Acid-Base Equilibria - PowerPoint PPT Presentation

laurel
acid base equilibria n.
Skip this Video
Loading SlideShow in 5 Seconds..
Acid-Base Equilibria PowerPoint Presentation
Download Presentation
Acid-Base Equilibria

play fullscreen
1 / 81
Download Presentation
Acid-Base Equilibria
143 Views
Download Presentation

Acid-Base Equilibria

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Acid-Base Equilibria AP Chem Unit 15

  2. Unit 15 Content • Solutions of Acids or Bases Containing a Common Ion • Buffered Solutions • Buffering Capacity • Titrations and pH Curves • Acid-Base Indicators

  3. Introduction Most chemistry in the natural world takes place in an aqueous solution. One of the most significant aqueous reactions is one with acids and bases. Most living systems are very sensitive to pH and yet are subjected to acids and bases. • The main idea of this unit is to understand the chemistry of a buffered system. Buffered systems contain chemical components that enable a solution to be resistant to change in pH.

  4. Solutions of Acids or Bases Containing a Common Ion 15.1

  5. Common Ions Solutions that contain a weak acid and the salt of it conjugate base create a common ion system. • A common ion system can also be a weak base and the salt of its conjugate acid. • Example: HF and NaFsolution or NH3 and NH4Cl solution.

  6. Common Ions • The soluble salts of conjugate acids and conjugate bases are typically strong electrolytes and therefore dissociate 100%. • This dissociation increases the concentration of the conjugate ion and a shift in equilibrium occurs according to Le Chatelier. • Example: • Shift Left!

  7. Common Ion Effect The shift left in the second equation is known as the common ion effect. • This effect makes a solution of NH3 and NH4Cl less basic than NH3 alone

  8. Common Ion Effect The shift left in the second equation is known as the common ion effect. This effect makes a solution of NaF and HF less acidic than a solution of HF alone.

  9. Common Ion Effect Calculations for common ion equilibria is similar to weak acid calculations except that the initial concentrations of the anion are not 0. • Initial concentrations for the equilibrium calculation must include the concentration of ions from the complete dissociation of the salt.

  10. Practice Problem #1 The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M NaF. Major species? ICE:

  11. Practice Problem #1 The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M NaF. ANS: .072%

  12. Buffered Solutions 15.2

  13. Buffered Solutions The most important application of a common ion system is buffering. • A buffered solution resists a change in its pH when either hydroxide ions or protons are added. • Blood is a practical example of a buffered solution; it is buffered with carbonic acid and the bicarbonate ion (among others).

  14. Buffered Solutions • A solution can be buffered at any pH by choosing the appropriate components. • Smaller amounts of H+ or OH- can be added with little pH effect.

  15. Practice Problem #2 A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and 0.50 M sodium acetate. Calculate the pH of this solution. pH = 4.74

  16. How Does Buffering Work? • When OH- ions are added to a solution of a weak acid, the OH- ions react with the best source of H+ (weak acid) to make water: • The net result is that OH- ions are replaced by A- ions and water until they are all consumed.

  17. How Does Buffering Work? • When H+ ions are added to a solution of a weak acid, the H+ ions react with the strong conjugate base. • The net result is that more weak acid is produced, but free H+ do not accumulate to make large changes in the pH.

  18. Buffered Solution Summary Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in Chapter 14. • When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilbrium calculations.

  19. Practice Problem #3 Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to the buffered solution in #2 (0.50 M acetic and 0.50 M sodium acetate). Compare this pH change with the original (pH=4.74) and the pH that occurs with 0.010 mol of solid NaOH is added to 1.0 L of water.

  20. Practice Problem #3 • Major species: • OH- +HC2H3O2 C2H3O2- + H2O • mole table:

  21. Practice Problem #3 • ICE table: pH= 4.76 (+.02 off original) pH of just OH-=12.00

  22. Henderson-Hasselbalch Equation Buffered solutions work well as long as the concentration of the salts and weak acids/bases in solution are much larger than the amount of OH- and H+ being added. • The amount of [H+] being in a buffered solution is often solved using a rearrangement of the equilibrium expression:

  23. Henderson-Hasselbalch Equation • This equation, , can be changed into another useful form by taking the negative log of both sides: This log form of the expression is called the Henderson-Hasselbalch Equation.

  24. Henderson-Hasselbalch Equation Henderson-Hasselbalch Equation: When using this equation it is often assumed that the the equilibrium concentrations of A_ and HA are equal to their initial concentrations due to the validity of most approximations. • Since the initial concentrations of HA and A- are relatively large in a buffered solution, this assumption is generally acceptable.

