Additional Aqueous Equilibria - PDF
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2025
S. Gulde
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These are notes and practice questions on additional aqueous equilibria, covering topics such as the common-ion effect and buffer solutions. It includes calculations for pH and solubility. The material is copyrighted to S. Gulde 2025.
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CHE-102 Lecture Outline Chapter 17 Additional Aqueous Equilibria Please watch PL-09 to fill in the beginning of these notes. Location: Lecture Recordings→CHE102-Recordings→...
CHE-102 Lecture Outline Chapter 17 Additional Aqueous Equilibria Please watch PL-09 to fill in the beginning of these notes. Location: Lecture Recordings→CHE102-Recordings→Dr. Gulde’s PL-Videos Common-Ion Effect  Shift Common ion effect – __________________ in equilibrium due to the presence of an ion _____________ involved in the equilibrium Le Chatelier's  AKA ______________________________________ Principle  I If add NaC2H3O2, what happens? ionic  B/c it is ________________, 100 % into __________ it breaks up__________ ions Nat Nat (____________ CHz0z & ____________________) C2H302 shift & -  More ___________________ causes equilibrium to _________________ dec Thus, [H3O+] _____________. & pH ______ basic inc (becomes more _______________) weak  Dissociation of a ___________________ decrease electrolyte (ie. HC2H3O2) will__________________ common ion when a strong electrolyte (ie. NaC2H3O2) containing a ________________________________ C2Mz0z (ie. _____________________) is added (W Acid). E  Which of the following will affect the pH of an HClO2 solution? (w acid). i. NaClO2 ii. NaBrO2 iii. LiClO3 & A. i only H((02(ag) H20(1)- H30 lag) Cloz (ag) + + + B. ii only C. iii only D. All will E. None will © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 1 CHE-102 Lecture Outline Chapter 17 Sample: pH HF 1. Calculate the pH of a 0.50M HF solution. Cu acid). HF = 0 50M Kat I (ag). 7 (ag) H0(l) =130"(ag) + + + 1 0 SOM g 0 = Y 6 8x10. -. C - X + x + Y ↑ appendix A3 0 +X 0 +X E 0 50 X = -. 6 8x10. Check 5 % Rule : ignorex % ) x100 6 8 x10"= x = 0. 018 M by. %) = 3 6%. log(0 018n) pH 1 74 (Assumption OK) pH = -. =. Sample: pH HF & NaF 2. Calculate the pH of a 0.50M HF solution when 0.10M NaF is added. (w acid). Cionic)  When 2 chemicals are involved, always write the equation for the _____________________________________! WEAK electrolyte Nat = 0 IOM. 0 IOM · F = (ag) = H20(l) H30 (ag) f(ag). + + = + 50 M HF = 0. o som O 0 10 M --. C X 6 8x10-4 + X. - ↓ X E 0 50 -X X 0 10.. + X 6 8x10-. _ C % 10 400 % = 0 68 %. ignore pH = - 10g (0 0034]. = 247 (CO) X 6 8x10 -" More basic 0T. = x= 0. Lecture Summary [HF] only :  Common Ion Effect: decrease  Will ___________________ the dissociation of a weak electrolyte pH = 1 74. 0.50M HF, pH = ________ 1 74. 0.50M HF & 0.10M NaF, pH = ________ 2 47. © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 2 CHE-102 Lecture Outline Chapter 17 Buffer Solutions  Buffers – contain a weak acid-base conjugate pair weak Made up of a ______________ acid/base & its _______________________ acid/base (__________) conjugate salt common , ion Should have ______________________ similar concentrations Sample Problem ______: #2 0.50M HF with 0.10M NaF  Examples: (W acid). (CB) HC2H3O2 & NaC2H3O2 NH3 & NH4Cl Question: Buffer  Which of the following is a buffer solution? electrolyte HNO3 & NaNO3 - weak ion HNO3 & NaOH - common HNO2 & NaNO3 O HNO2 & NaNO2 HNO2 & HNO3 Buffer Calculations:  When calculating the pH of a Buffer solution (2 chemicals involved) use: a) ICE table: Seen in problem #2 (pH=2.47) OR pH = pK a + log [base] [acid ] b) Henderson-Hasselbalch eq. (H-H eq):  Sample Problem #2 Revisited: Calculate pH of a 0.50M HF solution when 0.10M NaF is added. W acid. Cong. base Y Ka - = 6 8. x 10 0 IOM O SOM.. pH = -1096. 8x10" + log s = 3 16. , + 7- 0. 