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Chapter 16 Buffers & Solubility Equilibria

Chapter 16 Buffers & Solubility Equilibria. Generic weak acid — HA ↔ H + + A -. NaA ( aq ) → Na + ( aq ) + A - ( aq ) . What happens if Salt of weak acid added …. a) [HA]↓ b) [H + ]↓ c both of these.

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Chapter 16 Buffers & Solubility Equilibria

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  1. Chapter 16 Buffers & Solubility Equilibria Generic weak acid — HA ↔ H+ + A- NaA(aq) → Na+(aq) + A- (aq) What happens if Salt of weak acid added …. a) [HA]↓ b) [H+]↓ c both of these The common ion effect works via LeChatelier’s Principle — If the salt of a weak acid is added to the weak acid the [H+]↓ and the pH↑. Ka = [H+] [A-] [HA]

  2. The Common Ion Effect sample calculation — 0.10 M acetic acid + 0.15M sodium acetate Ka = 1.85 x 10-5 = [H+] [CH3COO-] [CH3COOH] CH3COOH ↔ H+ + CH3COO- I 0.10 0 0 C -x +x +x E 0.10 - x xx = 0.00134M & pH = 2.87 Add 0.15 M sodium acetate (fully dissociated) Will pH be … a) =2.87 b) < 2.87 c) 2.87 < pH < 7.0 d) > 7.0 CH3COOH ↔ H+ + CH3COO- 0.15 I 0.10 0 ??? C E pH = 4.91 -x +x +x 0.10 + x x 0.15 + x ??? = a) 0.00134 b) 0.15 c) 0 d) 0.10 Solve for x 1.85 x 10-5 = x (0.15 + x) (0.10 – x) x = (0.10 • 1.85 x 10-5) = 1.23 x 10-5 0.15

  3. The Henderson Hasselbach Equation A short cut to the last problem pH = pKa + log (b/a) Generic weak acid: HA ↔ H+ + A- Ka= [H+] [A-] = [H+] • [A-] [HA] [HA] take -log of both sides -log xy = -log x – log y - log Ka= - log [H+] - log ([A-]/[HA]) pKa= pH - log ([A-]/[HA]) pH= pKa + log ([A-]/[HA]) pH = pKa + log ([conjugate base]/[weak acid]) For acetic acid pKa = -log 1.85 x 10-5 = 4.73 Add0.10 Macetic acid with0.15 M sodium acetate pH = 4.73 + log (0.15/0.10) = 4.91

  4. pH = pKa + log (b/a)

  5. Enough HCl is added to pure water to make a 0.01 M HCL solution. The pH will drop from 7.0 to …. a) 2.0 b) 4.0 c) 6.0 d) 1.0 A solution contains equal amounts (0.1M) of acetic acid and sodium acetate. The pKa of acetic acid is 4.74. The pH of the solution is ….. a) 2.00 b) 3.54 c) 4.74 d) 7.00 If enough HCl is added to the acetic acid/sodium acetate solution. What will happen to the pH It will decrease to 2.00 b) it will decrease to 3.74 It will decrease but very little d) it will increase

  6. A buffer solution is a solution of: • A weak acid or a weak base and • The salt of the same weak acid or weak base • Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. CH3COOH ↔ H+ + CH3COO- add strong acid ….. HCl→ H+ + Cl- CH3COOH ↔ H+ + CH3COO- add strong base ….. As you add/remove H+ by adding acid or base the equilibrium will shift to remove/restore [H+] and the pH stays nearly the same.

  7. CH3COOH ↔ H+ + CH3COO- add strong acid ….. HCl→ H+ + Cl- As you add H+ by adding acid the equilibrium will shift to remove H+ and the pH decreases only slightly until the acid added exceeds the buffer available.

