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Adding NaOH to HCHO 2

added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96. added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22. initial HCHO 2 solution 0.00250 mol HCHO 2 pH = 2.37. added 5.0 mL NaOH 0.00200 mol HCHO 2 pH = 3.14. added 10.0 mL NaOH 0.00150 mol HCHO 2 pH = 3.56. added 25.0 mL NaOH

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Adding NaOH to HCHO 2

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  1. added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 initial HCHO2 solution 0.00250 mol HCHO2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56 added 25.0 mL NaOH equivalence point 0.00250 mol CHO2− [CHO2−]init = 0.0500 M [OH−]eq = 1.7 x 10-6 pH = 8.23 Adding NaOH to HCHO2 added 12.5 mL NaOH 0.00125 mol HCHO2 pH = 3.74 = pKa half-neutralization added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 15.0 mL NaOH 0.00100 mol HCHO2 pH = 3.92 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 added 20.0 mL NaOH 0.00050 mol HCHO2 pH = 4.34 Tro, Chemistry: A Molecular Approach

  2. Titrating Weak Acid with a Strong Base • the initial pH is that of the weak acid solution • calculate like a weak acid equilibrium problem • e.g., 15.5 and 15.6 • before the equivalence point, the solution becomes a buffer • calculate mol HAinit and mol A−init using reaction stoichiometry • calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init • half-neutralization pH = pKa Tro, Chemistry: A Molecular Approach

  3. Titrating Weak Acid with a Strong Base • at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established • mol A− = original mole HA • calculate the volume of added base like Ex 4.8 • [A−]init = mol A−/total liters • calculate like a weak base equilibrium problem • e.g., 15.14 • beyond equivalence point, the OH is in excess • [OH−] = mol MOH xs/total liters • [H3O+][OH−]=1 x 10-14 Tro, Chemistry: A Molecular Approach

  4. Ex 16.7a – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point HNO2 + KOH  NO2 + H2O Tro, Chemistry: A Molecular Approach

  5. Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH HNO2 + KOH  NO2 + H2O 0.00100 0.00300 Tro, Chemistry: A Molecular Approach

  6. Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. HNO2 + H2O  NO2 + H3O+ Table 15.5 Ka = 4.6 x 10-4 Tro, Chemistry: A Molecular Approach

  7. Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point HNO2 + KOH  NO2 + H2O at half-equivalence, moles KOH = ½ mole HNO2 0.00200 0.00200 Tro, Chemistry: A Molecular Approach

  8. Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. HNO2 + H2O  NO2 + H3O+ Table 15.5 Ka = 4.6 x 10-4 Tro, Chemistry: A Molecular Approach

  9. Titration Curve of a Weak Base with a Strong Acid Tro, Chemistry: A Molecular Approach

  10. Titration of a Polyprotic Acid • if Ka1 >> Ka2, there will be two equivalence points in the titration • the closer the Ka’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH Tro, Chemistry: A Molecular Approach

  11. Monitoring pH During a Titration • the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+] • using a probe that specifically measures just H3O+ • the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve • if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with anindicator Tro, Chemistry: A Molecular Approach

  12. Monitoring pH During a Titration Tro, Chemistry: A Molecular Approach

  13. Indicators • many dyes change color depending on the pH of the solution • these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution HInd(aq) + H2O(l) Ind(aq) + H3O+(aq) • the color of the solution depends on the relative concentrations of Ind:HInd • when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and HInd • when Ind:HInd > 10, the color will be mix of the colors of Ind • when Ind:HInd < 0.1, the color will be mix of the colors of HInd Tro, Chemistry: A Molecular Approach

  14. Phenolphthalein

  15. Methyl Red Tro, Chemistry: A Molecular Approach

  16. Monitoring a Titration with an Indicator • for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point • an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH • pKa of HInd ≈ pH at equivalence point Tro, Chemistry: A Molecular Approach

  17. Acid-Base Indicators

  18. Solubility Equilibria • all ionic compounds dissolve in water to some degree • however, many compounds have such low solubility in water that we classify them as insoluble • we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water Tro, Chemistry: A Molecular Approach

  19. Solubility Product • the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp • for an ionic solid MnXm, the dissociation reaction is: MnXm(s)  nMm+(aq) + mXn−(aq) • the solubility product would be Ksp = [Mm+]n[Xn−]m • for example, the dissociation reaction for PbCl2 is PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) • and its equilibrium constant is Ksp = [Pb2+][Cl−]2 Tro, Chemistry: A Molecular Approach

  20. Tro, Chemistry: A Molecular Approach

  21. Molar Solubility • solubility is the amount of solute that will dissolve in a given amount of solution • at a particular temperature • the molar solubility is the number of moles of solute that will dissolve in a liter of solution • the molarity of the dissolved solute in a saturated solution • for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq) Tro, Chemistry: A Molecular Approach

