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Ch 17: Additional Aspects of Aqueous Equilibria

Ch 17: Additional Aspects of Aqueous Equilibria. Brown, LeMay Ch 17 AP Chemistry Monta Vista High School. 17.1: Common Ion Effect. Addition of a “common ion”: solubility of solids decrease because of Le Châtelier’s principle. Ex: AgCl (s) ↔ Ag + (aq) + Cl - (aq)

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Ch 17: Additional Aspects of Aqueous Equilibria

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  1. Ch 17: Additional Aspects of Aqueous Equilibria Brown, LeMay Ch 17 AP Chemistry Monta Vista High School

  2. 17.1: Common Ion Effect • Addition of a “common ion”: solubility of solids decrease because of Le Châtelier’s principle. Ex: AgCl (s) ↔ Ag+ (aq) + Cl- (aq) • Addition of Cl- shifts equilibrium toward solid

  3. 17.4: Solubility Equilibria • Dissolving & precipitating of salts • Solubility rules discussed earlier are generalized qualitative observations of quantitative experiments. Ex: PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl- (aq) Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5 Ksp = solubility-product constant(found in App. D) • Recall that both aqueous ions and solid must be present in solution to achieve equilibrium • Changes in pH will affect the solubility of salts composed of a weak acid or weak base ion.

  4. Molar solubility:moles of salt per liter of solution in a saturated solution Ex: What is the molar solubility of Ag2SO4 at 25ºC? What mass of Ag2SO4 does that represent per liter? Ksp (Ag2SO4) = 1.4 x 10-5; MW Ag2SO4 = 311.80 g Ag2SO4 (s) ↔ 2 Ag+ (aq) + SO4-2 (aq) Ksp = [Ag+]2[SO4-2] (-x) + 2x M + x M X 2x M x M Ksp = [Ag+]2[SO4-2] = (2x)2(x) = 4x3 x = (1.4 x 10-5/4)1/3 = 0.015 M = [Ag2SO4] = [SO42-]

  5. Ex: What mass of Ag2SO4 does that represent per L? (MW Ag2SO4 = 311.80 g, molar solubility, [Ag2SO4] = 0.015 M) 4.7 g Ag2SO4

  6. Ex: 3.0 x 10-5 g Pb3(AsO4)2 will dissolve in 1.0 L water to make a saturated solution. What is Ksp of Pb3(AsO4)2? (MW of Pb3(AsO4)2 = 899.44 g/mol) Pb3(AsO4)2 (s) ↔ 3 Pb+2 (aq) + 2 AsO4-3 (aq) Ksp = [Pb2+]3[AsO4-3]2 [Pb3(AsO4)2]= (3.0 x 10-5 g)/(899.44 g/mol)/(1.0 L) = 3.3 x 10-8 M Pb3(AsO4)2 = x (-x) + 3x M + 2x M X 3x M 2x M Ksp=[Pb2+]3[AsO4-3]2 = [3(3.3 x 10-8)]3 [2(3.3 x 10-8)]2 = = 4.5 x 10-36

  7. Ex: What is the molar solubility of CaF2 in a 1.0 L solution of 0.10 M Ca(NO3)2? Ksp (CaF2) = 3.9 x 10-11 CaF2 (s) ↔ Ca2+ (aq) + 2 F- (aq) (-x) + x M + 2x M X (0.10 + x) M 2x M Assume x is negligible: x = 9.9x10-6 moles of CaF2 are soluble per L.

  8. Predicting whether a precipitate (ppt) will form depends on concentration of ions present and Ksp of all likely salts. • Compare the ion-product, Q, with Ksp • Q has the same mathematical form as Ksp • Q must be calculated for each example If: Q > Ksp precipitation occurs (continues until Q = Ksp) Q = Ksp solution is at equilibrium Q < Ksp solid dissolves (continues until Q = Ksp)

  9. Ex: Will a precipitate form when 0.10 L of 3.0 x 10-3 M Pb(NO3)2 is added to 0.400 L of 5.0 x 10-3 M Na2SO4? Possible new salts: PbSO4 (Ksp = 1.6 x 10-8) NaNO3 (Ksp = 112.12) Q = [Pb2+][SO42-] [Pb2+] = (3.0 x 10-3 M)(0.10 L) / (0.10 L + 0.400 L) = 3.0 x 10-4 mol / 0.50 L = 6.0 x 10-4 M Pb2+ [SO42-] = (5.0 x 10-3 M)(0.400 L) / (0.10 L + 0.400 L) = 2.0 x 10-3 mol / 0.50 L = 4.0 x 10-3 M SO42- Q = [Pb2+][SO42-] = (6.0x10-4)(4.0x10-3) = 2.4 x 10-6 • Since Q > Ksp, PbSO4 will precipitate.

