 3 V 0005 L 01 M NaOH From 4 01 molL 0005 L WW Jlt gtlt gt 1L 5x10 4 molL From the charge balance 5 Na H A6 OH 5 5x104 H Ac OH 1 Assume 5 x104 gtgt H and Ad gtgt OH 391 Then Ac 5 x104 p Ac 33 From the graph pH 47 M0005 L C V L 7 05 f C V 3 0 0 10 mol1L L 4 V 001 L 01 M NaOH From 4 CT N 1M 0 1m01L0 01 L 1 x 10 3 molL 39 1L Note that Mass Balance CTVAc HAc Ac 2 10 3 2 N51 From the charge balance 5 Na H AC OH 5 IOVAcidrBase Titration7309 Because CTVAc HAc Ac 2 10 3 Na HAc Ac H 2 Ac OH Assume HAc gtgt H Thus HAc OH From the diagram pH 78 01mol001L CV L C V 3 10 0 0 10 Lmol1L f 5 V 003 L 10 M NaOH 10 mol L 003 L 1 L CW Nu 3x10 2 molL From the charge balance 5 Na H AC OH 5 3 X 102 H 2 Ac OH Assume 3 X 10 2 gtgt H and Ac ltlt OH basic solution OH 39 3 x102 or pOH 15 From the diagram pH 125 01mol003 L C V L ziz zgo f C 0 V0 10 3 mol1L L 10AcidyBase Titration7309 Summary Cone of Volume of Equiv Fraction NaOH NaOH pH M L 01 0 0 36 H A0 Initial pH 01 00025 025 44 Na H A0 01 0005 05 47 AHo A0 Midpoint pKa pH 01 001 1 78 HAo OH 39 Equivalence point pH 10 003 3 125 Na OH39 f lt 05 Addition of base produce a solution which has pH between that of the original HAc solution and the PKa f 05 Mid point pH pKa The titration curve has an in ection point The solution is very well buffered I The change of pH for a given addition of strong base is small f 10 Equivalence point fgt10 pH is that of a 10 3 M NaAc solution The curve has an in ection point in the opposite direction to that at f 05 The solution is very poorly buffered I The resistance of the solution to pH change on the addition of strong base or strong acid is small The pH increases very slowly as it approaches the pH of the solution being added 10 Acid Base TitrationSO9 13 P 1 pC l I l f i P l h 2 l Y 39 3 o i LE 39 2 l t l 3 a I E A 4 gt R gt I I I 1 It K I I O 1 2 3 37 4 5 6 7 T8 9 10 I 12 13 14 Imm 4W E MVAQNQ m Pkck PH Pom w GMc S NAOH ml of 01 M UV IOVAcidrBase Titration7309 Approximation Weak acid pH during titration with strong base Initial pH pH Z 12 pKa pC Midpoint pH pH Z pKa Equivalence point pH pH Z 12 pKa pKW pC Weak base pH during titration with strong acid Initial pH pH Z pKW 12 pr pC Midpoint pH pH Z pKW pr Equivalence point pH pH Z 12 pKW pr pC Example Weak acid pH during titration with strong base Determine initial pH midpoint pH and equivalence point pH Given pKa 47 pC 3 Solution Initial pH f0 pH 12 pKa pC 12 47 3 385 Midpoint pH f 05 pH Z pKa 47 Equivalence point pH f 1 pH12pKa pKW pC 124714 3785 lerolubility7309 Solubility Chap 6 7 Precipitation and Dissolution S 243 Fundamental Concept No matter how insoluble all solids are soluble to some degree For example silver chloride AgCl barium sulfate BaSO4 and strontium sulfate SrSO4 are considered to be insoluble but they dissolve slightly in contact with water and form equilibria AgC1S lt gt Ag C17 Kssz50178x 103910 BaSO4S lt gt Ba2 so Ksszsoz 11 x 103910 SrSO4s lt gt 812 042 Ksszsoz 28 x107 In lime soda ash softening process we