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Acid – Base Equilibria

Acid – Base Equilibria. Buffer Solutions: Question: Was the ICE Problem set up needed? Answer: No. The assumption of x << [HA], [A - ] is valid for all “traditional” buffers Traditional Buffer Weak acid (3 < pK a < 11) Ratio of weak acid to conjugate base in range 0.1 to 10

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Acid – Base Equilibria

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  1. Acid – Base Equilibria Buffer Solutions: Question: Was the ICE Problem set up needed? Answer: No. The assumption of x << [HA], [A-] is valid for all “traditional” buffers Traditional Buffer Weak acid (3 < pKa < 11) Ratio of weak acid to conjugate base in range 0.1 to 10 mM concentration range

  2. Acid – Base Equilibria Buffer Solutions: Since ICE not needed, can just use Ka equation Ka = [H+][A-]/[HA] = [H+][A-]o/[HA]o (always valid) (valid for traditional buffer) But log version more common pH = pKa + log([A-]/[HA]) Also known as Henderson-Hasselbalch Equation

  3. Acid – Base Equilibria Buffer Solutions: Ways to make buffer solution: Mix weak acid and conjugate base Add strong base to weak acid (weak acid must be in excess) – this converts some of the weak acid to its conjugate base Add strong acid to weak base (weak base must be in excess) – this converts some of weak base to its conjugate acid

  4. Acid – Base Equilibria Example Problems: How many moles of hydroxyl ammonium chloride (HONH3+Cl-) needs to be added to 500 mL of 0.020 M HONH2 to obtain a buffer solution with a pH of 6.20? The pKa for HONH3+ is 5.96. What is the pH of a solution made from mixing 400 mL of 0.018 M CH3CO2H (pKa = 4.75) with 100 mL of 0.024 M NaOH?

  5. Acid – Base Equilibria Last Example Problem: How many mL of 0.0500 M NaOH should be added to 50.0 mL of 0.00850 M methyl ammonium chloride (CH3NH3+Cl-) in order to make a buffer with a pH of 10.80? The pKa for CH3NH3+ is 10.645.

  6. Acid – Base Equilibria Additional Questions: Which of the following will result in a traditional buffer? 0.00100 M HNO3 + 0.00200 M NaNO3 0.010 M NH4Cl + 0.003 M NH3 1.0 x 10-5 M CH3CO2H + 0.010 M NaCH3CO2 0.0020 M HCl + 0.010 M CH3CO2H 0.0020 M HCl + 0.010 M NaCH3CO2

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