  10. Ex: Will a precipitate form if 20.0 mL of 3.50 x 10-3 M Hg2(NO3)2 solution are mixed with 40.0 mL of 2.00 x 10-3 M NaCl solution? Ksp (Hg2Cl2) = 1.3 x 10-18 [Hg22+] = (3.50x10-3 M)(0.0200 L)/(0.0200 L + 0.0400 L) = 0.00117 M [Cl-] = (2.00 x 10-3 M)(0.0400 L)/(0.0200 L + 0.0400 L) = 0.00133 M Q = [Hg22+][Cl-]2 Q = (.00117)(.00133)2 = 2.07 x 10-9 • Since Q > Ksp, there will be a precipitate (Hg2Cl2)

  11. 17.2: Buffers: • Solutions that resist drastic changes in pH upon additions of small amounts of acid or base. • Consist of a weak acid and its conjugate base (usually in salt form) Ex: acetic acid and sodium acetate: HC2H3O2 + NaC2H3O2 • Or consist of a weak base and its conjugate acid (usually in salt form) Ex: ammonia and ammonium chloride: NH3 + NH4Cl

  12. Consider a weak acid/salt buffer: HA (aq) ↔ H+ (aq) + A- (aq) • Thus, the pH of a buffer is determined by Ka of weak acid and the ratio of the concentrations of an acid & its conjugate base.

  13. Henderson-Hasselbalch equation: Karl Hasselbalch(1874-1962) Lawrence Henderson(1878-1942) Use for all buffer solutions!

  14. Why/how do buffers work? • A small addition of [H+] will decrease the [A-] but increase the [HA]: H+ (aq) + A- (aq) ↔ HA (aq) pH decreases, but D pH will be small • A small addition of [OH-] will decrease the [HA] but increase the [A-]: • OH- (aq) + HA (aq) ↔ H2O(l) + A- (aq) pH increases, but D pH will be small

  15. In general, when choosing a buffer, select one in which the acid form has a pKa close to the desired pH. • If [base] > [acid], then pH > pKa • If [base] < [acid], then pH < pKa • If [base] = [acid], then pH = pKa and [H+] = Ka

  16. Buffer capacity: • Amount of acid or base a buffer can neutralize before the pH begins to change “significantly”. • The greater the concentrations of acid & base in the buffer, the greater the buffer capacity. • A “typical” buffer solution can hold the pH to ± 1.00 • Buffer solutions can also be considered common-ion solutions.

  17. Ex: Calculate the pH of a buffer made by adding 0.300 mol acetic acid and 0.300 mol sodium acetate to enough water to make 1.00 L. (Ka = 1.8 x 10-5) CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq) -x M + x M + x M x M (0.300 – x) M (0.300 + x) M

  18. Or, use Henderson-Hasselbalch: • Assumes that x is negligible

  19. Ex: Calculate the pH of this buffer after the addition of 0.020 mol NaOH (assume volume remains constant.) • Neutralization: calculate the concentrations after the initial reaction of the strong base with the weak acid part of the buffer. CH3COOH (aq) + OH- (aq) → CH3COO- (aq) + H2O(l) -0.020 mol - 0.020 mol + 0.020 mol 0.280 mol 0.320 mol 0 mol Then, consider how these amounts behave in equilibrium!

  20. Equilibrium: CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq) -x mol + x mol + x mol (0.280 – x) mol x mol (0.320 + x) mol

  21. Or, use Henderson-Hasselbalch: The pH increased, but not by much.

  22. Ex: Calculate the pH of the original buffer after the addition of 0.020 mol HCl (assume volume remains constant.) • Neutralization: calculate the concentrations after the initial reaction of the strong acid with the weak conjugate base. CH3COO- (aq) + H+ (aq) → CH3COOH (aq) -0.020 mol - 0.020 mol + 0.020 mol 0.280 mol 0.320 mol 0 mol Then, consider how these amounts behave in equilibrium!

