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Chapter 16: Aqueous Ionic Equilibrium

Chapter 16: Aqueous Ionic Equilibrium. Buffers. S olutions that resist changes in pH when an acid or base is added Act by neutralizing acid or base that is added to the buffered solution There is a limit to buffering capacity  eventually the pH changes Standard Buffer Solutions =

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Chapter 16: Aqueous Ionic Equilibrium

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  1. Chapter 16: Aqueous Ionic Equilibrium

  2. Buffers • Solutions that resist changes in pH when an acid or base is added • Act by neutralizing acid or base that is added to the buffered solution • There is a limit to buffering capacity  eventually the pH changes • Standard Buffer Solutions = • solution of a weak acid + solution of soluble salt containing the conjugate base anion

  3. Making an Acid Buffer

  4. How Acid Buffers Work:Addition of BaseHA(aq) + H2O(l) A−(aq) + H3O+(aq) • Buffers = Application of Le Châtelier’s Principle to weak acid equilibrium • Buffer solutions contain significant amounts of the weak acid molecules, HA • HA react with added A- to neutralize it HA(aq) + OH−(aq) → A−(aq) + H2O(l) • you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium

  5. H2O How Buffers Work new A− A− HA A− HA  H3O+ + Added HO−

  6. How Acid Buffers Work:Addition of AcidHA(aq) + H2O(l) A−(aq) + H3O+(aq) • The buffer solution also contains significant amounts of the conjugate base anion, A− • These ions combine with added acid to make more HA H+(aq) + A−(aq) → HA(aq) • After the equilibrium shifts, the concentration of H3O+ is kept constant

  7. H2O How Buffers Work new HA HA HA A− A−  H3O+ + Added H3O+

  8. Common Ion Effect HA(aq) + H2O(l) A−(aq) + H3O+(aq) • Adding NaA, a salt containing the A− • A− = conjugate base of the acid (the common ion) • This shifts the equilibrium to the left • The new pH is higher than the pH of the acid solution • Lower H3O+ ion concentration

  9. Common Ion Effect

  10. Practice − What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O  F + H3O+

  11. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O  F + H3O+ x +x +x x 0.14 x 0.071 + x

  12. Practice – What is the pH of a buffer that is 0.14 M and 0.071 M KF? Ka for HF = 7.0 x 10−4 pKa for HF = 3.15 0.14 x 0.071 +x

  13. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10−4 x = 1.4 x 10−3 the approximation is valid

  14. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? 0.14 x 0.071 + x x x = 1.4 x 10−3

  15. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?

  16. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10−4 the values are close enough

  17. Henderson-Hasselbalch Equation • Calculating the pH of a buffer solution can be simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation • The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate base • as long as the “x is small” approximation is valid

  18. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O  F + H3O+

  19. When should you use the “Full Equilibrium Analysis” or the “Henderson-HasselbalchEquation”? • The Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable • Generally, the “x is small” approximation will work when both of the following are true: • the initial concentrations of acid and salt are not very dilute • the Ka is fairly small • For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000x larger than the value of Ka

  20. How Much Does the pH of a Buffer Change When an Acid or Base Is Added? • While buffers resist change in pH when acid or base is added to them, their pH does change • Calculating the new pH after adding acid or base requires breaking the problem into two parts • a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other • added acid reacts with the A− to make more HA • added base reacts with the HA to make more A− • an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]

  21. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? F− + H3O+  HF + H2O

  22. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? F− + H3O+  HF + H2O

  23. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? F− + H3O+  HF + H2O

  24. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? F− + H3O+  HF + H2O −0.020 −0.020 +0.020 0.051 0 0.160 0.051 0 0.160

  25. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? HF + H2O  F + H3O+

  26. H2O(l) + NH3(aq) NH4+(aq) + OH−(aq) Basic BuffersB:(aq) + H2O(l)  H:B+(aq) + OH−(aq) • Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−

  27. Henderson-Hasselbalch Equation for Basic Buffers • The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product • The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product • B: + H2O  H:B+ + OH− • To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction • H:B+ + H2O  B: + H3O+ • this does not affect the concentrations, just the way we are looking at the reaction

  28. Relationship between pKa and pKb • Just as there is a relationship between the Ka of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base Ka Kb = Kw = 1.0 x 10−14 −log(Ka Kb) = −log(Kw) = 14 −log(Ka) + −log(Kb) = 14 pKa + pKb = 14

  29. Buffering Effectiveness • A good buffer should be able to neutralize moderate amounts of added acid or base • However, there is a limit to how much can be added before the pH changes significantly • The buffering capacityis the amount of acid or base a buffer can neutralize • The buffering rangeis the pH range the buffer can be effective • The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base

