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What is a buffer?

What is a buffer?. A solution containing a conjugate acid/base pair. A solution of a weak acid and the salt of a weak acid. A solution of a weak base and the salt of a weak base. A solution that resists changes in pH upon the addition of either strong acids or strong bases.

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What is a buffer?

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  1. What is a buffer? A solution containing a conjugate acid/base pair A solution of a weak acid and the salt of a weak acid. A solution of a weak base and the salt of a weak base. A solution that resists changes in pH upon the addition of either strong acids or strong bases.

  2. Example – Acetate buffer A solution is made 0.100M in acetic acid (CH3COOH) and 0.100M in sodium acetate (NaCH3COO). Sodium acetate is completely soluble …. before dissolution 0.100 M 0 M 0 M NaCH3COO  Na+ + CH3COO- After dissolution 0 M 0.001 M 0.001 M Before dissociation0.100 M CH3COOH + H2O H3O+ + CH3COO- After dissociation (0.100 – x) M x M (0.100 + x) M Ka = 1.8 x 10-5 = [H3O+][CH3COO-]/[CH3COOH]

  3. Conjugate base Generic weak acid Log xy = log x + log y The Henderson-Hasselbach Equation derivation HA + H2O H3O+ + A- Ka = [H3O+] [A-]/[HA] pKa = - log Ka = - log ([H3O+] [A-]/[HA]) pKa = - log [H3O+] - log ([A-]/[HA]) pH = pKa + log ([A-]/[HA]) pKa = pH - log ([A-]/[HA]) pH = pKa + log ([base]/[acid])

  4. pH = pKa + log ([base]/[acid]) Try different combinations of [CH3COO-]/[CH3COOH]! 0.10 M acid and 0.10 M base 0.10 M acid & 0.02 M base 0.10 M acid and 0.20 M base etc. Try a different acid/conjugate base combination. HCN/CN- Ka = 4.0 x 10-10 find pKa - insert [ ]s of acid and base into HH equation Try a weak base/conjugate acid combination. NH3/NH4+ Kb = 1.8 x 10-5 find Ka and pKa – insert [ ]s of acid and base into HH equation

  5. General Buffer Concepts • A buffer is most effective when pH = pK. • At this pH [acid] = [conjugate base]. • The buffer is equally effective against added acid or added base. • The buffer loses effectiveness when added acid/base • exceeds the [buffer]. Some Buffers: acetate: acetic acid + sodium acetate pKa = 4.74 phosphate: H2PO4- and HPO42- pKa2 = 7.21 ammonium: NH4Cl + NH3 pKa = 9.26

  6. Qualities of a good buffer solution. 1) The solution contains conjugate weak acid/base pair 2) The pH is within 1 pH unit of the pK for the buffer system. 3) The total [buffer] > [acid] or [base] that will be added. Preparing a buffer solution • Make a solution of a weak acid and the salt of the weak acid until • both the pH and the total [buffer] is the desired value 2) Make a solution of a weak acid and add a strong base until the pH is at the desired value.

  7. Preparing Buffer solutions 1. By adding the conjugate acid/base combination Make a 0.20 M ammonium buffer at pH 9.0? pH = pK + log ([NH3]/NH4+]) 9.00 = 9.26 + log {(0.2 – x)/x} x = 0.13 M = [NH4+] [NH3] = 0.2 – x = 0.07 M • By adding the strong acid a weak base (or salt of weak acid) • or adding strong base to weak acid (or salt of weak base)

  8. pH = pKa + log ([base]/[acid]) Acetate buffer – adding acid or base CH3COOH + H2O H3O+ + CH3COO- What happens if 0.050M HCl is added to a solution containing 0.10 M CH3COOHand0.10 M NaCH3COO? Qualitatively – use LeChatelier’s Principle Quantitatively – Use HH equation Assume all of acid reacts with conjugate base This will increase [acid] and decrease [base] by amount of HCl added

  9. Neutralization Reactions Strong Acid and strong base …… HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) What happens to the pH as a strong base solution is added to a strong acid?

  10. [H+] = 1 x 10-7 M [HCl] = [H+] = (0.01*0.1)/0.11 = 0.009 M pH = - log 0.009) = 2.04 pH = - log [H+] Begin with 0.1 L H2O- titrate with 0.1M HCl

  11. Equivalence point – amount acid added = amount of base. Begin with 0.1 L of 0.1M NaOH- titrate with 0.1M HCl

  12. 0.10 M NH3 – Kb = 1.8 x 10-5 (pKa = 9.26) pH = pKa + log ([NH3]/[NH4+]) Equivalence point – similar to having 0.01 mol of NH4+. Begin with 0.1 L of 0.1M NH3 - titrate with 0.1M HCl

  13. The salt of a weak acid is a base – 0.1M NaCH3COO- pH = pKa + log ([CH3COO-]/[CH3COOH]) Equivalence point – similar to having 0.01 mol of Acetic Acid. Begin with 0.1 L of 0.1M NaCH3COO- titrate with 0.1M HCl

  14. End point of titration neutral solution Strong acid added to strong base 13 7 1

  15. End point of titration pH of weak acid not neutral 9.2 Strong acid added to weak base or conjugate base of weak acid 13 7 Buffering zone Center = pKa of buffer. 1

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