  23. Equilibrium: CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq) -x mol + x mol + x mol (0.320 – x) mol x mol (0.280 + x) mol

  24. Or, use Henderson-Hasselbalch: The pH decreased, but not by much.

  25. 17.3: Acid-Base Titrations Titration curve: graph of pH vs. volume of titrant added • Each type of titration has a uniquely shaped titration curve Every titration consists of four regions: • The initial pH • Between the initial pH and the equivalence point: pH is determined by amount of solution not yet neutralized • The equivalence point: moles of acid = moles of base, leaving a solution of the salt produced in the acid-base neutralization • After the equivalence point: pH is determined by amount of excess titrant

  26. Strong Acid – Strong Base Titration Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) 14 pH 7 0 0 50 Volume of NaOH added (mL) • Region 1: Initial pH • pH starts very low (here, pH = -log 0.100 = 1) because a strong acid is being titrated.

  27. 14 pH 7 0 Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl 0 50 Volume of NaOH added (mL) • Region 2: Between initial pH and equivalence point • As base is added, pH increases slowly, then rapidly as it approaches the equivalence point. • Before equivalence point, moles acid > moles base added; pH is determined by amount of solution not yet neutralized

  28. 14 pH 7 0 Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl moles H+ from HCl = moles OH- added 0 50 Volume of NaOH added (mL) • Region 3: Equivalence point • Reached when moles H+ = moles OH- added; here, at pH = 7 because only species present are H2O and the salt (Na+1 and Cl-1), which do not hydrolyze to produce an acidic or basic solution, and at V = 50 mL because MAVA= MBVB = (0.100)(.0500) = (0.100)VB • Endpoint: point at which an indicator changes color; an approximation of the actual equivalence point

  29. 14 pH 7 0 Ex: 0.100 M NaOH added to 50.0 mL of 0.100 M HCl 0 50 Volume of NaOH added (mL) • Region 4: After equivalence point • As base is added, pH rises rapidly, then continues to increase slowly. • Moles acid < moles base added • pH eventually approaches an upper limit as base is added; here at pH = 13 (-log [OH-] = 1).

  30. pH Indicators: • Chemicals with acid and base forms with significant color differences • For a titration, choose an indicator with a range near the equivalence point Phenolphthalein: colorless (HPh: acid form above) to pink (Ph-, base form) at pH ≈ 8.5-9.5 * Methyl red: red (HMer, acid form) to yellow (Mer-, base form) at pH ≈ 4.2 – 6.3

  31. pH Indicators and Titrations • Phenophthalein is often used as an indicator in the titration of a strong acid with a strong base since it turns pink around pH 9, only a small addition of titrant beyond the equivalence point at pH 7.

  32. Weak Acid – Strong Base Ex: 0.100 M NaOH (aq) is added to 50.0 mL of 0.100 M CH3COOH (aq) Equivalence point: MAVA = MB VB (0.100 M)(0.0500 L) = (0.100 M)(VB) VB = 0.0500 L NaOH Equation 1: Neutralization: reaction goes to completion CH3COOH (aq) + OH-(aq) → H2O (l) + CH3COO-(aq) Equation 2: Equilibrium: extent of reaction based on Ka CH3COOH (aq) ↔ H+ (aq) + CH3COO- (aq)

  33. Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH 14 pH 7 0 Region 1: Initial pH • pH starts below pH 7 because a weak acid is being titrated. 0 50 Volume of NaOH added (mL)

  34. Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH 14 pH 7 0 Region 2: Between initial pH and equivalence point • Moles acid > moles base added • pH is determined by considering extent of neutralization and equilibrium reactions. 0 50 Volume of NaOH added (mL)

  35. Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH 14 pH 7 0 Region 3: Equivalence point • Equivalence point at pH > 7 because only species present are H2O and the salt (Na+ and CH3COO-); acetate ion is a weak base and hydrolyzes to produce a basic solution. moles H+ from CH3COOH = moles OH- added 0 50 Volume of NaOH added (mL)

  36. Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH 14 pH 7 0 Region 4: After equivalence point • Moles acid < moles base added • pH is determined by concentration of excess base in solution (as in strong acid/strong base); equilibrium can be ignored 0 50 Volume of NaOH added (mL)

  37. Ex: 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH; calculate pH at regular intervals of titration (Ka = 1.8 x 10-5): (similar to Exercise 17.7, p. 637) After 30.0 mL of 0.100 M NaOH has been added: mol OH- = (0.100 M)(0.0300 L) = 3.00 x 10-3 mol OH- mol CH3COOH = (0.100 M)(0.0500 L) = 5.00 x 10-3mol - 3.00 x 10-3 mol - 3.00 x 10-3 mol + 3.00 x 10-3 mol 2.00 x 10-3 mol 3.00 x 10-3 mol 0 [CH3COOH] = (2.00 x 10-3 mol)/(0.0800 L) = 0.0250 M [CH3COO-] = (3.00 x 10-3 mol)/(0.0800 L) = 0.0375 M