  25. Practice Problem #4 Calculate the pH of a solution containing 0.75 M lactic acid (Ka=1.4x10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion. pH = 3.38

  26. Reminder Buffered solutions can be formed from a weak base and the corresponding conjugate acid. In these solutions, the weak base B reacts with any H+ added: • The conjugate acid BH+ reacts with any added OH-:

  27. Practice Problem #5 A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution. pH=9.05

  28. Practice Problem #6 Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution from #5 (.25M NH3and .40 NH4Cl). pH= 8.73

  29. Summary • Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+. • When H+ is added to a buffered solution, it reacts essentially to completion with the weak base:

  30. Summary • When OH-is added to a buffered solution, it reacts essentially to completion with the weak acid present: • The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentration of the buffering materials (HA and A- or B and BH+) are large compared with the amounts of H+ or OH- added.

  31. Buffering Capacity 15.3

  32. Buffering Capacity The buffering capacity of a buffered solution represents the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. • The pH of a buffered solution is determined by the ratio [A-] / [HA]. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].

  33. Practice Problem #7 Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following solutions: • Soln A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 • Soln B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2 Ka=1.8x10-5. Soln A: pH=4.74 Soln B: pH=4.56

  34. Optimal Buffering The optimal buffering occurs when [HA] is equal to [A-]. • This ratio of 1 resists the most amount of pH change. • The pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH. • For example, if a buffered solution is needed with a pH of 4.0. The most effect buffering will occur when [HA] is equal to [A-] and the pKa of the acid is close to 4.0 or Ka=1.0x10-4.

  35. Practice Problem #8 A chemist needs a solution buffered at pH=4.30 and can choose from the following acids and their sodium salts. Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30. Which system will work best? • chloroacetic acid (Ka=1.35x10-3) • propanoic acid (Ka=1.3x10-5) • benzoic acid (Ka=6.4x10-5) • hypochlorous acid (Ka=3.5x10-8) benzoic acid

  36. Titrations and pH Curves 15.4

  37. Titration Curve An acid-base titration is often graphed by plotting the pH of the solution (y-axis) vs. the amount of titrant added (x-axis). • Point (s) is the equivalence point/stoichiometric point. This is where [OH-]=[H+].

  38. Strong Acid-Strong Base Titrations The net ionic reaction for a strong acid- strong base titration is: To compute [H+] at any point in a titration, the moles of H+ that remains at a given point must be divided by the total volume of the solution. • Titrations often include small amounts. The mole is usually very large in comparison. • Millimole (mmol) is often used for titration amounts.

  39. Molarity with mmols

  40. Strong Acid-Base Titration Example Calculation 50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. • NaOH has not been added. Since HNO3 is a strong acid (complete dissociation), pH = 0.699

  41. Strong Acid-Base Titration Example Calculation 50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. • 10.0 ml of 0.100 M NaOH has been added. H+ = 10 mmol – 1.0 mmol of OH- = 9.0mmol 9.0 mmol H+/ 60 ml = 0.15 M pH=0.82

  42. Strong Acid-Base Titration Example Calculation 50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. • 20.0 mL of 0.100 M NaOH has been added. pH=0.942

  43. Strong Acid-Base Titration Example Calculation 50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. • 50.0 mL of 0.100 M NaOH has been added. pH=1.301

  44. Strong Acid-Base Titration Example Calculation 50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. • 100.0 mL of 0.100 M NaOH has been added. The stoichiometric point/equivalence point is reached. pH=7

  45. Strong Acid-Base Titration Example Calculation 50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. • 150.0 mL of 0.100 M NaOH has been added. pH = 12.40

  46. Strong Acid-Base Titration Example Calculation 50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at the following points during the titration. • 200.0 mL of 0.100 M NaOH has been added. pH=12.40

  47. Strong Acid-Base Titration Example Calculation The results of these calculations are summarized by this graph: The pH changes very gradually until the titration is close to the equivalence point, then dramatic change occurs. Near the E.P., small changes produces large changes in the OH-/H+ ratio.

  48. Strong Acid-Base Summary • Before the equivalence point, [H+] can be calculated by dividing the number of millimoles of H+ remaining by the total volume of the solution in mL. • At the equivalence point, pH=7.0 • After the equivalence point, [OH-] can be calculated by dividing the number of millimoles of excess OH- by the total volume of the solution. [H+] is calculated from Kw.

  49. Titrations of Weak Acids with Strong Bases When acids don’t completely dissociate, calculations used previously need adjusted. • To calculate [H+] after a certain amount of strong base has been added, the weak acid dissociation equilibrium must be used. • Remember: that a strong base reacts to completion with a weak acid.

  50. Calculating the pH Curve for a Weak Acid-Strong Base Titration Calculating the pH Curve for a Weak Acid-Strong Base Titration is broken down into two steps: • A stoichiometric problem. The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined. • An equilibrium problem: The position of the weak acid equilibrium is determined, and the pH is calculated.