Applications of Aqueous Equilibria Chapter 8 E-mail: [email protected] 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 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 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 of water b. 0.5 mol H2S and 0.8 mol NaHS 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 mol LiOH 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 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 Applications of Aqueous Equilibria– ch. 8 Buffers are classic examples of Le Chatlier’s Principle 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 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 Applications of Aqueous Equilibria– ch. 8 Henderson-Hasselbalch (Buffer) Equation A– pH = pKa + log HA or − pH = pKa + log nnA HA 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 HCl is 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-] 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 nnA HA 2. Equivalence point ⇒ nweak = nstrong pH only depends on the conjugate of the weak acid or weak base H3O+ A– use Ka = HA OH− HA or Kb = A– to solve for unknown 3. Beyond the equivalence point ⇒ nweak < nstrong pH depends on the [xs strong acid] or [xs strong base] 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 pH <7 pH Volume of strong base Buffer Zone Volume of strong acid 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-] 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. 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? 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? 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) 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 Applications of Aqueous Equilibria– ch. 8 11. Determine the Ksp values for the following compounds: a. Al(OH)3 (solubility = 5 x 10-9 mol/L) b. MgF2 (solubility = 0.0735 g/L) 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? 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? 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? Applications of Aqueous Equilibria– ch. 8 You have completed ch. 8 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 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 of water Strong acid and it’s conjugate base is not a buffer b. 0.5 mol H2S and 0.8 mol NaHS 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 mol LiOH 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 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 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 –] [A pH = pKa + log [HA] pH = - log (1.9 x 10–5) + log 0.1 0.2 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 Hasselbalch eqn (HH)⇒ pH = pKa + log nnA− HA … continue to next slide 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 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 2.5 mmol H+ 5 mmol NO2– Reagent H+ NO2– I 2.5 5 0 ∆ -2.5 -2.5 +2.5 F 0 2.5 2.5 HNO2 After the neutralization rxn is complete there’s HNO2/NO2– present ⇒ buffer using the altered HH eqn ⇒ pH = pKa + log nnA− HA pH = - log(4 x 10–4) + log(2.5/2.5) pH = 3.4 note since the [HA] = [A–] ⇒ pH = pKa Answer Key – ch. 8 5. b. 50 mL of 0.1 M HCl /25 mL of 0.2 M KNO2 Note ⇒ Equivalence Point 5 mmol NO2– 5 mmol H+ H+ NO2– I 5 5 0 ∆ -5 -5 +5 F 0 0 5 HNO2 H2O ⇌ HNO2 H3O+ NO2– I 0.067 N/A 0 0 ∆ -x N/A +x +x E 0.067 N/A x x After the neutralization rxn is complete there’s only a weak acid (HNO2) in solution ⇒ change mmol to M [HNO2] = 5 mmol = 0.067 M 75 mL Use Ka to solve for (x) (x)(x) 4x10–4= 0.067 x = 0.00518 = [H3O+] pH = -log(0.00518) pH = 2.3 Answer Key – ch. 8 c. 60 mL of 0.1 M HCl/25 mL of 0.2 M KNO2 5 mmol NO2– 6 mmol H+ H+ NO2– I 6 5 0 ∆ -5 -5 +5 F 1 0 5 Limiting Reagent HNO2 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+] = 1 mmol = 0.0118 M 85 mL pH = - log(0.0118) pH = 1.9 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– Limiting Reagent OH– HClO I 8 ∆ F 20 mmol HClO H2O ClO– 20 N/A 0 -8 -8 N/A +8 0 12 N/A 8 After the neutralization rxn is complete there’s HClO/ClO– present ⇒ buffer using the altered HH eqn ⇒ pH = pKa + log nnA− HA pH = -log(3.5x10–8) + log 8 12 pH = 7.3 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– OH– 20 mmol HClO HClO H2O ClO– I 20 20 N/A 0 ∆ - 20 -20 N/A 20 F 0 0 N/A 20 ClO– H2O ⇌ OH– I 0.167 N/A 0 0 ∆ -x N/A +x +x F 0.167 N/A x x HClO After the neutralization rxn there’s only weak base (ClO–) in solution ⇒ change mmol to M [ClO–] = 20 mmol = 0.167M 120 mL Need Kb to solve for (x) 1x10–14 Kb = Kw = = 2.86x10–7 Ka 3.5x10–8 2.86x10–7= (x)(x) 0.167 –4 x = 2.18x10 = [OH–] pOH = -log(2.18x10–4) = 3.7 pH = 14 – 3.7 = 10.3 Answer Key – ch. 8 6. c. 100 mL of 0.25 M KOH/40 mL of 0.5 M HClO 25 mmol OH– OH– HClO 20 mmol HClO H2O ClO– I 25 20 N/A 0 ∆ - 20 -20 N/A 20 F 5 0 N/A 20 Limiting Reagent 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 ⇒ 5 mmol [OH–] = = 0.0357 M 140 mL pOH = - log(0.0357) = 1.45 pH = 14 – 1.45 pH = 12.5 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–) CH3CH2COO– H+ I 60 10 50 mmol ∆ - 10 - 10 + 10 F 50 0 60 CH3CH2COOH After the neutralization rxn is complete the solution is still a buffer using the altered HH eqn ⇒ pH = pKa + log nnA− HA pH = 4.87 + log 50 60 pH = 4.79 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) CH3CH2COO– CH3CH2COOH OH- I 50 mmol 4 mmol 60 mmol ∆ -4 -4 +4 F 46 0 64 After the neutralization rxn is complete the solution is still a buffer using the altered HH eqn ⇒ pH = pKa + log nnA− HA pH = 4.87 + log 64 50 pH = 4.98 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 nnA− HA 9.0 = -log(6.2 x 10-10)+log 1.5−1.5X => X=0.71 X Therefore the buffer is made by mixing 0.71 L of 1.0M HCN with 0.29 L of 1.5M NaCN 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? H2O HNO2 OH- NO2- I 15mmol x mmol N/A 0 Δ -x -x N/A +x F 15-x 0 N/A x pH = pKa + log nnA− HA -4 10 )+log x 4 = -log(4.0 x 1.5−x x = 1.2 mmol of NaOH Molar mass of NaOH = 40 g/mol (1.2 mmol)(40 g/mol) = 48 mg of NaOH 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 I ∆ Eq 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 2+ 2– BaCO3 ⇌ Ba CO3 1.6 x 10-9 = (x)(x) N/A 0 0 x = 4 x 10–5 since the molar ratio is 1:1 N/A +x +x the molar solubility of BaCO3 N/A x x = 4 x 10–5 mol/L or solubility = (4 x 10–5 mol/L)(197.34 g/mol) = 0.0079 g/L Answer Key – ch. 8 10. b. Ag2S (Ksp = 2.8 x 10-49) Ag2S ⇌ 2 Ag+ S2– I N/A 0 0 ∆ N/A +2x +x Eq N/A 2x x 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 10–17 mol/L)(247.8 g/mol) = 1 x 10–14 g/L Answer Key – ch. 8 11. Determine the Ksp values for the following compounds: a. Al(OH)3 (solubility = 5 x 10-9 mol/L) ⇒ the solubility tells us how much will dissolve in order to get to equilibrium Al(OH)3 ⇌ Al3+ 3 OH– I N/A 0 0 ∆ N/A + 5 x 10-9 +3(5 x 10-9 ) Eq N/A 5 x 10-9 1.5 x 10-8 Ksp = (5 x 10-9)(1.5 x 10-8)3 = 1.7 x 10–32 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 MgF2 ⇌ Mg2+ 2 F– I N/A 0 0 ∆ N/A + 0.00118 +2(0.00118 ) Eq N/A 0.00118 0.0024 Ksp = (0.00118)(0.0024)2 = 6.6 x 10–9 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 Ca3(PO4)2 ⇌ 3 Ca2+ 2 PO43- I N/A 0 0.02 ∆ N/A + 3x +2x Eq N/A 3x 0.02 + 2x Insignificantly small 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 10–18 M 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 Zn(OH)2 ⇌ Zn2+ 2 OH- I N/A 0 10–3 M ∆ N/A +x +2x Eq N/A x 10–3 + 2x Insignificantly small 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 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 BaCrO4 or 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 Answer Key – ch. 8

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