698) = 2 47. © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 3 CHE-102 Lecture Outline Chapter 17 Sample: Buffer Calc (w acid). Cionic)  Calculate the pH of a 0.40M HA solution when 0.10M CaA2 is added. (Ka of HA=5.1x10−5) - + (2) (0 10). HA(ag) 120(l) Hz0 (ag) (ag) A - > = A - + + Ka 0 40M. 0 20M. PH pKa= + base log BUffER ICE Or H-H = 4 2924. + log = 4. 2924 - 3010 = 3 9914. Buffer Purpose  Buffer Purpose: ____________________________ resist change when limited amounts of pH _________________ strong acid or base is added Natural buffers: human blood ________________________ maintains pH of 7.4 C C C _____________________ Seawater maintains pH of 8.1-8.3 Buffer Capacity  Buffer Capacity – amount of acid/base added before a significant pH-change ___________________________________ ________________ occurs  drops of GM In video: count _______________________ added a) ~1 drop: turned neutral water Very Acidic _____________________ b) ~10 drops: turned 0.1M buffer _____________________ Slightly more acidic c) ~10 drops added to 1.0M buffer _____________________ No effect  Capacity depends on ______________________ initial conc. of acid/base in buffer Which has a greater buffering capacity? 1 M HF & 1 M NaF - aa moreinitial materi acid/base 0.1 M HF & 0.1 M NaF © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 4 CHE-102 Lecture Outline Chapter 17 Buffer Range  Buffer range - ___________ PH range where buffer is __________________ effective  Want __________________ conc. of acid/base & conjugate to be ________________________ similar Ex. 0.50M HF & 0.10M NaF  If conc of 1 component in buffer is _______________ < 10X the conc of the other, buffering good action is __________________  Choose an acid whose ____________ pKa is close to the desired pH IIpH Range = _____________ unit of pKa Questions:  Which of the following will have the worst buffering Range? 0.1M HA and 0.1M A− 0.2M HA and 0.1M A− O 0.1M HA and 1.5M A− 1.0M HA and 1.0M A−  Which of the following would have the best buffering capacity? 0.1M HA and 0.1M A− 0.2M HA and 0.1M A− 0.1M HA and 1.5M A− O 1.0M HA and 1.0M A− more initial material  You wish to make a buffered solution with a pH = 7.0, which acid will be the best to use? i. HNO2: Ka = 4.5x10−4 pKa = 3.35 ↓ best ii. HClO: Ka = 3.0x10−8 pKa = 7.52 4 52. iii. H3PO4: Ka = 7.5x10−3 pKa = 2.12 iv. H2CO3: Ka = 4.3x10−7 pKa = 6.37 0 68. v. HBrO: Ka = 2.5x10−9 pKa = 8.60 A) ii only B) iii only C) iv only D) i & v s E) ii & iv © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 5 CHE-102 Lecture Outline Chapter 17 How a Buffer Works  Buffer Purpose: ____________________________ resist pH change when limited amounts of _________________ acid or base is added strong  How does this happen: The added acid or base _________________ REACTS buffer w/something in the _______________ limiting The added acid or base is ____________________  Ex. Buffer: ADDING something to a buffer: A. ACID added (ex. HCl): The _____ + As in the buffer & a REACTION occurs H (from HCl) is attracted to the ____ HA Producing more ___________ RXN: (ag) (ag) & H+ A HA(ag) - + - It HA d B. Base added (ex. NaOH): The ____ OH (from NaOH) is attracted to the _____ HA in the buffer & a REACTION occurs A- Producing more ___________ RXN: & OH Tag) HA(ag) add Hz0(l) A (ag) ①+ - - + still a buffer ADDING a Chemical to a Buffer  ALL used up (LIMITING) NOTE: if added acid/base is strong, it is _______________________________________________  Calculating the pH when a STRONG acid/base is added to a buffer (3 chemicals involved): a) Write _________________________ dissociation equation b) Write ___________ RXN for substance added to buffer, ICE table (must use ___________!) MOLES c) Then use H-H eq: (must use __________) M © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 6 CHE-102 Lecture Outline breaks% Chapter 17 100 ↓ up Sample: pH of HF & NaF ADDING Cionic) w. acid 3. A) Calculate the pH of a 1.0L solution containing 0.10ME NaF & 0.50M E HF, when 9.5mL of 2.1M −4 HCl is added. (Assume minimal/no volume change) Ka of HF =6.8x10 - a) Dissociation Eg : - (ag) + Hf (ag) = H30 f - (ag) H + + 120(l) + (2 10 M) > - 0 SOM DOM.. Nat F Hf Added H (ag) reacts with 10 10M). Co SOM). BUFFER ! b) RXN : [t(ag) f(aq) + -F (ag) - + smaller reactant value Change by ICE table RXN :LES I (ag) + H +- (aq) > - Hf(aq) 009) COOI (0. 