  22. Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) Ksp = [Pb2+][Cl−]2 Tro, Chemistry: A Molecular Approach

  23. Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C Ksp = [Pb2+][Cl−]2 Ksp = (S)(2S)2 Tro, Chemistry: A Molecular Approach

  24. Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M Tro, Chemistry: A Molecular Approach

  25. Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M PbBr2(s)  Pb2+(aq) + 2 Br−(aq) Ksp = [Pb2+][Br−]2 Tro, Chemistry: A Molecular Approach

  26. Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M Ksp = [Pb2+][Br−]2 Ksp = (1.05 x 10-2)(2.10 x 10-2)2 Tro, Chemistry: A Molecular Approach

  27. Ksp and Relative Solubility • molar solubility is related to Ksp • but you cannot always compare solubilities of compounds by comparing their Ksps • in order to compare Ksps, the compounds must have the same dissociation stoichiometry Tro, Chemistry: A Molecular Approach

  28. The Effect of Common Ion on Solubility • addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt • for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2 PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) addition of Cl− shifts the equilibrium to the left Tro, Chemistry: A Molecular Approach

  29. Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C CaF2(s)  Ca2+(aq) + 2 F−(aq) Ksp = [Ca2+][F−]2 Tro, Chemistry: A Molecular Approach

  30. Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C Ksp = [Ca2+][F−]2 Ksp = (S)(0.100 + 2S)2 Ksp = (S)(0.100)2 Tro, Chemistry: A Molecular Approach

  31. The Effect of pH on Solubility • for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide • and the lower the pH, the higher the solubility • higher pH = increased [OH−] M(OH)n(s)  Mn+(aq) + nOH−(aq) • for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq) H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l) Tro, Chemistry: A Molecular Approach

  32. Precipitation • precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound • if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur • Q = Ksp, the solution is saturated, no precipitation • Q < Ksp, the solution is unsaturated, no precipitation • Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate • some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturatedsolutions Tro, Chemistry: A Molecular Approach

  33. a supersaturated solution will precipitate if a seed crystal is added precipitation occurs if Q > Ksp Tro, Chemistry: A Molecular Approach

  34. Selective Precipitation • a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others • a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different Tro, Chemistry: A Molecular Approach

  35. Ex 16.13 What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater? precipitating may just occur when Q = Ksp Tro, Chemistry: A Molecular Approach

  36. Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M Tro, Chemistry: A Molecular Approach

  37. Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M Tro, Chemistry: A Molecular Approach

  38. Qualitative Analysis • an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysisscheme • wet chemistry • a sample containing several ions is subjected to the addition of several precipitating agents • addition of each reagent causes one of the ions present to precipitate out Tro, Chemistry: A Molecular Approach

  39. Qualitative Analysis Tro, Chemistry: A Molecular Approach

  40. Group 1 • group one cations are Ag+, Pb2+, and Hg22+ • all these cations form compounds with Cl− that are insoluble in water • as long as the concentration is large enough • PbCl2 may be borderline • molar solubility of PbCl2 = 1.43 x 10-2 M • precipitated by the addition of HCl Tro, Chemistry: A Molecular Approach

  41. Group 2 • group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+ • all these cations form compounds with HS− and S2− that are insoluble in water at low pH • precipitated by the addition of H2S in HCl

  42. Group 3 • group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides • all these cations form compounds with S2− that are insoluble in water at high pH • precipitated by the addition of H2S in NaOH

  43. Group 4 • group four cations are Mg2+, Ca2+, Ba2+ • all these cations form compounds with PO43− that are insoluble in water at high pH • precipitated by the addition of (NH4)2HPO4 Tro, Chemistry: A Molecular Approach

  44. Group 5 • group five cations are Na+, K+, NH4+ • all these cations form compounds that are soluble in water – they do not precipitate • identified by the color of their flame

  45. Complex Ion Formation • transition metals tend to be good Lewis acids • they often bond to one or more H2O molecules to form a hydrated ion • H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq) • ions that form by combining a cation with several anions or neutral molecules are called complex ions • e.g., Ag(H2O)2+ • the attached ions or molecules are called ligands • e.g., H2O Tro, Chemistry: A Molecular Approach

  46. Complex Ion Equilibria • if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H2O)2+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) + 2 H2O(l) • generally H2O is not included, since its complex ion is always present in aqueous solution Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) Tro, Chemistry: A Molecular Approach

  47. Formation Constant • the reaction between an ion and ligands to form a complex ion is called a complex ionformation reaction Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) • the equilibrium constant for the formation reaction is called the formation constant, Kf Tro, Chemistry: A Molecular Approach

  48. Formation Constants Tro, Chemistry: A Molecular Approach

  49. Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Tro, Chemistry: A Molecular Approach

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