  38. After 30.0 mL of 0.100 M NaOH has been added: - x + x + x 0.0250 - x + x 0.0375 + x Or use H-H since its a buffer solution:

  39. After 50.0 mL of 0.100 M NaOH has been added: • Initial: mol OH- = (0.100 M)(0.0500 L) = 5.00 x 10-3 mol OH- mol CH3COOH = (0.100 M)(0.0500 L) = 5.00 x 10-3mol - 5.00 x 10-3 mol - 5.00 x 10-3 mol + 5.00 x 10-3 mol 0 5.00 x 10-3 mol 0 [CH3COO-] = (5.00 x 10-3 mol)/(0.100 L) = 0.0500 M

  40. After 50.0 mL of 0.100 M NaOH has been added: - x + x + x 0.0500 - x x + x

  41. After 70.0 mL of 0.100 M NaOH is added: mol OH- = (0.100 M)(0.0700 L) = 7.00 x 10-3 mol OH- mol CH3COOH = (0.100 M)(0.0500 L) = 5.00 x 10-3 mol mol OH- in excess = (7.00 x 10-3 – 5.00 x 10-3) = 2.00 x 10-3 mol OH- [OH-] = (2.00 x 10-3 mol)/(0.1200 L) = 0.0167 M OH- pOH = 1.778 and pH = 12.22 (pH + pOH = 14.00)

  42. Weak Base – Strong Acid Ex: 0.100 M HCl (aq) is added to 40.0 mL of 0.0750 M NH3 (aq) Equivalence point: MAVA = MB VB (0.100 M)(VA) = (0.0750 M)(0.0400 L) VA = 0.0300 L HCl Equation 1: Neutralization: reaction goes to completion NH3 (aq) + H+ (aq) → NH4+ (aq) Equation 2: Equilibrium: extent of reaction based on Kb NH3 (aq) + H2O (l) ↔ OH- (aq) + NH4+ (aq)

  43. Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq) 14 pH 7 0 Region 1: Initial pH • pH starts above pH 7 because a weak base is being titrated. 0 30 Volume of HCl added (mL)

  44. Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq) 14 pH 7 0 Region 2: Between initial pH and equivalence point • Moles base > moles acid added • pH is determined by considering extent of neutralization and equilibrium reactions. 0 30 Volume of HCl added (mL)

  45. Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq) 14 pH 7 0 Region 3: Equivalence point • Equivalence point at pH < 7 because only species present are H2O and the salt (NH4+ and Cl-); ammonium ion is a weak acid and hydrolyzes to produce an acidic solution. moles NH3= moles H+ added 0 30 Volume of HCl added (mL)

  46. Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq) 14 pH 7 0 Region 4: After equivalence point • Moles base < moles acid added • pH is determined by concentration of excess acid in solution (as in strong acid/strong base); equilibrium can be ignored 0 30 Volume of HCl added (mL)

  47. Ex: 0.100 M HCl is added to 40.0 mL of 0.0750 M NH3 (aq); calculate pH at regular intervals of titration. (Kb = 1.8 x 10-5) After 15.0 mL of 0.100 M HCl has been added: mol H+ = (0.100 M)(0.0150 L) = 1.50 x 10-3 mol H+ mol NH3 (aq) = (0.0750 M)(0.0400 L) = 3.00 x 10-3mol - 1.50 x 10-3 mol - 1.50 x 10-3 mol + 1.50 x 10-3 mol 1.50 x 10-3 mol 1.50 x 10-3 mol 0 [NH3] = (1.50 x 10-3 mol)/(0.0550 L) = 0.0273 M [NH4+] = (1.50 x 10-3 mol)/(0.0550 L) = 0.0273 M

  48. After 15.0 mL of 0.100 M HCl has been added: - x + x + x 0.0273 - x + x 0.0273 + x Or use H-H since its a buffer solution:

  49. After 30.0 mL of 0.100 M HCl has been added: • Initial: mol H+ = (0.100 M)(0.0300 L) = 3.00 x 10-3 mol H+ mol NH3 = (0.0750 M)(0.0400 L) = 3.00 x 10-3mol - 3.00 x 10-3 mol - 3.00 x 10-3 mol + 3.00 x 10-3 mol 0 3.00 x 10-3 mol 0 [NH4+] = (3.00 x 10-3 mol)/(0.0700 L) = 0.0429 M

  50. After 30.0 mL of 0.100 M HCl has been added: - x + x + x 0.0429- x x + x

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