  8. A buffer solution is a solution of: • A weak acid or a weak base and • The salt of the same weak acid or weak base • Both must be present! Which of the following are buffer systems? (a) KF/HF , (b) KBr/HBr, (c) Na2CO3/NaHCO3 d) both a & c (a) KF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is its conjugate acid buffer solution

  9. Calculate the pH of the 0.35 MCH3NH2/0.19 MCH3NH3Cl buffer system. (CH3NH2 Kb = 4.4 x 10-4) methylammonium chloride is soluble (100%) CH3NH3Cl → CH3NH3+ + Cl- Therefore [CH3NH3+] = 0.19 M CH3NH2 + H2O ↔ CH3NH3+ + OH- I 0.35M 0.19 M pH = pKa + log ([base]/[weak acid]) pH = pKa + log ([CH3NH2]/[CH3NH3+]) (CH3NH3+ Ka = Kw/Kb = 1 x 10-14/ (4.4 x10-4) = 2.3 10-11 pKa = - log Ka = 10.64 pH = 10.64 + log (0.35/0.19) = 10.91

  10. Calculate the pH of the 0.35 MCH3NH2/0.19 MCH3NH3Cl buffer system. (CH3NH2 Kb = 4.4 x 10-4) methylammonium chloride is soluble (100%) CH3NH3Cl → CH3NH3+ + Cl- Therefore [CH3NH3+] = 0.19 M CH3NH2 + H2O ↔ CH3NH3+ + OH- I 0.35M 0.19 M pH = pKa + log ([base]/[weak acid]) pH = pKa + log ([CH3NH2]/[CH3NH3+]) (CH3NH3+ Ka = Kw/Kb = 1 x 10-14/ (4.4 x10-4) = 2.3 10-11 pKa = - log Ka = 10.64 pH = 10.64 + log (0.35/0.19) = 10.91

  11. Calculate the pH of the 0.35 MCH3NH2/0.19 MCH3NH3Cl buffer system. (CH3NH3 Kb = 4.4 x 10-4) CH3NH2 + H2O ↔ CH3NH3+ + OH- pH = 10.64 + log (0.35/0.19) = 10.91 pKa for CH3NH3+ If I add 20.0 ml of 0.10M HCl to this buffer what will happen? a) pH drops a little b) pH drops a lot c) pH increases a little d) pH increases a lot Does the Molarity of methylamine CH3NH2 ….. ? a) increasse b) decrease c) stay the same Why does it decrease ….. ? a) the solution volume increases b) the methylamine reacts with H+ c) both of the above c) none of the above ↓CH3NH2 + H+ ↔ ↑CH3NH3+ ↑

  12. Calculate the pH of the 0.35 MCH3NH2/0.19 MCH3NH3Cl buffer system. What is the pH after the addition of 20.0 mL of 0.10 MHClto 80.0 mL of the buffer solution? (CH3NH3 Kb = 4.4 x 10-4) CH3NH2 + H+ ↔ CH3NH3+ b a pH = 10.64 + log ([CH3NH2]/[CH3NH3+]) moles base (initial) = 0.35 M • 0.08 L = 0.028 mols (base) moles acid (initial) = 0.19 M • 0.08 L = 0.0152 mols (acid) moles acid (added) = 0.10 M • 0.02 L = 0.0020 mols (acid) moles base (final) = 0.028 mols – 0.002 = 0.026 moles acid(final) = 0.0152 mols + 0.002 = 0.0172 pH = 10.64 + log (0.026/0.0172) = 10.82

  13. Calculate the pH of the 0.35 MCH3NH2/0.19 MCH3NH3Cl buffer system. What is the pH after the addition of 20.0 mL of 0.10 MHClto 80.0 mL of the buffer solution? (CH3NH3 Kb = 4.4 x 10-4) CH3NH2 + H+ ↔ CH3NH3+ pH = 10.64 + log ([CH3NH2]/[CH3NH3+]) moles base (initial) = 0.35 M • 0.08 L = 0.028 mols (base) moles acid (initial) = 0.19 M • 0.08 L = 0.0152 mols (acid) moles acid (added) = 0.10 M • 0.02 L = 0.0020 mols (acid) pH = 10.64 + log (0.026/0.0172) = 10.82 If the moles of HCl added was greater than 0.028 mols, what would happen to the pH then? a) pH drops a little b) pH drops a lot c) pH increases a little d) pH increases a lot

  14. Alternative Method of Equivalence Point Detection monitor pH

  15. Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

  16. Titration Produce buffers Strong acid with strong base weak acid with strong base Strong base with strong acid weak base with strong acid Equivalence point – occurs when the moles base/acid added is equal to the moles of acid/base initially present in solution. Titrant – solution that is added – typically from a burette. Titration plot: Plot pH vs. volume of titrant added

