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Applications of Aqueous Equilibria Electrolyte Effect

Chapter 8. Applications of Aqueous Equilibria Electrolyte Effect. -. -. -. -. -. -. -. -. -. -. -. -. +. +. +. +. +. +. +. +. +. +. +. +. Ideal vs. real ionic solutions. Ideal solution: Cations and anions do not interact. Real solution: Electrostatic interactions. .

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Applications of Aqueous Equilibria Electrolyte Effect

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  1. Chapter 8 Applications of Aqueous Equilibria Electrolyte Effect

  2. - - - - - - - - - - - - + + + + + + + + + + + + Ideal vs. real ionic solutions • Ideal solution: Cations and anions do not interact. • Real solution: Electrostatic interactions. ideal Shaded region: Solvation cage real

  3. Ionic Strength The effect of added electrolyte on equilibria is independent of the chemical nature of electrolyte, but depends on a property of the solution called the ionic strength. [A], [B], [C]…: molar species concentration species ZA, ZB, ZC..: ionic charge

  4. Molality Seawater (SW)Lake Water (LW) Na+ 0.49 0.2 x 10-3 Mg2+ 0.053 0.14 x 10‑3 Ca2+ 0.010 0.22 x 10-3 K+ 0.010 0.03 x 10-3 Cl- 0.57 0.09 x 10-3 SO42- 0.028 0.102 x 10-3 HCO3- 0.002 0.816 x 10-3 ISW = 1/2 (mNa × 12 + mMg× 22 + mCa × 22 + mK × 12 + mCl × 12 + mSO4 × 22 + mHCO3× 12) = 0.72 mol kg-1 ILW = 0.0015 = 1.5 × 10-3 mol kg‑1

  5. Salt Effect • The consequence of this effect is a decrease in overall attraction between barium and sulfate ions and an increase in solubility. BaSO4 dissolved in NaCl(aq) Na+ Na+ Cl- Cl- SO4-2 Na+ Cl- Ba+2 Na+ Cl- Cl- Na+ Na+ Cl- Small net negative charge Small net positive charge

  6. The effect of added NaCl to increase the size of the Ksp for BaSO4 • At 0 M NaCl, Ksp has a value of 1.1 x 10-10. • At 1 x 10-3 M NaCl, Ksp has a value of approximately 1.8 x 10-10. • At 1 x 10-2 M NaCl, Ksp has a value of approximately 2.85 x 10-10.

  7. The effect of added NaCl to increase the size of the Ka for acetic acid • At 0 M NaCl, Ka has a value of 1.75 x 10-5. • At 1 x 10-2 M NaCl, Ka has a value of approximately 2.1 x 10-5. • At 1 x 10-1 M NaCl, Ka has a value of approximately 2.7 x 10-5.

  8. Activity Coefficients • The activity, or effective concentration, of species X depends on the ionic strength of the medium and is defined as: aX=gx[X] aX:activity gx: activity coefficient The aX & gx vary with ionic strength. ex. XmYn precipitate

  9. Properties of Activity Coefficients • In very dilute solutions, where the ionic strength is minimal, the effectiveness becomes constant. • gX→1 and aX→[X], then K’sp→Ksp • As ionic strength increases, an ion loses some of its effectiveness and its activity coefficient decreases. • At high ionic strength, m > 0.1m, g often increases and may even becomes greater than unity.

  10. Properties of Activity Coefficients • In solutions that are not too concentrated, the g for a given species is independent of the nature of the electrolyte and dependent only on the ionic strength. • The activity coefficient of an unchanged molecule is approximately unity, regardless of ionic strength.

  11. gX=activity coefficient ZX=charge m=ionic strength aX=effective diameter of the hydrated ion X in nm Debye-Hückel Equation

  12. General Rules for Activity Coefficients • i 1 as m 0 i.e. activity = concentration at infinite dilution • iis decreased as mincreased i.e., the free ion activity coefficient decreases with ionic strength • 2+ < +, the activity corrections decrease with increasing charge

  13. Solutions of Acids or Bases Containing a Common Ion • Common ion effect: The shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction.

