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2-4-2011

2-4-2011. common ion. The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with one of the products. The presence of a common ion suppresses the ionization of a weak acid or a weak base.

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2-4-2011

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  1. 2-4-2011

  2. common ion The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with one of the products. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid). CH3COONa (s) Na+ (aq) + CH3COO- (aq) CH3COOH (aq) D H+ (aq) + CH3COO- (aq)

  3. CH3COOH (aq) + H2O (l)H3O+ (aq) + CH3COO- (aq) Adding the conjugate base(a stress!) to the equilibrium system of an acid dissociation shows the common-ion effect, where the addition of a common ion causes the equilibrium to shift. This is an example of Le Chatalier’s Principle. Addition of the weak base to the acid dissociation

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  5. Buffers • This is a solution that resists changes in pH upon the addition of small amounts of either acid or base.. • Very important in nature (blood = pH 7.4) • Prepared by adding a weak acid or base to the salt of its conjugate • Examples: NH4Cl / NH3, H2PO4-/HPO42-, HOAc/OAc-, HCO3-/CO32-, etc • We can choose the right components to form a buffer at any pH.

  6. A buffer solution is a solution containing: • A weak acid or a weak base and • The salt of the weak acid or weak base • Both must be present! Consider an equal molar mixture of CH3COOH and CH3COONa Add strong acid H+ (aq) + CH3COO- (aq) g CH3COOH (aq) Add strong base OH-(aq) + CH3COOH(aq) g CH3COO-(aq) + H2O(l)

  7. HCl H+ + Cl- HCl + CH3COO- CH3COOH + Cl-

  8. Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (a) HF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is its conjugate acid buffer solution

  9. Example: Water – H2O If add 1 drop of 10 M HCl to 10 mL of water the pH goes from 7 to 1 If add 1 drop of 10 M NaOH to 10 mL of water the pH goes from 7 to 13 Water is not a buffer; the ΔpH was 6 in both cases; or the Δ[H+] = 106 M change!

  10. Example: 1 M Acetic Acid and 1 M Sodium Acetate - HA & NaA If add 1 drop 10 M HCl to 10 mL of the above, the pH goes from 4.8 to 4.7 If add 1 drop 10 M NaOH to 10 mL of the above, the pH goes from 4.8 to 4.9 The ΔpH = 0.1 in both cases, or the Δ[H+] = 10-1 M change!

  11. Why a buffer resists pH changes? Assume acetic acid / acetate buffer: HOAc D H+ + OAc- OAc- + H2O D HOAc + OH- Acetate from buffer will make the 1st equilibrium proceed to left. Therefore, HOAc will not dissociate to any appreciable extent. At the same time, HOAc from buffer will drive acetate dissociation in the reverse direction> Therefore, both equilibria will not proceed in the forward direction to any appreciable extent.

  12. If more acid (H3O+) or base (OH-) is added to the system, the system has enough of the original acid and conjugate base molecules in the solution to react with the added acid or base, and so the new equilibrium mixture will be very close in composition to the original equilibrium mixture.

  13. Equation Derivation HA D H+ + A- Ka = [H+] [A-] [HA] [H+]=Ka[HA] [ A-] Adding acid or base causes little change in ratio because of equilibrium

  14. NaA (s) Na+(aq) + A-(aq) HA (aq) H+(aq) + A-(aq) [H+][A-] Ka [HA] Ka = [H+] = [HA] [A-] -log [H+] = -log Ka - log [HA] [A-] [A-] -log [H+] = -log Ka + log [HA] [HA] [A-] pH = pKa + log [conjugate base] pH = pKa + log [acid] Consider mixture of salt NaA and weak acid HA. pKa = -log Ka

  15. Henderson-Hasslbalch Equation Put previous equation in terms of pH pH = pKa + log {[base]/[acid]} Half way to the stoichiometric equivalence point, [HA] = [A-] What happens to equation pKa = pH

  16. [A-] [HA] pH = pKa + log pKa = -log Ka For a solution of weak acid and its conjugate base to act as a buffer, the ratio between the acid and its conjugate base should be between 0.1 and 10, otherwise it will not behave like a buffer. Under these conditions we have: pH = pKa+ 1 This means that a buffer solution is effective within two pH units only.

  17. A buffer works best when [HA]/[A-] = 1:1 = 1.00 When [HA] = [A-] = small • The buffer is poor when the ratio of HA to A- is far from 1:1. • When generating a buffer, choose a buffer with a pKa within ±1.0 of the desired acidity. • Which acid will buffer best at pH=4? HF (Ka = 2x10-4) or HCN (Ka = 5x10-10)

  18. 2) A quick lab way to get pKa is to measure pH of a 1:1 mix of M HA & A-. pH = pKa + log [A- ]/[HA] pH = pKa + log 1/1 pH = pKa + 0 pH = pKa

  19. 3) Note: Can use either moles or M in equilibrium constant equation; OK since volume cancels out. 4) A weak base and its salt such as NH3 & NH4Cl (NH4+) is also a buffer.

  20. 5) Buffer Problem Solving Method when adding strong H+ or OH- : a) allow the added strong acid or base to react completely with appropriate component of the buffer b) calculate moles (or M) of HA & A- left over after the reaction c) plug into the equilibrium constant Equation & solve for pH

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