  17. Strong Acid-Strong Base Titrations 25 ml of 0.1 M HCl titrated with 0.1M NaOH NaOH(aq) + HCl(aq) → H2O (l) + NaCl(aq) OH-(aq) + H+(aq) →H2O (l) [H+] = (mols acid – mols base) L solution (total) pOH = - log [OH-] mols acid = mols base pH = - log [H+] [OH-] = (mols base – mols acid) L solution (total)

  18. Weak Acid-Strong Base Titrations: 25 ml of 0.10 M HAc with 0.10M NaOH CH3COOH (aq) + NaOH(aq)→ CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH-(aq)→ CH3COO-(aq) + H2O (l) strong base added becomes weak base At equivalence point (pH > 7): CH3COO-(aq) + H2O (l)↔ OH-(aq) + CH3COOH (aq) D C B A

  19. Strong Acid-Strong Base Titrations 25 ml of 0.1 M HCl titrated with 0.1M NaOH NaOH(aq) + HCl(aq) → H2O (l) + NaCl(aq) OH-(aq) + H+(aq) →H2O (l) [H+] = (mols acid – mols base) L solution (total) pOH = - log [OH-] mols acid = mols base pH = - log [H+] [OH-] = (mols base – mols acid) L solution (total)

  20. Weak Acid-Strong Base Titrations: 25 ml of 0.10 M HAc with 0.10M NaOH CH3COOH (aq) + NaOH(aq)→ CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH-(aq)→ CH3COO-(aq) + H2O (l) strong base added becomes weak base At equivalence point (pH > 7): CH3COO-(aq) + H2O (l)↔ OH-(aq) + CH3COOH (aq) D C B A

  21. Weak Acid-Strong Base Titrations: 25 ml of 0.10 M HAc with 0.10M NaOH Formula unit eq.: CH3COOH (aq) + NaOH(aq)→ CH3COONa (aq) + H2O (l) Net ionic eq: CH3COOH (aq) + OH-(aq)→ CH3COO-(aq) + H2O (l) At equivalence point (pH > 7): CH3COO-(aq) + H2O (l) ↔ OH-(aq) + CH3COOH (aq) All of the strong base added has reacted to form acetate ions (conjugate weak base) After equivalence point: [OH-] = (mols base – mols acid) L solution (total) pOH = - log [OH-] Use weak acid ICE chart Use pH = pKa + log(b/a) Use salt of weak acid ICE chart

  22. Strong Acid-Weak Base Titrations NH3 Kb = 1.8 x 10-5 NH4+Ka = 5.6 x 10-10 HCl(aq) + NH3(aq) → NH4Cl (aq) H+(aq) + NH3(aq)→ NH4Cl (aq) At equivalence point (pH < 7): NH4+(aq) + H2O (l) ↔ NH3(aq) + H+(aq) Use weak base ICE chart Use weak acid ICE chart account for volume change Use pH = pKa + log(b/a) After equivalence point: [H+] = (mols acid – mols base) L solution (total)

  23. Acid-Base Indicators HIn(aq) H+(aq) + In-(aq) [HIn] [HIn]  10  10 [In-] [In-] Color of acid (HIn) predominates Color of conjugate base (In-) predominates

  24. pH Solutions of Red Cabbage Extract

  25. The titration curve of a strong acid with a strong base.

  26. Which indicator(s) would you use for a titration of HNO2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein

  27. Solubility Guidelines (abbreviated) Insoluble compounds: e.g. AgCl, PbCl2, BaSO4, Ca3PO4, etc. are not 100% insoluble – they all release small amounts of ions into solution. Ca(NO3)2(aq) + Na2CO3(aq)→ CaCO3(s) + 2NaNO3(aq)

  28. You have a solution that contains 0.010 M Cu+ and 0.010M Au+ What % of the gold can you recover (pure) by adding solid NaBr? Data AuBrKsp = 5.0 x 10-17 CuBrKsp = 5.3 x 10-9 a) Which cation will precipitate first? Strategy: 1) Find the [Br -] when CuBr first precipitates. 2) Find the [Au+] at this [Br -]

