Buffers

1
Buffers

• Buffers are composed of a weak acid and its salt
  or a weak base and its salt.

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- ----gt 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-Hasselbalc
  h 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)
• 0.20 mmol ml base (1 mmol/ml)
• ml base 20 ml

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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.

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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.

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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- ----gt HPO4 H pKa 6.8
  conj. acid (HA) conj. base (A-)

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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.

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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 -----gt 2 NaH2PO4
• This is due to the fact that H3PO4 is the only
  other starting material for the buffer.

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Amount of Na2HPO4 needed

• The 21.3 mmol of HA needed are obtained by
  combining 10.65 mmol of H3PO4 and 10.65 mmol of
  Na2HPO4.
• H3PO4 Na2HPO4 -----gt 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.

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Amount of H3PO4 needed

• 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.

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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.

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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 ?

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Example 3, Effect of NaOH on the Glycine Buffer

• Five ml of 1 M NaOH (5 ml)(1M) 5 mmol NaOH
• NaOH NH3CH2COO- ----gt 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.)

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Example 4

• Glucose-6-phosphate
• Glucose-6-phosphate exhibits two ionizations
• Glucose-6-OPO3H2 lt gt Glucose-6-OPO3H- H
  pK1 0.94
• Glucose-6-OPO3H- lt gt 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.
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End of Buffers