50) (1 0) 0 50 mol.. - 0199 -. 0199 +. 0199 E g , 0 08. mol 0. 51qmol - 1. 0L 1 OL. c) H - H(M) PH = pla logus + PH = - 10g6 10g8. 8x10* + 16) 0 82 pH 3 -. =. 2 3 Buffer only : pH =. pH = 2 47. © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 7 CHE-102 Lecture Outline Chapter 17 3. B) Calculate the pH of a 1.0L solution containing 0.10M NaF & 0.50M HF, when 9.5mL of 2.1M NaOH is added (assume minimal volume change). Ka of HF =6.8x10−4 - L a) dissociation OH (2 1M) Hz0(t) Hz0 (aq) + + f - (aq) Hf(ag). + O IOM. o. SOM Added : OH-(ag) reacts wh ICE table oxn (moles) b)RXN & Hf(ag) -H20(l) I f OH (aq) + (ag) - + -2 2 1x0 0095. 10 0 1.. x O I 0199 mol somol. 0 O mil. 0 10.. C 0 0199 0 0199 + 0 0199. - -.. ⑧ usmol 0. 11 mol O E. c) H H- eg(M) pH pKa = + logus buffer only PH = - 10g6 8x10-2. + log 3 16 , -0 60 pH 56. pH = 2. =. © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 8 CHE-102 Lecture Outline Chapter 17 pH Comparison 1 74 1. 0.50M HF: pH = ___________. 2 47 2. Buffer 0.50M HF & 0.10M NaF: pH = _____________. inc (more basic  pH ______________________________________. due presence of a common ion 3.A) Buffer 0.50M HF & 0.10M NaF w/added 9.5mL of 2.1M HCl: pH = ___________ 2 3. dec (more acidic)  pH _________________________________________. acid due addition of ____________ 2 56 3. B) Buffer 0.50M HF & 0.10M NaF w/added 9.5mL of 2.1M NaOH: pH = ____________. inc (more basic)  pH ________________________________________. base due addition of _____________ Summary: ICE Table © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 9 CHE-102 Lecture Outline Chapter 17 Sample: Base Problems  Calculate the pH for the following: amine-N- 1. A 2.00L* buffer solution containing 0.105M of the weak base methylamine, - CH3NH2, and - 0.125M of CH3NH3ICl. The Kb of methylamine is 4.4x10.− 4 (2 chemicals) & CHyNHzt B(ag) Hc0(l) + = BH (ag) OH (aq) - + + 0 105M. 0 125M. pH = 10. 63s + log o calc pKa : PH = pKa + 10g base ( 0 0757) pH 10 638 + -. =. (ka)(kb) Kw = pH = 10 56. a (basic - Kw = 1 0. x 10 4) + (a) (4 4x10 4) (1 0x10 -. =. Ka = 2 3x10-1. pKa = -log2 3 x10-1. pKa = 10. 63s OR pKa pkb + = 14 00. 2. A 2.00L buffer solution containing 0.105M of the weak base methylamine, CH3NH2, and - 0.125M of CH3NH3Cl after adding 0.100moles of HCl. The Kb of methylamine is 4.4x10−4. - - (3 chemicals) a) Dissociation : Blag) H20(l) = BH (ag) + + + OH - (aq) 0 105M 0 125M.. C) H H - b) RXN : PH e logebase / + H (ag) + B(ag) - BH (ag) + = + - -o Osm)(2002) (0 125 M)(2 001).. do me 0 210 mol 0 250 mol Ka 2 3 x 10-1 :. =.. - 0 100. - 0 100. + 0 100. pH = -log2 3x10-11. + log pH = 10 638 + ( - 0. 5027) 0 350 mol. E 8 0 110. mol. 10 14 pH =. 0 0550M. 0 175M. © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 10 CHE-102 Lecture Outline Chapter 17 Base Problems Cont’d  Calculate the pH of for the following: 3. A 2.00L buffer solution containing 0.105M of the weak base methylamine, CH3NH2, and - 0.125M of CH3NH3Cl after adding 0.150moles of NaOH. The Kb of methylamine is - - 4.4x10−4. (3 chemicals) a) Dissociation : B(ag) Hz0(l) + = BH + (aq) + OH - (ag) 0 105M. 0 125 M. b) RXN : I OH-(ag) (ag) H20(l) B(ag) + + BH - + - 10 125 M) (2.. L) Co 10SM) (2. 2) mol. I 0 150. 0. 250mol 0 210 mol. - 0 150 - 0 150 + 0 150 C... some 0. 100 mol O - E 2 0. 0500 M 0 180 M. C) H H (a) ( kb) - = Kw PH = pKa + logase (4 4x10 4) (kb) D. - =. 0 x 10 - 14) pKa = 10. 63g = 10 630. + log = 10. 638 + (0 556a). 