intend to convert soluble hardness Ca2 and Mg to insoluble hardness CaC03 30 mgL as CaC03 or 06 meqL MgOH2 10 mgL as CaC03 or 02 meqL Limiting total hardness is 40 mgL as CaC03 or 08 meqL Solubility Rules 1 All salts of the alkali metals are soluble eg NaCl NaHC03 KCl 2 All ammonium salts are soluble eg NH4ZSO4 3 All salts containing the anions nitrate N037 chlorate C1037 perchlorate C104 and acetate C2H302 are soluble Exceptions AgC2H302 and KClO4 are slightly soluble 4 All chlorides Cl bromides Br and iodides I are soluble Exceptions those of Ag Pb and Hg are slightly soluble Note that mercury in the 1 oxidation state exists as the ion ng 2 eg PbClz is slightly soluble 5 All sulfates 8042 are soluble Exceptions are those of Pb 812 and Ba The sulfates of Ca2 and Ag are slightly soluble 6 All metal oxides are insoluble Exceptions those of the alkali metals and Ca2 Sr2 and Ba2 are soluble Metal oxides when they dissolve react with the solvent to form hydroxides e g Ca0 H20 9 Ca2 20H 7 All hydroxides are insoluble e g MgOH2 Exceptions those of the alkali metals Ba2 and 812 CaOH2 is slightly soluble 8 All carbonates phosphates sulfides and sulfites are insoluble Exceptions those of NH4 and the alkali metals 127301ubility7309 Principles of Solubility Butler p 174 FampC p 106 Table 36 SJ p 449 Tablebrl Two phase equilibria solid and liquid phases Solubility Product Ksp K50 Aa EMS lt gt a Ab b Ba Note ab ba Aba Ba b Ama 131b Ksp K50 Solubility Product 13141339 Note the activity of the solid phase is 1 by convention a b a b Am Bk 2 Aha Am 78 Bk 2 KW KSp A 133 B 2 b KSp wt 3 x M b For simplicity let us neglect the ionic strength effect and assume all activity coefficients vi 1 so that cK51 KSp Example a Calcite limestone CaC03 pKsz 834 Table 6 1 S 449 CaC03S lt gt Ca2 C03 Ca2 COf K CazCO32 10 83934557X10 9at25 C Ca 3s 1 b MgOH2s PKsp1074 MgOH2S lt gt Mg2 on Mg2 0H 12 Mg2 OH2 182 x 10 11 at 25 C 127301ubjlity7309 c Dolomite CaMgC032S CaMgC032 lt gt Ca2 Mg2 2C032 Ca2 Mg2CO32 12 Keq Ca2 Mng CO32 2 2 1017390 at 25 C Table 36 Ca C0s 1 320 d Albite NaA1s1308s NaAlsi308s H2CO3 H20 lt gt N3r HCO37 12A12S1205OH4S kaolinjte 1 W Hcog H4s10412 AlzsyxasFOH45J NaAl 85 H2C03 1M 1 Na HCO3 H4SiO42 H2C03 AS Pcoz thus H2C03T NaT HC037 T H4SiO4T 127301ubility7309 Solubility of Ionic Salt in Pure Water Solubility 7 the mass of the mineral that will dissolve in a unit volume of solution a If S moles of solid ABS dissolved in pure water to form 1 liter of saturated solution ABS lt gt A B7 S S Ksp N B s s s2 s Kspm b If s mole of solid AaBbs dissolved in pure water to form 1 liter of saturated solution A313 lt gt aAb bBa as bs K5 Aba Ba b asabsb then solve fors Example Calculate the solubility of calcium uoride Can in pure water at 25 C knowing that KSp of Can 146 X 1010 Neglecting ionic strength effect Solution In water saturated with Can Can lt gt Ca2 213 s 2s Note s moles of Can that dissolves in pure water produces s moles of Ca2 and 2s moles of F K5 Ca2 13 2 s 2s2 