  29. 30.0 ml of an unknown HNO2 solution is titrated with 0.100M NaOH solution. The endpoint is reached after 32.7 ml of NaOH has been added. For HNO2Ka = 4.5 x 10-4pKa = 3.35 Kb NO2- = 2.2 x 10-11 What is the concentration of the unknown HNO2 solution prior to the titration? Moles NaOH = moles acid initial = 0.100 M • 0.0327 = 0.00327 moles 0.00327 moles  0.0300 L = 0.109 M What was the pH of the original HNO2 solution? HNO2↔ H+ + NO2-4.5 x 10-4 = x2/(0.109 – x) I 0.109 M 0 0 use quadratic C -x +x +x x = 0.00678 M H+ (0.0070) E 0.109 – x x xpH = 2.17 (2.16 if x << 0.109)

  30. What was the pH of the solution after 10.0 ml of NaOH were added? Moles base added = .100 M • 0.010 L = 0.001 moles base Moles acid remaining = 0.00327 – 0.001 = 0.00227 moles pH = 3.35 + log(0.001/0.0027) = 2.92 What is the pH at the equivalence point? Moles NO2- = moles base added = 0.00327 0.00327  0.0627 L = 0.0522 M NO2- + H2O ↔ OH-+ HNO2 2.2 x 10-11 = x2/(0.0522 – x) I 0.0522 M 0 0 C -x +x +x x = 1.07 x 10-6 M OH- E 0.0522 – x x xpOH= 5.97 & PH = 8.03

  31. What is the pH after 35.0 ml of NaOH have been added? Moles base added = .100 M • 0.035 L = 0.0035 moles base Moles base excess = 0.0035 – 0.00327 = 0.00023 moles [OH-] = 0.00023 moles  0.065 L = 0.00354 M OH- pOH = 2.45 and pH = 11.55

  32. Insoluble compounds: e.g. AgCl, PbCl2, BaSO4, Ca3PO4, etc. are not 100% insoluble – they all release small amounts of ions into solution. Solubility Equilibria AgCl(s) ↔ Ag+(aq) + Cl-(aq) Ksp = [Ag+] [Cl-] = 1.6 x 10-10 K = [Ag+] [Cl-] [AgCl(s)] Ksp = solubility product constant Molar Solubility = moles of compound that will dissolve per L of solution (mol/L) Solubility = grams of compound that will dissolve per L of solution (g/L)

  33. Molar Solubility = moles of compound that will dissolve per L of solution (mol/L) this value is based on the compound not on a specific cation/anion Let x = Molar solubility of AgCl Ksp = [Ag+] [Cl-] = 1.6 x 10-10 = x2 x = (1.6 x 10-10)½ = 1.3 x 10-5 M Solubility = grams of compound that will dissolve per L of solution (g/L) Solubility = Molar solubility • FW = mol/L • g/mol = g/L AgCl solubility = 1.3 x 10-5 • 143.3 = 1.9 x 10-3 g/L

  34. PbCl2(s) ↔ Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+] [Cl-]2 = 2.4 x 10-4 PbCl2 = 278.1 g/mol Let x = Molar solubility of PbCl2 PbCl2(s) ↔ Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+] [Cl-]2 = 2.4 x 10-4 Ksp = x • (2x)2 = 2.4 x 10-4 = x • 4x2 = 4x3 Molar solubility = x = (Ksp/4)⅓ = (2.4 x 10-4/4)⅓ = 3.9 x 10-2 M solubility = 3.9 x 10-2 • 278.1 = 11 g/L Ca3(PO4)2(s) ↔ 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3 [PO43-]2 = 1.1 x 10-26 Ksp = (3x)3 • (2x)2 = 1.1 x 10-26 = 27x3 • 4x2 = 108x5 Molar solubility = x = (Ksp/108)1/5 = (1.1 x 10-26/108)1/5 = 2.5 x 10-6 M solubility = 2.5 x 10-6 • 310.2 = 7.8 x 10-4 g/L Which is more soluble? a) lead chloride b) calcium phosphate

