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Buffers

Buffers. Buffers are composed of a weak acid and its salt or a weak base and its salt. HA < ==> A - + H + This weak ionization has HA available to react with added base or A - to react with acid which leaves H + relatively constant in either case. Buffers.

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Buffers

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  1. Buffers Buffers are composed of a weak acid and its salt or a weak base and its salt. HA <==> A- + H+ This weak ionization has HA available to react with added base or A- to react with acid which leaves H+ relatively constant in either case.

  2. Buffers Examples of preparation and use of buffers. Example 1: Using Glycine Example 2: Using phosphate Example 3: Resistance to acid or base Example 4: Glucose-6-phosphate

  3. Example 1, Glycine Buffer Prepare 500 ml of 0.2 M glycine buffer at pH 9.2 from crystalline zwitterionic glycine, +NH3CH2COO-, and 1 M NaOH. Glycine pKas are 2.4 and 9.8 and it has a molecular weight of 75 gm/mol.

  4. Titration curve for Ala

  5. Equilibrium involved Since the second pKa is within one pH unit of the desired buffer pH (9.2), the ionization involved in this buffer is: +NH3CH2COO- ----> NH2CH2COO- + H+ conj. acid (HA) conj. base (A-) pKa 9.8

  6. Total amount of Glycine needed Total mmol glycine needed: (500 ml)(0.2 mmol/ml) = 100 mmol Total grams of glycine needed: (75 mg/mmol)(100 mmol) = 7500 mg = 7.5 gram (Note: this is all HA) If the mmol of each HA and A- are known then one can determine the volume of 1 M NaOH needed to convert part of the initial 100 mmol of glycine (HA) to its conj. base.

  7. Amounts of HA and A- needed The amounts of the conj. acid and conj. base needed are obtained using the Henderson-Hasselbalch equation. The Henderson-Hasselbalch Equation: pH = pKa + log (A-) / (HA) 9.2 = 9.8 + log (A-) / (HA) Solving this gives: (A-)/(HA) = 0.25

  8. Amounts of HA and A- Since the ratio of (A-)/(HA) = 0.25 The fraction of total glycine as: A- = (0.25/1.25) = 0.20 and HA = (1.0/1.25) = 0.80 So, the mmol of A- = 0.2 (100mmol) = 20 mmol and the mmol of HA = 0.80 (100mmol) = 80 mmol

  9. Amount of NaOH needed Therefore, 20 mmol of 1 M NaOH are needed to convert 20 mmol of HA to A-. Volume of base needed is: mmol = (ml)*(M) 20 mmol = ml base (1 mmol/ml) ml base = 20 ml

  10. Example 1Buffer Preparation To prepare this buffer one would weigh out 7.5 gm of glycine, transfer it to a 500 ml volumetric flask and dissolve it in some water. Then add 20 ml of 1 M NaOH, fill to the mark with water and mix.

  11. Example 2, Phosphate Buffer Prepare 250 ml of 0.2 M phosphate buffer at pH 6.93. Available are 2.8 M H3PO4 (phosphoric acid) and crystalline Na2HPO4 (MW = 142 gm/mol). Phosphate pKas are 2.1, 6.8 and 12.3.

  12. Equilibrium involved Which phosphate equilibrium is involved here? Since the second pKa is within one pH unit of the desired buffer pH (6.93), the ionization involved in this buffer is: H2PO4-----> HPO4=+ H+ pKa 6.8 conj. acid (HA) conj. base (A-)

  13. Total amount of Phosphate needed Total mmol phosphate needed: (250 ml)(0.2 mmol/ml) = 50 mmol (Which is the sum of H2PO4- and HPO4=) If the mmol of each HA and A- are known then one can determine the volume of 2.8 M H3PO4 needed and the weight of Na2HPO4 needed to prepare the buffer. Use the Henderson-Hasselbalch equation to do this.

  14. Amounts of HA and A- needed pH = pKa + log (A-) / (HA) 6.93 = 6.8 + log (A-) / (HA) and solving this gives: (A-) /(HA) = 1.349 The fraction of A- is: (1.349/2.349 = 0.574 and the fraction as HA is: (1.0/2.349) = 0.426 So, the mmol of A- = 0.574 (50mmol) = 28.7mmol The mmol of HA = 0.426 (50mmol) = 21.3 mmol

  15. Producing HA A- (HPO4=) is available as Na2HPO4 but HA (H2PO4-) must be made by the equation below: H3PO4 + Na2HPO4 -----> 2 NaH2PO4 This is due to the fact that H3PO4 is the only other starting material for the buffer.

  16. Amount of Na2HPO4needed The 21.3 mmol of HA needed are obtained by combining 10.65 mmol of H3PO4 and 10.65 mmol of Na2HPO4. H3PO4 + Na2HPO4 -----> 2 NaH2PO4 10.65 10.65 21.3 In addition, another 28.7 mmol of Na2HPO4 are needed to provide A-. Total mmol Na2HPO4 required = 10.65 + 28.7 = 39.35 mmol.

  17. Amount of H3PO4needed Total grams of Na2HPO4 required = 39.35 mmol (142 mg/mmol) = 5587.7 mg = 5.588 gm The volume of H3PO4 needed: 10.65 mmol = ml H3PO4 (2.8 mmol/ml) so the ml of H3PO4 = 3.80 ml.

  18. Example 2Buffer Preparation To prepare this buffer one would weigh out 5.588 gm of Na2HPO4, transfer it to a 250 ml volumetric flask and dissolve it in some water. Then add 3.80 ml of 2.8 M H3PO4, fill to the mark with water and mix.

  19. Example 3, Effect of NaOH on the Glycine Buffer In example 1, 500 ml of pH 9.2 glycine buffer was prepared. If 5 ml of 1 M NaOH are added to this buffer, what will be the new pH ? How much of a change in pH does this represent ?

  20. Example 3, Effect of NaOH on the Glycine Buffer Five ml of 1 M NaOH = (5 ml)(1M) = 5 mmol NaOH NaOH + +NH3CH2COO- ----> NH2CH2COO- + H2O Initial: (HA) = 80 mmol (A-) = 20 mmol Final: (HA) = 75 mmol (A-) = 25 mmol pH = pKa + log (A-) / (HA) pH = 9.8 + log (25/505) / (75/505) = 9.8 - 0.5 pH = 9.3 (A change of 0.1 pH unit.)

  21. Example 4 Glucose-6-phosphate: Glucose-6-phosphate exhibits two ionizations: Glucose-6-OPO3H2 < == > Glucose-6-OPO3H- + H+ pK1 = 0.94 Glucose-6-OPO3H- < == > Glucose-6-OPO3= + H+ pK2 = 6.11 1. Which equilibrium predominates at pH 7.1 ? The second equilibrium (pK2).

  22. Example 4 Glucose-6-phosphate: 2. What fraction of each of the forms in the second equilibrium exist at pH 7.1 ? pH = pKa + log (A-) / (HA) 7.1 = 6.11 + log (A-) / (HA) Solving this gives: (A-)/(HA) = 9.75 So, the fraction of A- = 9.75/10.75 = 0.91 and the fraction of HA = 1/10.75 = 0.09.

  23. End of Buffers

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