1 / 64

Chapter 15 Applications of Aqueous Equilibria

Chapter 15 Applications of Aqueous Equilibria. HC 2 H 3 O 2 ( aq ) + H 2 O ( l ). H 3 O + ( aq ) + C 2 H 3 O 2 − ( aq). The Common-Ion Effect. Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

pepper
Download Presentation

Chapter 15 Applications of Aqueous Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 15Applications of Aqueous Equilibria

  2. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) The Common-Ion Effect • Consider a solution of acetic acid: • If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

  3. The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

  4. Ka = [H3O+] [F−] [HF] = 6.8  10-4 The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Ka for HF is 6.8  10−4.

  5. HF(aq) + H2O(l) H3O+(aq) + F−(aq) The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

  6. (0.10) (x) (0.20) 6.8  10−4 = (0.20) (6.8  10−4) (0.10) The Common-Ion Effect = x 1.4  10−3 = x

  7. The Common-Ion Effect • Therefore, [F−] = x = 1.4  10−3 [H3O+] = 0.10 + x = 1.01 + 1.4  10−3 = 0.10 M • So, pH = −log (0.10) pH = 1.00

  8. Acidic Solutions Containing Common Ions In section 14.5 we found that the equilibrium concentration of H+ in a 1.0 M HF solution is 2.7  10−2, & the % dissociation of HF is 2.7%. Calculate the [H+] & the % dissociation of HF in a solution that is 1.0 MHF (Ka for HF is 7.2  10−4) & 1.0 M NaF.

  9. Buffers: • Solutions of a weak conjugate acid-base pair. • They are particularly resistant to pH changes, even when strong acid or base is added.

  10. Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

  11. Buffers If acid is added, the F− reacts to form HF and water.

  12. Ka = [H3O+] [A−] [HA] HA + H2O H3O+ + A− Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

  13. [A−] [HA] [H3O+] Ka = base [A−] [HA] −log [H3O+] + −log −log Ka = pKa acid pH Buffer Calculations Rearranging slightly, this becomes Taking the negative log of both side, we get

  14. [base] [acid] pKa = pH − log [base] [acid] pH = pKa + log Buffer Calculations • So • Rearranging, this becomes • This is the Henderson–Hasselbalchequation.

  15. Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  10−4.

  16. [base] [acid] (0.10) (0.12) pH = pKa + log pH = −log (1.4  10−4) + log Henderson–Hasselbalch Equation pH = 3.85 + (−0.08) pH = 3.77

  17. The pH of a Buffered Solution I A buffered solution contains 0.50 M acetic acid (HC3H5O3 ) (Ka for HC3H5O3 is 1.8  10−5) and 0.50 M sodium acetate. Calculate the pH of the solution.

  18. pH Changes in Buffered Solutions Calculate the pH change that occurs when 0.010 mol of solid NaOH is added to 1L of the buffer in the previous probIem. Compare this change to the change in pH when 0.010 M NaOH is added to 1.0 L of water.

  19. pH Changes in Buffered Solutions Calculate the pH change that occurs when 0.010 mol of solid NaOH is added to 1L of the buffer in the previous probIem. Compare this change to the change in pH when 0.010 M NaOH is added to 1.0 L of water.

  20. The pH of a Buffered Solution II Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4  10−4) & 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biological systems. For example, it is found in milk & present in human muscle tissue during exertion.

  21. The pH of a Buffered Solution III A buffered solution contains 0.25 M NH3 (Kb = 1.8  10−5) & 0.40 M NH4Cl. Calculate the pH of this solution.

  22. pH Range • The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH.

  23. When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

  24. Addition of Strong Acid or Base to a Buffer • Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. • Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

  25. Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8  10−5) = 4.74

  26. Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)

  27. (0.320) (0. 200) pH = 4.74 + log Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + 0.06 pH = 4.80

  28. Adding a Strong Acid to a Buffered Solution Calculate the pH of this solution that results when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered solution from the previous example.

  29. Adding a Strong Acid to a Buffered Solution II Calculate the pH of this solution that results when 0.10 mol of gaseous HCl is added to 1.0 L of the following solutions (Ka for acetic acid is 1.8  10−5) : Solution A: 5.00 M HC2H3O2 & 5.00 M NaC2H3O2 Solution B: 0.050 M HC2H3O2 & 0.050 M NaC2H3O2

  30. Preparing a Buffer A chemist needs a solution buffered at a pH of 4.30 & can choose from the following acids (& their sodium salts): • Chloroacetic acid (Ka = 1.35  10−3) • propanoic acid (Ka = 1.3  10−5) • Benzoic acid (Ka = 6.4  10−5) • Hypochlorous acid (Ka = 3.5  10−8) Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30. Which system works the best?

  31. Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

  32. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

  33. Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

  34. Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

  35. Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

  36. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

  37. The titration of a strong base with a strong acid Calculate the pH at the following points of a titration of 50.0 ml of 0.200 M NaOH with 0.100 M HNO3 • No HNO3 added • 20.0 ml HNO3 added • 50.0 ml HNO3 added

  38. The titration of a strong base with a strong acid Calculate the pH at the following points of a titration of 50.0 ml of 0.200 M NaOH with 0.100 M HNO3 • 100.0 ml HNO3 added • 150.0 ml HNO3 added

  39. Titration of a Weak Acid with a Strong Base • Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. • The pH at the equivalence point will be >7. • Phenolphthalein is commonly used as an indicator in these titrations.

  40. Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

  41. Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

  42. Titration of a Weak Base with a Strong Acid • The pH at the equivalence point in these titrations is < 7. • Methyl red is the indicator of choice.

  43. The titration of a weak acid with a strong base Hydrogen cyanide gas (HCN), a powerful respiratory inihbitor, is highly toxic. It is a very weak acid (Ka = 6.2 x 10-10) when dissolved in water. If a 50.0 ml sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution. • After 8.00 ml NaOH has been added

  44. The titration of a weak acid with a strong base b. At the halfway point of the titration c. At the equivalence point of the titration.

  45. The titration of a weak base with a strong acid If a 100.0 ml sample of 0.050 M NH3 is titrated with 0.100 M HCl, calculate the pH of the solution. • Before the addition of any HCl • After 10.0 ml NaOH has been added

  46. The titration of a weak base with a strong acid b. At the halfway point of the titration c. At the equivalence point of the titration.

  47. Calculating Ka A chemist has synthesized a monoprotic weak acid & wants to determine its Ka value. To do so, the chemist dissolves 2.00 mmol of the solid acid in 100.0 ml water & titrates the resulting solution with 0.0500 M NaOH. After 20.0 ml NaOH has been added, the pH is 6.00. What is the Ka value for the acid.

  48. Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.

  49. BaSO4(s) Ba2+(aq) + SO42−(aq) Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

  50. Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42−] where the equilibrium constant, Ksp, is called the solubility product. Solubility is often expressed as grams of solute per liter of of solution (g/L). Molar Solubility is the number of moles per liter that will dissolve (mol/L).

More Related