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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Chapter 8 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/. Applications of Aqueous Equilibria– ch. 8. 1. What is the pH at the equivalence point for the following titrations? a. NaOH with HBr

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria Chapter 8 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

  2. Applications of Aqueous Equilibria– ch. 8 1. What is the pH at the equivalence point for the following titrations? a. NaOH with HBr pH = 7 pH > 7 pH < 7 b. HCl with NaCH3COO pH = 7 pH > 7 pH < 7 c. KOH with HCN pH = 7 pH > 7 pH < 7

  3. Applications of Aqueous Equilibria– ch. 8 2. Which of the following will result in a buffer solution upon mixing? a. 0.1 mol HClO3 and 0.1 mol NaClO3 are put into 1L ofwater b. 0.5 mol H2S and 0.8 molNaHS are put into 1L of water c. 0.03 mol KHC2O4 and 0.02 mol K2C2O4 are put into 1L of water d. 0.1 molLiOH and 0.2 mol H3PO4 are put into 1L of water e. 0.1 mol HNO3 and 0.04 mol NH3 are put into 1 L of water

  4. Applications of Aqueous Equilibria– ch. 8 Buffers⇒ Solutions that can resist change in pH Composed of a weak acid and it’s conjugate base 3 ways to make a buffer: 1. Mix a weak acid and it’s conjugate base 2. Mix a strong base with a weak acid – the strong base must be the limiting reagent 3. Mix a strong acid with a weak base – the strong acid must be the limiting reagent

  5. Applications of Aqueous Equilibria– ch. 8 Buffers are classic examples of Le Chatlier’s Principle

  6. Applications of Aqueous Equilibria– ch. 8 3. Which of the following buffer solutions can “absorb” the most acid without changing the pH? (All solutions have the same volume) a. 0.4 M HF /0.5 M NaF b. 0.8 M HF /0.7 M NaF c. 0.3 M HF/0.7 M NaF

  7. Applications of Aqueous Equilibria– ch. 8 4. Calculate the pH of the following: a. A 35 mL solution that contains 0.2 M HN3 (Ka = 1.9 x 10-5) and 0.1 M NaN3 b. A solution prepared by mixing 30 mL of 0.2 M HN3 with 80 mL of 0.1 M NaN3

  8. Applications of Aqueous Equilibria– ch. 8 Henderson-Hasselbalch (Buffer) Equation pH = pKa + log or pH = pKa + log

  9. Applications of Aqueous Equilibria– ch. 8 5. Calculate the pH for the following titrations: a. 25 mL of 0.1 M HCl is mixed with 25 mL of 0.2 M KNO2 (Kb = 2.5 x 10-11) b. 50 mL of 0.1 M HCl is mixed with 25 mL of 0.2 M KNO2 c. 60 mL of 0.1 M HClis mixed with 25 mL of 0.2 M KNO2 d. Draw a titration curve and label 1) the equivalence point, 2) the region with maximum buffering, 3) where pH=pKa, 4) the buffer region, 5) where the pH only depends on [HA] and 6) where the pH only depends on [A-]

  10. Applications of Aqueous Equilibria– ch. 8 3 possible scenarios when titrating a weak acid with a strong base or titrating a weak base with a strong acid 1. Buffer ⇒nweak > nstrong use pH = pKa + log 2. Equivalence point ⇒nweak= nstrong pH only depends on the conjugate of the weak acid or weak base use Ka = or Kb = to solve for unknown 3. Beyond the equivalence point ⇒nweak< nstrong pH depends on the [xs strong acid] or [xs strong base]

  11. Applications of Aqueous Equilibria– ch. 8 Titration Curves Titration of a weak acid with a strong base Titration of a weak base with a strong acid >7 <7 pH pH Buffer Zone Volume of strong base Volume of strong acid

  12. Applications of Aqueous Equilibria– ch. 8 6. Calculate the pH of the following titrations: a. 40 mL of 0.20 M KOH is mixed with 40 mL of 0.5 M HClO (Ka = 3 x 10-8) b. 80 mL of 0.25 M KOH is mixed with 40 mL of 0.5 M HClO c. 100 mL of 0.25 M KOH is mixed with 40 mL of 0.5 M HClO d. Draw a titration curve and label 1) the equivalence point, 2) the region with maximum buffering, 3) where pH=pKa, 4) the buffer region, 5) where the pH only depends on [HA] and 6) where the pH only depends on [A-]

  13. Applications of Aqueous Equilibria– ch. 8 7. Calculate the pH for the following: a. 10 mL of 1 M HCl is added to a 100 mL buffer with 0.5 M CH3CH2COOH (Ka = 1.34 x 10-5) and 0.6 M CH3CH2COONa. b. 20 mL of 0.2 M KOH is added to a 100 mL buffer with 0.5 M CH3CH2COOH and 0.6 M CH3CH2COONa.