  35. Ksp = [Ag+] [Cl-] = 1.6 x 10-10 = x2 x = (1.6 x 10-10)½ = 1.3 x 10-5 M Solubility = grams of compound that can dissolve per L of solution (g/L) Solubility = Molar solubility • FW = mol/L • g/mol = g/L AgCl solubility = 1.3 x 10-5• 143.3 = 1.9 x 10-3 g/L What will you observe if you add 0.0050 g (5 mg) AgCl into 0.10L of water? a) unsaturated solution b) saturated solution c) precipitate forms What will you observe if you add 0.0050 g AgCl into 5.0 L of water ? No precipitate Unsaturated solution Qsp< Ksp Qsp= Ksp Saturated solution Precipitate will form Qsp> Ksp Supersaturated solution

  36. Al(OH)3 solubility = 2.23 x 10-7 g/L Al(OH)3 FW = 78.0 g/mol Al(OH)3(s) ↔ Al3+(aq) + 3OH-(aq) Ksp = [Al3+] [OH-]3 = x • (3x)3 = 27x4 Molar solubility = solubility (g/L) ÷ FW (g/mol) = 2.23 x 10-7 ÷ 78.0 = 2.86 x 10-9 Ksp = [Al3+] [OH-]3 = x • (3x)3 = 27x4 = 27 • (2.86 x 10-9)4 = 1.81 x 10-33

  37. PbI2(s) ↔ Pb2+(aq) + 2I-(aq) Ksp = [Pb2+] [I-]2 = 1.4 x 10-8 PbI2 = 461 g/mol Let x = Molar solubility of PbI2 1.4 x 10-8 = a) 4x3 b) 2x3 c) 4x2 d) 2x2. Molar solubility = x = (1.4 x 10-8/4)⅓ = 1.52 x 10-3 M solubility = 1.5 x 10-3 • 461 = 0.70 g/L Will the Molar solubility of PbI2 in 0.10 M NaI a) increase b) decrease c) stay the same 1.4 x 10-8 = x • (0.1 + 2x)2 assume x < 0.10 …… 1.4 x 10-8 = x • 0.12 & Molar solubility = x = 1.4 x 10-6 solubility = 1.4 x 10-6 • 461 = 6.5 x 10-4 g/L The presence of a common ion suppresses solubility – an example of LeChatelier’s Principle Try to find the Molar solubility of PbI2 in 0.20M PbNO3.

  38. Equations: Kw = 1.0 x 10-14 = [H3O+] [ OH-] Kw = Ka Kb pH = - log [H3O+] • pKa = - log Ka pH = pKa + log ([base]/[acid]) • Benzoic acid reacts as follows ….. • C6H5COOH + H2OC6H5COO- + H3O+Ka = 6.3 x 10-5pKa = 4.20 • 50.0 ml of 0.100 M benzoic acid is titrated with 0.120 M NaOH • Moles acid initial = 0.100 M • 0.050L = 0.0050 moles • (2) write the reaction between benzoic acid and OH- • C6H5COOH + OH-C6H5COO- + H2O • (2) T/F After 25 ml of NaOH has been added? • moles NaOH added = .120 M • 0.025L = 0.0030 moles • __T_ a) The solution is a buffer at this point. • moles b = 0.0030 • moles a = 0.0050 – 0.0030 = 0.0020 • __T_ b) The pH > 4.20 more base than acid

  39. (3) The equivalence point occurs when 41.7 ml NaOH has been added. • What is the [C6H5COO-] and the pH at this point? (Kb for benzoate is 1.59 x 10-10) • moles NaOH added = .120 M • 0.0417L = 0.0050 moles • this forms 0.0050 moles acetate • [C6H5COO-] = 0.0050 moles  0.0917 L = 0.0545 M • C6H5COO- + H2O  C6H5COOH + OH- • I 0.00545 0 0 • C -x +x +x • E 0.00545 – x x x • 1.59 x 10-10 = x2/(0.00545 – x) ignore x in denominator • x = [OH-] = 5.4 x 10-5 M pOH = 4.27 pH = 9.73

  40. 4. (3) CaF2 (FW = 78.0 g/mol), has a Ksp value of 4.0 x 10-11. What is the Molar solubility and solubility of CaF2? CaF2 Ca2+ + 2F- x 2x 4.0 x 10-11 = x • (2x)2 = 4x3 Molar Solubility = x = (4.0 x 10-11/4)1/3 = 2.15 x 10-4 Solubility (g/L) = 2.15 x 10-4 M • 78.0 g/mol = 0.017 g L-1

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