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Buffer This

Buffer This. The Common Ion Effect and Buffers. There are two common kinds of buffer solutions: Solutions made from a weak acid plus a soluble ionic salt of the weak acid. Solutions made from a weak base plus a soluble ionic salt of the weak base

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Buffer This

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  1. Buffer This

  2. The Common Ion Effect and Buffers • There are two common kinds of buffer solutions: • Solutions made from a weak acid plus a soluble ionic salt of the weak acid. • Solutions made from a weak base plus a soluble ionic salt of the weak base • Solutions made of weak acids plus a soluble ionic salt of the weak acid • One example of this type of buffer system is: • The weak acid - acetic acid CH3COOH • The soluble ionic salt - sodium acetate NaCH3COO

  3. Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases • One example of the type I of buffer system is: • The weak acid - acetic acid CH3COOH • The soluble ionic salt - sodium acetate NaCH3COO CH3COOH H3O+ + CH3COO- ~100% NaCH3COO →Na+ + CH3COO- • This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of H+ and the pH of a solution that is 0.15 M in both acetic acid sodium acetate yields:

  4. Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases • Alternatively you might have noticed: • [base]/[acid] = 1 = 100 • log 100 = 0 • pH = pKa + 0 = pKa= 4.74 • [H+] = Ka = 1.8E-5 • [H+] is ~90 times greater in pure acetic acid than in buffer solution. • Note that the pH of the buffer equals the pKa of the buffering acid.

  5. The general expression for the ionization of a weak monoprotic acid is: The generalized ionization constant expression for a weak acid is: The Common Ion Effect and Buffers

  6. If we solve the expression for [H+], this relationship results: By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to: The Common Ion Effect and Buffers

  7. The Common Ion Effect and Buffers • The relationship developed in the previous slide is valid for buffers containing a weak monoprotic acid and a soluble, ionic salt. • If the salt’s cation is not univalent the relationship changes to:

  8. The Common Ion Effect and Buffers • Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation. The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid.

  9. Henderson-Hasselbalch - Caveats and Advantages • Henderson-Hasselbalch equation is valid for solutions whose concentrations are at least 100 times greater than the value of their Ka’s • The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid. • A special case exists for the Henderson-Hasselbalch equation when [base]/[acid] = some power of 10, regardless of the actual concentrations of the acid and base, where the Henderson-Hasselbalch equation can be interpreted without the need for calculations: • [base]/[acid] = 10x • log 10x = x • in general pH = pKa + x • Examples: • when [base] = [acid], [base]/[acid] = 1 or 100, log 1 = 0, pH = pKa, • (corresponds to the midpoint in the titration of a weak acid or base) • 2. when [base]/[acid] = 10 or 101, log 10 = 1 then pH = pKa + 1 • 3. when [base]/[acid] = .001 or 10-2, log 10 = -2 then pH = pKa -2

  10. Buffer Solutions There are two common kinds of buffer solutions: • Commonly, solutions made from a weak acid plus a soluble ionic salt of the conjugate base of the weak acid. • Less common, solutions made from a weak base plus a soluble ionic salt of the conjugate acid of the weak base. Both of the above may also be prepared by starting with a weak acid (or weak base) and add half as many moles of strong base (acid)

  11. Buffer Solutions: Weak Bases Plus Salts of Their Conjugate Acids • This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of OH- and the pOH of the solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in ammonium nitrate, NH4NO3 yeilds: • One example of the type II of buffer system is: • The weak base – ammonia NH3 • The soluble ionic salt – ammonium nitrate NH4NO3 NH3 NH4 ++OH- NH4NO3→NH4 ++NO3- ~100% • Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

  12. The Common Ion Effect and Buffers • We can derive a general relationship for buffer solutions that contain a weak base plus a salt of a weak base similar to the acid buffer relationship. • The general ionization equation for weak bases is: • Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation.

  13. The Common Ion Effect and Buffers • A comparison of the aqueous ammonia concentration to that of the buffer described above shows the buffering effect. • The [OH-] in aqueous ammonia is 180times greater than in the buffer.

  14. Buffering Action • If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl. • Calculate the pH of the original buffer solution. NH4NO3→NH4 ++NO3- ~100% NH3 NH4 ++OH- • Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

  15. Buffering Action • Next, calculate the concentration of all species after the addition of the gaseous HCl. • The HCl will react with some of the ammonia and change the concentrations of the species. • This is another limiting reactant problem. HCl→H++Cl- ~100% NH3 + H+ NH4 +

  16. Buffering Action • Using the concentrations of the salt and base and the Henderson-Hassselbach equation, the pH can be calculated. • Finally, calculate the change in pH.

  17. Buffering Action • If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH. NaOH →Na++OH- ~100% NH4 + + OH- NH3

  18. Buffering Action • Using the concentrations of the salt and base and the Henderson-Hassselbach equation, the pH can be calculated. • Finally, calculate the change in pH.

  19. Notice that the pH changes only slightly in each case. Buffering Action

  20. Preparation of Buffer Solutions • Calculate the concentration of H+ and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. • Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction. • NaOH and CH3COOH react in a 1:1 mole ratio. • After the two solutions are mixed, Calculate total volume. • The concentrations of the acid and base are: • Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.

  21. Preparation of Buffer Solutions • For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Starting with a solution that is 0.100M in aqueous ammonia prepare 1.00L of a buffer solution that has a pH of 9.15 using ammonium chloride as the source of the soluble ionic salt of the conjugate weak acid. • The Henderson-Hasselbalch equation is used to determine the ratio of the conjugate acid base pair • pOH can be determined from the pH: • pKb can be looked up in a table: • [base] concentration is provided: • Solve for [acid]: • Does this result make sense? H2O ~100% NH4Cl NH4 + + Cl- NH3 NH4++ OH-

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