11 19 pH =. © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 11 CHE-102 Lecture Outline Chapter 17 Please watch PL-10 to fill in the beginning of these notes. Location: Lecture Recordings→CHE102-Recordings→Dr. Gulde’s PL-Videos Acid/Base Titration (pH curves)  Introduced in CHE101  Mol:mol calculations color  change Indicators – chemicals that __________________________________ when a solution turns from acidic to basic  Phenolphthalein clear Acid = ________________ Base = ________________ pink volume  pH Titration Curve – _________ vs. ____________________ of titrant added buret  Titrant – liquid in _______________________ stoichiometric  Equivalence Point – ____________________________ H+ # of moles of acid ___________ OH- # mols base _____________ reacted Dead center ________________________ of vertical rise  Endpoint – where indicator ______________________ color permanently changes Just after ______________________ the equivalence point vertical (still on _________________ portion) endpt - W & equiv pt. pH at Various Titration Points  2 types of titrations 1. Strong/Strong: _______________________________________ 2. Weak/Strong: ________________________________________  What is happening to pH @ 4 different points NOTE: Substance mentioned _______ 1st A. Initial pH is always in the _____________! BEAKER B. Before equivalence pt C. @ equivalence pt D. After equivalence pt  & THEORY important, you don’t need to calculate, math used to explain theory © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 12 CHE-102 Lecture Outline Chapter 17 buret 1. Strong Acid - Strong Base: (NaOH added to HCl) y inbedker before A. Initial pH: ________________ adding base No  _______ NaOH added, only ___________ HC present acidic  Very ___________________  pH = _____________ 0 30. react B. Before equivalence pt: HCl & NaOH _______________ Base  ____________ Acid limiting & __________________ excess basic  pH gets more _____________, acid b/c ____________________ less in excess in © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 13 CHE-102 Lecture Outline Chapter 17 react C. AT equivalence pt: HCl & NaOH _______________ + H  Moles acid ____________ OH moles base ________ Neutralized Completely ______________________________  pH = ____________. 00 7 Nac & H20  How get pH when only _______________________________ present? water ______________ dissociates react D. After equivalence pt: HCl & NaOH _______________ Acid  ____________ Base limiting & ____________________ excess  pH gets ____________________________ more basic D - & O - - - C © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 14 CHE-102 Lecture Outline Chapter 17 What does a Strong Base – Strong Acid Titration look like? y F buret beaker 2 Lecture Summary 1. Titration Curves: 1. Strong/Strong A. Initial: No NaOH added, only HCl present pH = Very _________________ acidic B. Before equivalence: HCl & NaOH react (_________ Base limiting) pH gets more ___________, basic b/c ___________________ less acid in excess C. @ equivalence: HCl & NaOH react (Moles acid H+ = Moles of base OH−) __________________________ Neutralized Completely D. After equivalence: HCl & NaOH react (Acid limiting) pH gets more ______________ basic © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 15 CHE-102 Lecture Outline Chapter 17 beaker S dinburet 2. Weak Acid & Strong Base (NaOH is added to HC2H3O2) before A. Initial pH: _______________ adding base No  _________ HC2H302 NaOH added, only ________________________ present acidic  pH = _________________ ! No calculations react B. Before equivalence pt: HC2H3O2 & NaOH ______________ Base  _____________ Acid limiting & _______________ excess basic  Gets more _________________, less acid b/c ________________________ in excess  A _____________________ BUFFER soln!  _____________________________ weak acid HC2H3O2 & __________________________________ base conjugate (C2H3O2−) present No calculations ! in © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 16 CHE-102 Lecture Outline Chapter 17 C. At equivalence pt: HC2H3O2 & NaOH _______________ react  Moles acid _____________ Ht moles base __________ - OH- ______________________________________ NOT neutralized!! completely Have (CH302) ______________________________ CB  pH ___________ > 7 & No calculations D. After equivalence pt: HC2H3O2 & NaOH _______________ react Acid  ____________ Base limiting & ___________________ excess more basic  pH = _________________________ D - ------ C © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 17 CHE-102 Lecture Outline Chapter 17 Titration Curve Comparisons 1. Initial pH: lower  S/S is ________________, acidic more _______________ Dissociates __________________________ completely + H more ____________ in soln 2. After Equivalence:  Curves are the _________________ same Amount of _________________ excess OH- is the same g 3. @ Equivalence point: C longer vertical *  S/S _____________________________ rise ------- --------- different  pH ___________________________ S/S __________ 7 00. W/S _________ 37 conjugate base − b/c ______________________ present 4. Before Equivalence point: flatter  S/S is _______________  W/S is a _____________________ BUFFERED soln! __________ Acid & _________________ conjugate ______________ base present © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 18 CHE-102 Lecture Outline Chapter 17 Compare Different Weak Acid Titrations  Weaker acids have: 1. a more ___________ basic starting pH basic 2. a more ___________ equivalence point 3. a more ____________________ buffered region prior to equivalence needs to = o sloping log1 = 0 e pH pKa # Ka =  Can determine ________ from a titration curve! buffered  Look in ______________________ area  Where does [CB] ____ I " equivalence volume [Weak Acid]? ____________________________________________ As move left → right: [Acid] ________ dec & [CB] ____________. inc vol pKa 25mL vol= equirsome pH pKa = * 8 pka = - logka PH S = - ! O 8 = - 10gka j Ka = 10-s Ka 1x10-0 = © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 19 CHE-102 Lecture Outline Chapter 17 Titration of Polyprotic Acids  Ionization occurs ________________________ 1 H atom multiple at a time, thus __________________ equivalence pts exist  Ex: H3PO3: Has _________ 3 donatable H’s Has _________ 3 equivalence pts weak Is ___________, 3 ka's thus __________________ 3rd Equiv d Most east Mostly Ist Equiv Mostly H2POy ↓ Kaz mostly HzPOs Kay ↓ © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 20 CHE-102 Lecture Outline Chapter 17 Question-Titration Curves. 1. Which titration curve would be representative of NH3 titrated with HCl? O E. None of these 2. Which titration curve would be representative of H2SO4 in NaOH? ↑ w. acid , 1st H equivalence 7 00. O F. None of these Titrating w/Indicators  How do you know what indicator to use? at/just after  Ideally it should change ________________________ equivalence pt. the ___________________________  Since pH changes so fast, as long as the indicator changes anywhere on ___________________________________ ____________________________ rapid rise of titration curve, all is ok © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 21 CHE-102 Lecture Outline Chapter 17 Titrating w/Indicators Cont’d  Are the following good indicators for the titration shown? 1. Phenolphthalein: 2. Methyl Red: yes NO  What if not given a picture diagram for color change?  Indicators are acids thus have a Ka I  Indicators change color when ________ pH equals of the system ____________ their ____________! pKa of equivalence pH Want pKa of Indicator to be as close to ____________________________________ Cor _______________________ after just as possible Questions:  Which indicator(s) would work for the titration shown? Phenolphthalein only Methyl Red only Both work O Neither work b/c both on rapid rise Methyl , red better by after equivalence  A laboratory worker is titrating an unknown solution of a weak acid with a strong base. Which indicator should they choose given the following titration curve? Want IND pla near equir point. Pentamethoxy red: pKa = 1.8 8 5 =. Congo red: pKa = 4.0 Azolitmin: pKa = 6.5 O Phenolphthalein: pKa = 9.0 Trinitrobenzoic acid: pKa = 12.7 © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 22 CHE-102 Lecture Outline Chapter 17 Solubility Equilibria  insoluble Some ions when combined form _____________________________ salts  ___________________________ Rxns (Solubility Rules) precipitation  Saturated