Question

As a technician in a large pharmaceutical research firm, you need to produce 450. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.91. 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.)

Answer 1

Sol :-

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

pH = pKa + log ( Base / Acid )

Here, pH is given as 6.91

pKa is 7.21

Acid is KH2PO4 and base is K2PO4

After substituting the values, we get

6.91 = 7.21 + log ( K2HPO4 / KH2PO4)

-0.30 = log ( K2HPO4 / KH2PO4)

10^-0.30 = ( K2HPO4 / KH2PO4)

( K2HPO4 / KH2PO4) = 0.5012

In terms of moles , above equation can be expressed as

moles of K₂HPO₄ = 0.5012 * moles of KH₂PO₄ ..............................(1)

The volume required is 450 mL which is 0.45 L

Let's assume volume of KH2PO4 as x L

Therefore volume of K2HPO4 would be 0.45 - x L

Molarity = moles / L

Moles = Molarity * L

Molarity of both the solutions is 1 M

Therefore, moles of KH2PO4 = x mol

Moles of K2HPO4 = 0.45 - x

Let's plug these moles in equation 1

moles of K2HPO4 = 0.5012 * moles of KH2PO4

0.45 - x = 0.5012 * x

1.5012 x = 0.45

x = 0.300 L = 300 mL

So,

300 mL of KH₂PO₄ is required to prepare the given buffer