483 146 x 10 10 at 25 C s 146 x10 104 3 3317 x1041 molL or 3317 x 10 4 molL78 x 103 mgmol 2587 mgL 127301ubjlity7309 Example 1 Calculate the Ksp of magnesium hydroxide MgOH2 in pure water knowing that Ksp of MgOH2 561 x 10 12 at 25 C 2 Calculate the Mg2 at pH 4 pH 7 and pH 11 Solution In MgOH2 saturated water MgOH2S lt gt Mg2 ZOH s 2s Ksp Mg2 OH 2 s 2s2 4 s3 561 x 10 12 at 25 C s3 561x10 124 140255 x10 12 s 140255 x1042 3 1119 x10 4 molL a At pH 4 pOH 10 thus OH 1010 KSp Mg2 OH 2 561 x 103912 561 x 103912 561x103912 Mg2 561x108 OHi2 107102 b At pH 7 pOH 7 thus OH 10397 561x10 12 561x10 12 Mg2 561x102 OH 2 1072 c At pH 11 pOH 3 thus OH 10 3 561x103912 561x103912 Mg2 561 x10 6 OH 2 1032 This is the reason why we raise the pH to remove Mg hardness in a lime soda ash water softening process 127301ubjlity7309 The Common Ion Effect Definition Decreased solubility of an ion due to an excess of another ion in the KSp expression Example SMc p33 Barium sulfate BaSO4 and barium chloride BaC12 BaSO4 is insoluble KSp 12 X 10 10 BaClz is very soluble 10 X 10 5 M of BaClz is added to a saturated BaSO4 solution Consider a solution that has been saturated with barium sulfate BaSO4 BaSO4S lt gt Ba2 so4239 S S 1 Determine solubility s Ksp Ba2 SO42 s s s2 12 x1010 s 12x 1040 m 11x10395 BaSO4 S lt gt Ba2 sof s s 11x10395 11x10395 2 If the barium ion Bah concentration is increased by an addition of BaClz the concentration of sulfate ion SO42 must decrease and the amount of precipitation BaSO4 must increase in order for KSp to remain the same If 10 X 10 5 molesL of BaClz is added to the above solution BaClz lt gt Ba2 2 C17 10 x105 10 x105 210 x 105 Very soluble The addition of BaClz will result in the formation of additional moles of precipitate BaSO4 S and establish new equilibrium with new solubility BaSO4 S lt gt Ba2 0 4 11X10 510X10 5 P 11X10 5P lerolubiJitySO9 According to the solubility product principle Ksp 1323 SO42 12 x 10 10 12 x1010 11 x105 10 x10395 P 11 x 105 P Thus P 098 x 10395 molL Ba2111X 10 5 10x 10 5 7 098 x105 101 x 10 5 SO4211X10 57098 x105 012 x105 By solving for P it is found that an additional 098 X10 5 molL of precipitated BaSO4 is formed and the new equilibrium concentrations of Bah 101 X 10 5 and SO42 012 X 10 5 molL This calculation indicated that SO42 is reduced considerably 11 X 10 5 9 012 X 10 5 Diverse Ion Effect Definition The adverse effect of unrelated ions on the solubility of some relatively insoluble substances a The solubility of slightly soluble salts can be increased when other salts that do not have an ion in common with the slightly soluble substances are introduced b The introduced salts interfere or decrease the effective concentration activity of the slightly soluble substance c Theoretically such ions play no part in the chemical equilibrium involved but often increase the solubility of desired precipitates to such an extent that quantitative results cannot be obtain d The explanation for this effect is that the