  14. Applications of Aqueous Equilibria– ch. 8 8. How would you prepare 1.0 L of a buffer at pH = 9.0 from 1.0 M HCN (Ka = 6.2 x 10-10) and 1.5 M NaCN?

  15. Applications of Aqueous Equilibria– ch. 8 9. How many grams of NaOH need to be added to a 50 mL of 0.3 M HNO2 (Ka = 4.0 x 10-4) to have a solution with a pH = 4?

  16. Applications of Aqueous Equilibria– ch. 8 10. Determine the solubility in mol/L and g/L for the following compounds: a. BaCO3 (Ksp = 1.6 x 10-9) b. Ag2S (Ksp = 2.8 x 10-49)

  17. Applications of Aqueous Equilibria– ch. 8 Solubility ⇒the maximum amount of solute that can dissolve into a given amount of solvent at any one temperature at this point the solution is said to be saturated and in equilibrium

  18. Applications of Aqueous Equilibria– ch. 8 11. Determine the Ksp values for the following compounds: a. Al(OH)3 (solubility = 5 x 10-9mol/L) b. MgF2 (solubility = 0.0735 g/L)

  19. Applications of Aqueous Equilibria– ch. 8 12. What is the solubility of Ca3(PO4)2 (Ksp = 1 x 10-54) in 0.02M solution of Na3PO4?

  20. Applications of Aqueous Equilibria– ch. 8 13. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution with pH of 11?

  21. Applications of Aqueous Equilibria– ch. 8 14. Will BaCrO4 (Ksp = 8.5 x 10-11) precipitate when 200 mL of 1 x 10-5 M Ba(NO3)2 is mixed with 350 mL of 3 x 10-5 M KCrO4?

  22. Applications of Aqueous Equilibria– ch. 8 You have completed ch. 8

  23. Answer Key – ch. 8 1. What is the pH at the equivalence point for the following titrations? a. NaOH(strong base) with HBr(strong acid) pH = 7 pH > 7 pH < 7 b. HCl(strong acid) with NaCH3COO (weak base) pH = 7 pH > 7 pH < 7 c. KOH (strong base) with HCN (weak acid) pH = 7 pH > 7 pH < 7

  24. Answer Key – ch.8 2. Which of the following will result in a buffer solution upon mixing? a. 0.1 mol HClO3 and 0.1 mol NaClO3 are put into 1L ofwater Strong acid and it’s conjugate base is not a buffer b. 0.5 mol H2S and 0.8 molNaHS are put into 1L of water weak acid and it’s conjugate base is a buffer c. 0.03 mol KHC2O4 and 0.02 mol K2C2O4 are put into 1L of water weak acid and it’s conjugate base is a buffer d. 0.1 molLiOH and 0.2 mol H3PO4 are put into 1L of water strong base and weak acid will be a buffer as long as the strong base is the limiting reagent e. 0.1 mol HNO3 and 0.04 mol NH3 are put into 1 L of water strong acid and weak base could be a buffer if the strong as is limiting – however in this case the weak base is limiting so no buffer

  25. Answer Key – ch. 8 3. Which of the following 50 mL solutions can absorb the most acid without changing the pH? a. 0.4 M HF /0.5 M NaF b. 0.8 M HF /0.7 M NaF c. 0.3 M HF/0.7 M NaF In order for a buffer to absorb an acid you want [base] to be high and the [acid] to be low and vice versa if absorbing base

  26. Answer Key – ch. 8 4. Calculate the pH of the following. a. 0.2 M HN3 (Ka = 1.9 x 10–5) /0.1 M NaN3 HN3/N3 – ⇒weak acid/conjugate base ⇒ buffer pH = pKa + log pH = - log (1.9 x 10–5) + log pH = 4.4 b. 30 mL of 0.2 M HN3/ 80 mL of 0.1 M NaN3 mixing solutions causes dilution so we can use M1V1 = M2V2 to get the new concentrations or we can alter the Henderson Hasselbalcheqn (HH)⇒ pH = pKa + log … continue to next slide

  27. Answer Key – ch. 8 4. b. …continued nA- ⇒ (80 mL)(0.1 mol/L) = 8 mmol nHA⇒ (30 mL)(0.2 mol/L) = 6 mmol pH = - log(1.9 x 10–5) + log(8/6) pH = 4.8

