Question

Buffer preparation

How can I prepare 100mL potassium phosphate buffer (10mM containing 0.9% NaCl)? Procedure needed.

Answer 1

You haven’t mentioned the pH of the buffer; however, I will show you a particular example. Suppose we wish to prepare a phosphate buffer at pH 7.4, which is close to the physiological pH of 7.0.

Consider the ionization of phosphoric acid (H₃PO₄) as below.

H₃PO₄ (aq) <=====> H⁺ (aq) + H₂PO₄^- (aq); pK_a1 = 2.15

H₂PO₄^- (aq) <=====> H⁺ (aq) + HPO₄^2- (aq); pK_a2 = 7.20

HPO₄^2- (aq) <=====> H⁺ (aq) + PO₄^3- (aq); pK_a3 = 12.35

Since we wish to prepare a phosphate buffer at pH 7.40, we shall choose potassium dihydrogen phosphate and dipotassium hydrogen phosphate as the components of our buffer. Use the Henderson-Hasslebach equation to find out the ratio of the weak acid (potassium dihydrogen phosphate, KH₂PO₄) and the conjugate base (dipotassium hydrogen phosphate, K₂HPO₄).

pH = pK_a2 + log [K₂HPO₄]/[KH₂PO₄]

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

===> 0.20 = log [K₂HPO₄]/[KH₂PO₄]

===> [K₂HPO₄]/[KH₂PO₄] = antilog (0.20) = 1.5848

===> [K₂HPO₄] = 1.5848*[KH₂PO₄] …..(1)

The Henderson-Hasslebach equation gives the ration of the weak acid and the conjugate base required to form the buffer. Again, it is given that,

[K₂HPO₄] + [KH₂PO₄] = 10 mM

===> 1.5848*[KH₂PO₄] + [KH₂PO₄] = 10 mM

===> 2.5848*[KH₂PO₄] = 10 mM

===> [KH₂PO₄] = (10 mM)/(2.5848) = 3.8688 mM.

Therefore, [K₂HPO₄] = 1.5848*[KH₂PO₄] = 1.5848*(3.8688 mM) = 6.1312 mM.

Thus, we have obtained the concentrations of the weak acid and the conjugate base required to prepare the buffer. The volume of the buffer solution is given; the molar masses of KH₂PO₄ and K₂HPO₄ are known. Thus, we can obtain the masses of solid KH₂PO₄ and K₂HPO₄ required preparing the buffer.

Molar mass of KH₂PO₄ = 136.086 g/mol; molar mass of K₂HPO₄ = 174.2 g/mol.

Mass of KH₂PO₄ required = (100 mL)*(1 L/1000 mL)*(3.8688 mM)*(1 M/1000 mM)*(136.086 g/mol)*(1000 mg/1 g) = 52.6489 mg ≈ 52.65 mg.

Mass of K₂HPO₄ required = (100 mL)*(1 L/1000 mL)*(6.1312 mM)*(1 M/1000 mM)*(174.2 g/mol)*(1000 mg/1 g) = 106.80 mg.

Therefore, dissolve 52.65 mg KH₂PO₄ and 106.80 mg K₂HPO₄ with deionized water in a 100 mL volumetric flask and make upto the mark with deionized water. The buffer solution is 0.9% in NaCl which simply means that 100 mL of the solution will contain 0.9 g NaCl. Therefore, add 0.9 g NaCl to the prepared buffer solution and obtain the desired buffer solution (ans).