  30. Effect of Relative Amounts of Acid and Conjugate Base A buffer is most effective with equal concentrations of acid and base Buffer 1 0.100 mol HA & 0.100 mol A− Initial pH = 5.00 Buffer 2 0.18 mol HA & 0.020 mol A− Initial pH = 4.05 pKa (HA) = 5.00 HA + OH− A + H2O after adding 0.010 mol NaOH pH = 5.09 after adding 0.010 mol NaOH pH = 4.25

  31. Effect of Absolute Concentrations of Acid and Conjugate Base A buffer is most effective when the concentrations of acid and base are largest Buffer 1 0.50 mol HA & 0.50 mol A− Initial pH = 5.00 Buffer 2 0.050 mol HA & 0.050 mol A− Initial pH = 5.00 pKa (HA) = 5.00 HA + OH− A + H2O after adding 0.010 mol NaOH pH = 5.02 after adding 0.010 mol NaOH pH = 5.18

  32. Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer

  33. Effectiveness of Buffers • A buffer will be most effective when the [base]:[acid] = 1 • equal concentrations of acid and base • A buffer will be effective when 0.1 < [base]:[acid] < 10 • A buffer will be most effective when the [acid] and the [base] are large

  34. Buffering Range • We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 • Substituting into the Henderson-Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pH Highest pH Therefore, the effective pH range of a buffer is pKa± 1 When choosing an acid to make a buffer, choose one whose is pKa closest to the pH of the buffer

  35. Practice – What ratio of NaC7H5O2 : HC7H5O2 would be required to make a buffer with pH 3.75? Benzoic acid, HC7H5O2, pKa = 4.19 to make a buffer with pH 3.75, you would use 0.363 times as much NaC7H5O2 as HC7H5O2

  36. Buffering Capacity • Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH • The buffering capacity increases with increasing absolute concentration of the buffer components • As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves • Buffers that need to work mainly with added acid generally have [base] > [acid] • Buffers that need to work mainly with added base generally have [acid] > [base]

  37. Titration • In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete • when the reaction is complete we have reached the endpoint of the titration • An indicator may be added to determine the endpoint • an indicator is a chemical that changes color when the pH changes • When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point

  38. Titration

  39. Titration Curve • A plot of pH vs. amount of added titrant • The inflection point of the curve is the equivalence point of the titration • Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH • The pH of the equivalence point depends on the pH of the salt solution • equivalence point of neutral salt, pH = 7 • equivalence point of acidic salt, pH < 7 • equivalence point of basic salt, pH > 7 • Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH

  40. Titration Curve:Unknown Strong Base Added to Strong Acid

  41. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes After Equivalence (excess base) Equivalence Point equal moles of HCl and NaOH pH = 7.00  Before Equivalence (excess acid)

  42. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH added 5.0 mLNaOH 5.0 x 10−4 mole NaOH added HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3 Before equivalence point 42

  43. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq) • To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10−3 moles • At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over • Because the NaCl is a neutral salt, the pH at equivalence = 7.00

  44. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) • Initial pH = −log(0.100) = 1.00 • Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3 • At equivalence point added 25.0 mLNaOH 2.5 x 10−3 mole NaOH added

  45. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) • Initial pH = −log(0.100) = 1.00 • Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3 • After equivalence point added 30.0 mLNaOH 3.0 x 10−3 mole NaOH added

  46. added 30.0 mLNaOH 0.00050 mol NaOH pH = 11.96 added 35.0 mLNaOH 0.00100 mol NaOH pH = 12.22 added 5.0 mLNaOH 0.00200 mol HCl pH = 1.18 added 10.0 mLNaOH 0.00150 mol HCl pH = 1.37 added 25.0 mLNaOH equivalence point pH = 7.00 25.0 mL 0.100 M HCl 0.00250 mol HCl pH = 1.00 Adding 0.100 M NaOH to 0.100 M HCl added 40.0 mLNaOH 0.00150 mol NaOH pH = 12.36 added 15.0 mLNaOH 0.00100 mol HCl pH = 1.60 added 50.0 mLNaOH 0.00250 mol NaOH pH = 12.52 added 20.0 mLNaOH 0.00050 mol HCl pH = 1.95

  47. Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • Before equivalence point added 10.0 mL NaOH

  48. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO3 to reach equivalence • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • At equivalence point: moles of NaOH = 1.25 x 10−2

  49. Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • After equivalence point added 100.0 mL NaOH

  50. Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • After equivalence point added 100.0 mL NaOH

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