forces of attraction caused by the charge on the unrelated ions decrease the activity or effective concentration of the slightly soluble ions Example 1 Nitrate ion has such an effect on silver chloride 2 Salinity increases the solubility of gypsum 2 Comparison of the solubilities of gypsum in pure water versus water with high salt content indicates that the salinity increases the solubility of gypsum CaSO42H20S lt gt Ca2 0 2H20 KSp10 45 Table 6 1 S 449 Gypsum 1 Ca sof mm Caso zom 1 KSp 127301ubjlity7309 Ksp Ca2 042410745 Ksp y Ca2 Ca2 y SO42 0421 For example i at low salinity u 0001 V cm 087 and y 3042 087 Ksp Y Ca2 C32l 39 Y so424 50427 Ksp 087Ca2 087 8042 103945 3162 x 10395 Ksp Ca2 031 3162 x 105 0872 418 x 10395 CaSO42HZOS lt gt Ca2 042 2H20 S S KSp Ca2 8042 s s 32 418 x 10395 s 418 x 105 2 647 x 10393 molL s 647 x 10393 molL ii at high salinity u 1 1 cm 2 024 and y 3042 008 Ksp 024 Ca2 008 8042 1045 3162 x 10395 cKsp Ca2 8042 3162 x 10 5 024 008 165 x 10 3 Ksp Ca2 8042 s s s2 165 x 103 s 165 x 103 2 40 x 10392 molL s 40 x 10392 molL Thus gypsum is more soluble in water with high salt content high 11 647 x103 M at 11 0001 a 40 x102 M at 11 1 Explanation The solubility of gypsum is increased by decreases in actiVity coefficients as a result of increased ionic strength lerolubility7309 Conditional Solubility Product Ps Ps CTM 110 where PS 2 conditional solubility product CTVM total concentration of the metal ion M in all of its complex forms CTVA total concentration of the anion A in all of its forms The value of Ps may change as a function of solution properties eg pH concentration of complex forming species because the fractions of CTVM and CTVA present as the free cation and anion may vary as a function of these properties Example A1OH3S lt gt Al3 30H Ksp A13 0H3 1033 If we measure Al3 other complexes Therefore the A such as OH Thus in solution we may actually measure Al3 AlOH2 AlOH2 13 in solution id dependent on the conditions of the solution Ps CTM Cm Ion Activity Product IAP A313 lt gt aAb bBa For unknown saturation condition Aba Barb Z Aba Barry If IAP Ksp saturated solution If IAP lt Ksp unsaturated solution If IAP gt Ksp supersaturated solution Saturation Index SI SI log K51 If SI 2 0 saturated solution If SI 2 unsaturated solution If SI 2 supersaturated solution lerolubility7309 Separation of Compounds by Precipitation separating soluble ions by precipitating one first Example Industrial wastewater contains 001 M BaClz and 001 M SrClz The industry wants to recover Ba2 and Sr2 separately from the waste stream We precipitate Ba2 and Sr2 as sulfates see rule 5 exceptions by the addition of sodium sulfate NaZSO4 BaClz NazSO4 lt gt BaSO4S SrClz Nast4 lt gt SISO4S 2 1 BaSO4s lt gt Ba2 so4239 KSPBaZso423911x103910 2 SrSO4S lt gt sf so4239 1szslr2so423928x10397 a Which metal will precipitate first with the addition of NaZSO4 Ba2 will precipitate first as BaSO4S because BaSO4S has smaller solubility smaller Ksp b Which one Ba2 or Sr