  28. Answer Key – ch.8 5. Calculate the pH for the following: In all 3 scenarios we’re adding a strong acid (HCl) to a salt with a weak base (NO2–) ⇒ since we have a strong substance (HCl) we can assume the neutralization reaction goes to completion ⇒ work stoichiometrically in moles a. 25 mL of 0.1 M HCl /25 mL of 0.2 M KNO2 Limiting Reagent 2.5 mmol H+ 5 mmol NO2– After the neutralization rxnis complete there’sHNO2/NO2–present ⇒ buffer using the altered HH eqn⇒ pH = pKa + log pH = - log(4 x 10–4) + log(2.5/2.5) pH = 3.4 note since the [HA] = [A–] ⇒ pH = pKa

  29. Answer Key – ch. 8 5. b. 50 mL of 0.1 M HCl /25 mL of 0.2 M KNO2 5 mmol H+ 5 mmol NO2– Note ⇒ Equivalence Point • After the neutralization rxn • is complete there’s only a • weak acid (HNO2) in • solution ⇒ change mmol to M • [HNO2] = = 0.067 M • Use Ka to solve for (x) • 4x10–4= • x = 0.00518 = [H3O+] • pH = -log(0.00518) • pH = 2.3

  30. Answer Key – ch. 8 c. 60 mL of 0.1 M HCl/25 mL of 0.2 M KNO2 6 mmol H+ 5 mmol NO2– After the neutralization rxn is complete there is strong acid (H+) and weak acid (HNO2) present A weak acid in the presence of a strong acid becomes insignificant [H+] = = 0.0118 M pH = - log(0.0118) pH = 1.9 Limiting Reagent

  31. Answer Key – ch. 8 6. Calculate the pH of the following: In all 3 scenarios we’re adding a strong base (OH–) to a weak acid (HClO)⇒ since we have a strong substance (OH–) we can assume the neutralization reaction goes to completion ⇒ work stoichiometrically in moles a. 40 mL of 0.2 M KOH/40 mL of 0.5 M HClO 8 mmol OH– 20 mmolHClO Limiting Reagent After the neutralization rxn is complete there’s HClO/ClO–present ⇒ buffer using the altered HH eqn⇒ pH = pKa + log pH = -log(3.5x10–8) + log pH = 7.3

  32. Answer Key – ch. 8 6. b. 80 mL of 0.25 M KOH/40 mL of 0.5 M HClO Note ⇒ Equivalence Point 20 mmol OH– 20 mmolHClO • After the neutralization rxn • there’s only weak base (ClO–) in • solution ⇒ change mmol to M • [ClO–] = = 0.167M • Need Kbto solve for (x) • Kb= = = 2.86x10–7 • 2.86x10–7= • x = 2.18x10–4= [OH–] • pOH = -log(2.18x10–4) = 3.7 • pH = 14 – 3.7 = 10.3

  33. Answer Key – ch. 8 6. c. 100 mL of 0.25 M KOH/40 mL of 0.5 M HClO 25 mmol OH– 20 mmolHClO • After the neutralization rxn is • complete there’s strong base • (OH–) and weak base (ClO–) • present ⇒ A weak base • is insignificant in the presence • of a strong base ⇒ • [OH–] = = 0.0357 M • pOH = - log(0.0357) = 1.45 • pH = 14 – 1.45 • pH = 12.5 Limiting Reagent

  34. Answer Key – ch. 8 7. Calculate the pH of the following: a. 10 mL of 1 M HCl is added to a 100 mL buffer with 0.5 M CH3CH2COOH (pKa = 4.87) and 0.6 M CH3CH2COONa. A strong acid (HCl) will react with the base in the buffer (CH3CH2COO–) After the neutralization rxn is complete the solution is still a buffer using the altered HH eqn⇒ pH = pKa + log pH = 4.87 + log pH = 4.79

  35. Answer Key – ch. 8 7 b. 20 mL of 0.2 M KOH is added to a 100 mL buffer with 0.5 M CH3CH2COOH and 0.6 M CH3CH2COONa.The strong base (OH–) will react with the acid in the buffer (CH3CH2COOH) After the neutralization rxn is complete the solution is still a buffer using the altered HH eqn⇒ pH = pKa + log pH = 4.87 + log pH = 4.98

