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 = 7.05. 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

We will need 147.772 mL of KH2PO4 to make this solution

Step-by-step explanation:

For this case we can give the following equation:

H2PO4 - ⇄ H+ + HPO42-

With following pH- equation:

pH = pKa + log [HPO42-]/[H2PO4-]

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

-0.16 = log [HPO42-]/[H2PO4-]

10^-0.16 = [HPO42-]/[H2PO4-]

0.6918 = [HPO42-]/[H2PO4-]

Let's say the volume of HPO42-= x then the volume of H2PO4- will be 250 mL - x

Since both have a concentration of 1M = 1 mol /L

If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]

0.6918 = x / (250 - x)

0.6918*250 - 0.6918x = x

172.95 = 1.6918x

x = 102.228 mL

The volume of HPO42- = 102.228 mL

Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL

To control this we can plug this in the pH equation

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

7.05= 7.21 + log (102.228 / 147.772) = 7.05

We will need 147.772 mL of KH2PO4 to make this solution