Question

As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.94. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.) I do not know how to do this could i get the base to acid ratio on how i would do this??

Answer 1

The pH of a buffer can be calculated using Henderson equation which is given below:

pH = pK_a + log ( Base / Acid )

Here, pH is given as 6.94

pK_a is 7.21

Acid is KH₂PO₄ and base is K₂HPO₄

After substituting the values, we get

6.94 = 7.21 + log ( K₂HPO₄ / KH₂PO₄ )

-0.27 = log ( K₂HPO₄ / KH₂PO₄ )

10^-0.27 = ( K₂HPO₄ / KH₂PO₄ )

( K₂HPO₄ / KH₂PO₄ ​) = 0.537

In terms of moles, above equation can be expressed as:

Moles of K₂HPO₄ = 0.537*Moles of KH₂PO₄ ...................... Equation (1)

The volume required is 250 mL which is 0.25 L

Let's assume volume of KH₂PO₄ as x L

Therefore volume of K₂HPO₄ would be (0.25 - x) L

Molarity = Moles / Volume

Moles = Molarity * Volume

Molarity of both the solutions is 1 M

Therefore, Moles of KH₂PO₄ = x mol

And, Moles of K₂HPO₄ = 0.25 - x

Let's plug these moles in equation 1:

Moles of K₂HPO₄ = 0.537*Moles of KH₂PO₄

0.25 - x = 0.537*x

1.537x = 0.25

x = 0.16265 L = 162.65 mL

Therefore, 162.65 mL of KH₂PO₄ is required to prepare the given buffer.