is more soluble in a solution in equilibrium with both solids BaSO4S and SISO4S In equilibrium Sr2 so42 28 x 10 7 Ba2 804239 11 x 1010 Sr2 is more soluble because 812 is larger than Bah by 2500 times c What is Bah when Sr2 begins to precipitate Sr2 28 x 10 7 25 x 103 When 812 becomes less and less Bah becomes less and less keeping the same ratio When Sr2 begins to precipitate Sr2 001 M lerolubility7309 001 25 x103 Bah Bah 00125 x103 4x 10 6M 635 Logarithmic Concentration Diagram SJ 254 7 Graphical Method Example BaSO4ltSgt H Ba 5042 Ksp Ba2 042 11 x 1040 1 SISO4S lt gt Sr2 SO42 Ksp Sr2 SO42 X From 1 Bah KSp SO42 log Bah log Ksp log SO42 pBa2 p SO42 log 11 x 1040 pBa2 p SO42 996 From 2 812 Ksp SO42 log 812 log Ksp log SO42 p Sr2 p SO42 log 28 x 10 p 812 p SO42 655 From the logarithmic concentration diagram you will see a Ba2 precipitates first b Ba2 precipitates as BaSO4S at P504 8 or SO42 108 When Sr2 begins to precipitate pBa2 54 Bah 1054 4 x 10396 12 5 Is or ZZLLZM4MI 1 79 II bu I 1 IL A 3 quot7 4w n a f 12 Solubility7809 127301ubjlity7309 Example SJ 254 Logarithmic Concentration Diagram CaOH2S lt gt Ca2 20H szp 53 KSp 1053 MgOH2S lt gt Mg2 20H szp 1074 Ksp 104074 Ca2OH 2 KSp flogCa2 72 log OH flongp flogCa2 2 pOH szp pCa2 72 pOH 53 53 0 03 25 Mg2OH 2 Ksp 710gMg2 72 log OH 40ng 710gMg2 2 pOH szp pMg2 72 pOH 1076 1076 0 5 076 See Fig 6 3 S1255 127301ubjlity7309 Example 6 6 SJ 256 Find the solubility of l CaOH2S and 2 MgOH2S in distilled water Solution Step 1 Determine pH or pOH of the CaOH2S solution Step 2 Determine the solubility CaOH2S lt gt Cu2 20H H20 lt gt H OH Note that solubility of Ca2 is in uenced by OH of water Charge balance 2Ca2 H OH Assumption H ltlt 2Ca2 because OH is added as CaOH2S Thus 2Ca2 OH 710g 2 710g Ca2 710g OH39 pOH 710g Ca2 03 21 Since pKw 2 pH pOH pOH pKw 7 pH b a and b must satisfy See S Fig 6 3 Plotting the line flog Ca2 03 we find this equation is satisfied at a point pOHl7 Note that the assumption that H ltlt 2Ca2 is valid at this point When pOH 17 we find 710g Ca2 2 Ca2 10 2 M Thus solubility s of Ca s 1x102 M ilgOH2S lt gt Mg2 20H H20 lt gt H OH Charge balance 2Mg2 H OH Assumption H ltlt 2Mg2 Thus 2Mg2 OH 710g 2 710g Mg2 710g OH39 pOH 710g Mg2 7 03 a lZ SOlubilitySO9 Similarly plotting the line log Mg2 03 we find this equation is satisfied at a point pOH35 When pOH 35 we find log Mg2 38 Mg2 10 3 8 M 16 x 10 4 M The solubility of Mg2 s 16 x 104 M 15 127301ubjlity7309 127301ubjlity7309 Note MW of Mg 2 243 gmol Eq wt of Mg 2 2432 2 1215 mgmeq Previously we calculated the solubility of Mg s Mg2 16 X 10 4 M 16 x104 mol 243 g 1000 mg 3888 mgL solubility of Mg L mol g 3888 mgL 032 meqL 1215 mgmeq 032 meqL50 mgmeq CaC03 16 mgL as CaC03 Mg2 is not totally removed by lime softening practically 10 mgL as CaC03 Example Gypsum CaSO4S CaSO4S lt gt Ca2 0 KSp Ca2 SO42 10396 Ca2 Ksp 804239 10gCaz flog KSp log SO42 pCa2 szp 7 p 5042 pCa2 7 p SO42 6 Silver sulphate AgZSO4S AgZSO4S lt gt 2Ag 042 Ksp Ag2 SO42 10748 A512 Ksp