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A common buffer used in biochemistry is a pH 7.4 phosphate buffer. Such a buffer costs...

A common buffer used in biochemistry is a pH 7.4 phosphate buffer. Such a buffer costs $53.00 for 100.00 mL from a common chemical supply house. Therefore, it is useful to know how to prepare such a buffer yourself! Imagine that you have available 0.1 M KH2PO4 and solid K2HPO4. Describe in detail how you would prepare 100.00 mL of a pH 7.4 phosphate buffer using these two reagents. **Hint: You need to identify what is your acid and what is your conjugate base to solve this problem.

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Expert Solution

Consider the ionization of phosphoric acid, H3PO4.

H3PO4 (aq) ---------> H+ (aq) + H2PO4- (aq); pKa1 = 2.15

H2PO4- (aq) --------> H+ (aq) + HPO42- (aq); pKa2 = 7.20

HPO42- (aq) --------> H+ (aq) + PO43- (aq); pKa3 = 12.35

Since we wish to prepare a buffer at pH 7.40, we shall choose the combination of KH2PO4 and K2HPO4. Use the Henderson-Hasslebach equation to get the ratio of the concentrations as

pH = pKa + log [K2HPO4]/[KH2PO4]

====> 7.40 = 7.20 + log [K2HPO4]/[KH2PO4]

====> log [K2HPO4]/[KH2PO4] = 0.20

====> [K2HPO4]/[KH2PO4] = antilog (0.20) = 1.5849

====> [K2HPO4] = 1.5849*[KH2PO4] ……(1)

The total phosphate concentration is given as 0.1 M; therefore,

[K2HPO4] + [KH2PO4] = 0.10 M

====> 1.5849*[KH2PO4] + [KH2PO4] = 0.10 M

====> 2.5849*[KH2PO4] = 0.10 M

====> [KH2PO4] = (0.10 M)/(2.5849) = 0.03868 M

Therefore, [K2HPO4] = 1.5849*[KH2PO4] = 1.5849*(0.03868 M) = 0.06130 M.

K2HPO4 and KH2PO4 are both available as solids; hence, we can prepare the buffer by directly weighing out the solids and dissolving in 100.0 mL water. Since the volume of the solution is 100.0 mL, we can find the number of moles of each.

Moles of K2HPO4 in 100.0 mL buffer = (100.0 mL)*(1 L/1000 mL)*(0.06130 M)*(1 mol/L/1 M) = 0.00613 mole.

Moles of KH2PO4 in 100.0 mL buffer = (100.0 mL)*(1 L/1000 mL)*(0.03868 M)*(1 mol/L/1 M) = 0.003868 mole ≈ 0.00387 mole.

Molar mass of K2HPO4 = 174.2 g/mol; mass of K2HPO4 = (0.00613 mole)*(174.2 g/mol) = 1.0678 g (ans).

Molar mass of KH2PO4 = 136.086 g/mol; mass of KH2PO4 = (0.00387 mole)*(136.086 g/mol) = 0.5266 g (ans).

Weigh out 0.5266 g KH2PO4 and 1.0678 g K2HPO4 in a 100.0 mL volumetric flask and fill upto the mark with DI water to prepare the buffer.

queen_honey_blossom answered 2 years ago
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