  36. Answer Key – ch. 8 8. How would you prepare 1.0 L of a buffer at pH = 9.0 from 1.0 M HCN (Ka = 6.2 x 10-10) and 1.5 M NaCN? The total volume of the buffer has to be 1.0 L => so if we designate X as the volume of HCN then the volume of NaCN must be 1-X => so the moles of HCN are (1.0 M)(X L) = X moles and the moles of NaCN are (1.5 M)(1-X) = 1.5-1.5X moles => now we can plug these into pH = pKa + log 9.0 = -log(6.2 x 10-10)+log => X=0.71 Therefore the buffer is made by mixing 0.71 L of 1.0M HCN with 0.29 L of 1.5M NaCN

  37. Answer Key – ch. 8 9. How many grams of NaOH need to be added to a 50 mL of 0.3 M HNO2 (Ka = 4.0 x 10-4) to have a solution with a pH = 4? pH = pKa + log 4 = -log(4.0 x 10-4)+log x = 1.2 mmol of NaOH Molar mass of NaOH = 40 g/mol (1.2 mmol)(40 g/mol) = 48 mg of NaOH

  38. Answer Key – ch. 8 10. Determine the solubility in mol/L and g/L for the following compounds: a. BaCO3 (Ksp = 1.6 x 10-9) ⇒ the solubility is defined as the maximum amount of solute that can be dissolved in a particular amount of solvent at any one temperature or the saturation point ⇒ the solute is at equilibrium for saturated solutions Use Ksp to solve for x 1.6 x 10-9 = (x)(x) x = 4 x 10–5 since the molar ratio is 1:1 the molar solubility of BaCO3 = 4 x 10–5 mol/L or solubility = (4 x 10–5 mol/L)(197.34 g/mol) = 0.0079 g/L

  39. Answer Key – ch. 8 10. b. Ag2S (Ksp = 2.8 x 10-49) Use Ksp to solve for x 2.8 x 10-49 = (2x)2(x) x = 4.14 x 10–17 Molar solubility = 4.14 x 10–17 M or solubility = (4.14 x 10–17mol/L)(247.8 g/mol) = 1 x 10–14 g/L

  40. Answer Key – ch. 8 11. Determine the Ksp values for the following compounds: a. Al(OH)3 (solubility = 5 x 10-9mol/L) ⇒the solubility tells us how much will dissolve in order to get to equilibrium Ksp = (5 x 10-9)(1.5 x 10-8)3 = 1.7 x 10–32

  41. Answer Key – ch. 8 11.b. MgF2 (solubility = 0.0735 g/L)⇒ first we need the molar solubility (0.0735 g/L)/(62.31 g/mol) = 0.00118 mol/L Ksp = (0.00118)(0.0024)2 = 6.6 x 10–9

  42. Answer Key – ch. 8 12. What is the solubility of Ca3(PO4)2 (Ksp = 1 x 10-54) in 0.02M solution of Na3PO4? The solute and the solution have the phosphate ion in common ⇒ this will result in lower solubility than if the solute were to be dissolved in pure water aka the common ion effect Use Ksp to solve for x 1 x 10-54 = (3x)3(0.02)2 x = 4.5 x 10–18 molar solubility = 4.5 x 10–18M Insignificantly small

  43. Answer Key – ch. 8 13. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution with pH of 11? The solute and the solution have the hydroxide ion in common ⇒ this will result in lower solubility than if the solute were to be dissolved in pure water aka the common ion effect ⇒ since the pH is 11 the pOH is 3 ⇒ [OH-] = 10–3 M Use Ksp to solve for x 4.5 x 10-17= (x)(10–3)2 x = 4.5 x 10-11 molar solubility = 4.5 x 10-11M Insignificantly small

  44. Answer Key – ch. 8 14. Will BaCrO4 (Ksp = 8.5 x 10-11) precipitate when 200 mL of 1 x 10-5 M Ba(NO3)2 is mixed with 350 mL of 3 x 10-5 M KCrO4? The solution is saturated with BaCrO4 if [Ba2+][CrO42–] = 8.5 x 10-11 however if [Ba2+][CrO42–] > 8.5 x 10-11 the solution is supersaturated and will precipitate out some BaCrO4or if [Ba2+][CrO42–] < 8.5 x 10-11 the solution is unsaturated and there will be no noticeable change We can use M1V1 = M2V2 to get the new concentrations [Ba2+] = (1 x 10–5 M)(200 mL)/(550 mL) = 3.64 x 10–6 M [CrO42–]= (3 x 10-5 M)(350 mL)/(550 mL) = 1.91 x 10 -5 M [Ba2+][CrO42–] = (3.64 x 10–6 M )(1.91 x 10 -5 M)= 6.95 x 10 – 11 the solution is unsaturated and no precipitate will form

  45. Answer Key – ch. 8

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