SO42 127301ubjlity7309 7210g Ag 710g Ksp 10g SO42 710g Ag 1210g SO42 7 1210g Ksp 12 p SO42 12 szp 12 p SO42 12 48 P Ag 7 12 pSO42 24 Note valence changes slope and intercept Precipitation Titrations Example Butler 186 50 mL of 010 M NaCl is titrated with 010 M AgN03 1 Calculate Ag as a function of volume of AgN03 added and plot the titration curve 127301ubility7309 01 M AgNo8 VmL Na Ag NaCl AgNo3 C1 gt gt 50 mL 01 M NaCI Overall NaC1 Ag NOg lt gt AgC1S Na S olubility product AgC1S lt gt Ag Cl KspAgC1 178x1010 Mass balance 01molLVmL Cm AgP 50VmL 01molL50mL 50VmL Cm Cl P We have 3 eqns and 3 unknowns Ag Cl P a Initially Ag ltlt Cl SO V 50V SO V 50V 3 2 Cl Ag01 Since Ag ltlt Cl Cf 01 Substitute 5 into 1 1 2 3 4 5 127301ubjlity7309 Ksp Ag Cl Ag01 5 4 178 x103910 50V Solving for Ag gives Agl78x10710 SO V 6 0 150V b Before and near the equivalence point Ag and C1 are on the same order V 50 2 73 Ag Cr 01W 4 crAg01V 50 7 V50 Substitute 7 into 1 50 Ksp AgC1 AglAg 0l8260 178 x1010 Solving for Ag V 50 50 KW AgAg01 Ev50Ag2 01Ag E50 Ag2 01 V SO V50AgKsp0 01w 01 ll 50 4 Ksp V 50 V 50 Ag 1 2 V 50 V 50 2 0 1 V50 01 V50 4K A81 2 l 5P 20 127301ubjlity7309 plCl39l AgNOs added V mL Example 34 SJ 72 Temperature effect on solubulity A municipal water supply enters a residence at 15 C and the water is heated to 60 C in the home water heater If the water is just saturated with CaC03S at 25 C what will be the condition of the water ie oversaturated or undersaturated with CaC03S i as it enters the residence and ii as it leaves the water heater Solution Dissolution of CaC03S CaC03S Haq lt gt HCO3raq Ca2aq A 28845 0 16518 42977 kcalmol 25 C 86 26978 0 44031 13218 kcalmol 25 C AH 16518 7 12977 7 28845 65 kcalmol Exotherrnal reaction AG 14031 7 13218 7 26978 271 kcalmol 21 127301ubjlity7309 CaC03s Haq lt gt HCO3raq Ca2aq HC03 Ca2 HC03 Ca2 Km 3993111111 H Define the equilibrium constant Keq with respect temperature 1K1 Keq25C at T125 C 298 K 2 K2 Keq15C ath 15 C 288 K 3 K3 Keq60c at T3 60 C 333 K 1 At T1 25 C 298 K the concentration of Ca2 HCOg and H are at saturation AG 7R1 1n Ksq 1n K1 AG I7RT 271 kcalmol 199 x 10 3 kcalmol K298K 4593 K1 e439593 9881 2 At 12 15 C 288 K 1n AH or anl an2AH K2 R T2 T1 R T2 T1 4593ln K2 657ltcal lmol 1 1 199x10 kcalmol K k288K 298K 1n K2 2 49755 K2 2 e 49755 1448 3At 13 60 C 333 K 4593 1nK3 65 kcalmol 1 1 199x10 3kcalmol K k333K 298K 1n K3 3435 K3 e 3435 2 310 22 127301ubjlity7309 Surnrnary 15 C K2 2 1448 25 C K1 9881 60 C K3 2 310 Solution Dissolution of CaC03S CaC03S Haq lt gt HCO3raq Ca2aq HC03 Ca2 HC03 Ca2 Roxanna Ir 1 Note Ion Activity Product IAP A313 lt gt aAb bBa For unknown saturation condition IAP Aba B3113 Z Aba B3113 If IAP Ksp IAP KSp 1 saturated solution If IAP lt Ksp IAP KSp lt1 unsaturated solution If IAP gt Ksp IAP KSp gt1 supersaturated solution IAP 988 At 25 C K1 25c IAP 1 saturation K1 9881 IAP 988 At 15 C K215c gt Keq 25c lt 1 under saturation K2 1448 IAP 988 At 60 C K3 50c lt Keq 25c gt 1 super saturation K3 310 23