  14. 1 0 0 0 1 1 1 0 0 1-x x x

  15. Buffered Solutions • A buffered solution is one that resists a change in pH when either hydroxide ions or protons are added. • A buffered solution may contain a weak acid and its salt or a weak base and its salt. (HF+NaF, NH3+NH4Cl)

  16. The equilibrium concentration of H+ and the pH are determined by the ratio [HA]/[A-]. How Do the H+/OH- Ions Work in Buffered Solutions

  17. When OH- are added, HA is converted to A- , causing the ratio [HA]/[A-] to decrease. If the amount of HA and A- originally present are very large compared with the amount OH- added, the change in [HA]/[A-] ratio is small. The Effect of Added Bases

  18. When protons are added to a buffered solution, the added H+ ions react with A- to form the weak acid. If [HA] and [A-] are large compared with the [H+] added, only a slight change in the pH occurs. The Effect of Added Acids

  19. Henderson-Hasselbalch Equation

  20. Exact Treatment of Buffered Solutions charge balance

  21. material balance

  22. [HCOOH]=0.4 M, [HCOONa]=1.0 M Ka=1.8x10-4

  23. Kb=1.75X10-5 [NH3]=0.2 M, [NH4Cl]=0.3 M

  24. Unique Properties of Buffer Solution • The Effect of Dilution - The pH of a buffer solution remains essentially independent of dilution. • The Effect of Added Acids and Bases -A buffer solution resists pH change after addition of small amounts of strong acids or bases • Buffer Capacity

  25. The Effect of Dilution

  26. Buffer Capacity • The number of moles of a strong acid or a strong base that causes 1L of buffer to undergo a 1 unit change in pH. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].

  27. Calculate Buffer Capacity 1. Calculate the buffer capacity (B) for a mixture of 0.01 moles of acetic acid and 0.03 moles of NaOAc in 100 mL of total solution. 2. Calculate the buffer capacity (B) for a mixture of 0.03 moles of acetic acid and 0.03 moles of NaOAc in 100 mL of total solution.

  28. 1. B=2.303× (0.1×0.3)/(0.1+0.3) =0.172 mol/L per pH 2. B=2.303×(0.3×0.3)/(0.3+0.3) =0.345 mol/L per pH Solution

  29. Prepare an Optimum Buffered Solution • The optimum buffering will occur when [HA] is equal to [A-]. Reasons: (1) The pKa of the weak acid selected for the buffer should be as close as possible to the desired pH. (2) It can provide best buffer capacity.

  30. Titrations and pH Curves • The equivalence point is defined by stoichiometry, no by the pH • The pH value at equivalence point is affected by the acid strength or base strength. • The strength of a weak acid or weak base have significantly effect on the shape of pH curves.

  31. Titrations of Weak Acid with Strong Bases Titration Curve Calculations • A stoichiometry problem: The reaction of hydroxide ion with the weak acid is assumed to run completion. • An equilibrium problem: The position of the weak acid equilibrium is determined, and the pH is calculated.

  32. The titration of 50ml of 0.1M acetic acid (Ka=1.8×10-5) with 0.1M NaOH 1. No NaOH has been added. 2. 10mlof 0.1M NaOH has been added. 3. 25ml of 0.1M NaOH has been added. 4. 40ml of 0.1M NaOH has been added. 5. 50ml of 0.1M NaOH has been added. 6. 60ml of 0.1M NaOH has been added. 7. 75ml of 0.1M NaOH has been added.

  33. Titration of Polyprotic Acids

  34. Titration of Triprotic Acids If the above reaction is the only important reaction involving these species, then [H3A]=[HA-2]

  35. Titration of Polyprotic Acids Amphiprotic Salts

  36. Kb2>>Ka2, the solution is basic

  37. Titration of Diprotic Acids • Point A: Treat the system as if it contained a single monoprotic acid. (Ka1>>Ka2) • Region B: Treat the system as if a simple buffer solution consisting of the H2A and NaHA. • Point C: Use equation • Region D: The second buffer solution consisting of the HA- and Na2A. • Point E: Use equation • Region F: The excess NaOH dominated the pH.

  38. Question • A 100.0-mL sample of the weak acid H3A (0.100 M) is titrated with 0.100 M NaOH. What are the major species after 40.0 mL of 0.100 M NaOH is added in the titration (water is always assumed to be a major species)? • H3A • H2A–, HA2– • H3A, H2A– • HA2– • H2A–

  39. Answer • c) H3A, H2A– • Section 8.7, Titration of Polyprotic Acids • The concentrations are equal, so the first equivalence point is at 100.0 mL of NaOH.

  40. Question • A 100.0-mL sample of the weak acid H3A (0.100 M) is titrated with 0.100 M NaOH. What are the major species after 170.0 mL of 0.100 M NaOH is added in the titration (water is always assumed to be a major species)? • H3A • H2A–, HA2– • H3A, H2A– • HA2– • H2A–

  41. Answer • b) H2A–, HA2– • Section 8.7, Titration of Polyprotic Acids • The concentrations are equal, so the second equivalence point is at 200.0 mL of NaOH.

  42. Determine the Equivalence Point of an Acid-Base Titration • Use a pH meter to monitor the pH and then plot a titration curve. • Use an acid-base indicator, which marks the endpoint of a titration by changing color.

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