solution – ________ max amount dissolved, if add more, will contain undissolved solute ______________  Previously, insoluble ____________ 20 19 dissolved (in 100g H2O). Some does  _________________________ dissolve (________________!) SLIGHTLY  Solubility – max __________________ grams dissolved to form a saturated solution grams/L  Units = __________________  Molar Solubility – max _____________ moles dissolved to form a saturated solution  Units = ______________ moes/L  Solubility-Equilibrium Product Constant, Ksp – equilibrium constant for a ________________ slightly nearly soluble (or _______________ insoluble) ionic compound  ___________________ Degree to which a solid is soluble in water ________________ Larger Ksp, the _______________ more the solid dissolves! solids  ____________ solids or liquids in equilibrium expression! At Appendix ______ Ksp [Pb2][Cl-J 2 - = = 1 7x10. Question: Most soluble  Which of the following is the most soluble salt? BaSO4 Ksp = 1.08x10−10 O BaSO3 Ksp = 5.00x10−10 Ba3(PO4)2 Ksp = 3.40x10−23 BaCrO4 Ksp = 1.17x10−10 Larger K , more ions © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 23 CHE-102 Lecture Outline Chapter 17 Sample: PbI2 only 1. IfC 0.0012 moles PbI2(s) dissolves in 1.0 L water, what is its Ksp value? Salt y all I Ksp [pb2 ] [I-] + 2 Assume 2(s) Pb +Fast = + e (aq) = ment disso 10 0012M g [0 0012](0 0024] 2 Ksp. =.. 0 001 + 0 0012 C-.. Ksp = 6. 4 x 1059 E OM 0 0024M ! really small 0 0012 M.. (insoluble satt) Sample: Molar Solubility initial max, e salt & −6 2. If the Ksp of BaF2 is 1.7x10 , calculate the molar solubility. all dissolves (s) (ag) Bari a- 2 + 10-3 [Baz ] (f -J + X 1 7 x = &. C Y X - [x][2x]2 -" EOM + 1 7. x10 = 1 7x. 10 - 3 4x3 = X = 7. 5x10 - 3 M Question: Solubility  Calculate the MOLAR solubility of Mg3(PO4)2, whose Ksp is 1.04x10−24. I [Mg2 J [PO - 1 04 X10-24 Ms(POn) Ksp + >3Mg2 = = (ag). POn(ag + 2 + 2 1. 04 x 10 - 24 = (3x]3(2x] 2 X 1 04 x10 - 24 (27x3)(4x4) = 1. 1 04. x10-24 = 108XS C - X + 3X + 2x 9 62x10. - 27 = XS ↑ 9 3M 6 26x/0 - X =. E O 3x 2x 10 (g) 16lity) mol = © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 24 CHE-102 Lecture Outline Chapter 17 Question: LeChat BaF2  If F− is added to the following equilibrium, what direction does the rxn shift? & Left Right It does not shift Lechat & Factors Affecting Solubility decreases 1. Common Ion Effect – Adding a common ion ______________________ the solubility of the partially soluble salt Le Chatelier's  AKA _________________________ Principle  Bart  Adding BaCl2 (_________) f or NaF (___________) - shifts equilibrium ___________ Sample: Molar Solubility in NaF  What is the- say molar solubility of BaF2 in 0.15M NaF? (Ksp=1.7x10−6) I sat Bafz(s) - Baz (ag) 25 ) ! + 10 - 0 [Bazt] [f-]z - 1 7 x = 0 0 15. X. z 1 7x0-0 = [X][0. 15+ 2x] 2x. + +Y - - X 1.7 x 10 3 [x] (0 15]2 - =. X 0 15 2x + small OM. E SM - 6X10 X= 7. Decreased (BAF, only = 1. 5x10-3 M) © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 25 CHE-102 Lecture Outline Chapter 17 Factors Affecting Solubility Cont’d will dissolve 2. pH: Almost any ionic compound __________________________ if the pH of the solution is altered basic anion a) If a salt has a ___________________________, acidic it will be more soluble in an _______________ solution  EX: BaF2 dissolves more when HNO3 is added ↓ d ⑳one ________ from HNO3 reacts w/______ F- from BaF2 H (ag) (ag) > Hf(ag) + + f - - 2H + (ag) + 2F - (ag) = CHF (ag) - +f(ag) + Bafz(s) + 21 (ag) = Baz (ag) + + 2 > - adding HT  Adding more ___________________________ shifts equilibrium ____________ acidic cation b) If a salt has an ___________________________, it will be more soluble in an _______________ basic solution © 2025 S. Gulde. Reproducing and distributing this material is prohibited. 26 CHE-102 Lecture Outline Chapter 17 Sample:  Which salts are more soluble in acidic solutions? - Tasic a) CaCO3 ↓ X Ca(OH)2 S. Base basic HCOz W soluble. Acid Had2) + Hoa) more adding acid pushes rxh to right sacidreauup) b) AgCl ↓ X H + (aq) & HCI + c) -