Question

As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 M potassium phosphate buffer solution of pH = 6.97. 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? Express your answer to three significant digits with the appropriate units.

Answer 1

Answer:

You have to add 285.642 mL of KH2PO4

Explanation:

Step 1: What is a buffer

A buffer is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Buffers work by reacting with any added acid or base to control the pH.

pH of a buffer = pKa + log [conj base] / [weak acid]

Step 2: The ph of the buffer

6.97 = 7.21 =log [K2HPO4] / [KH2PO4]

-0.24 = log [K2HPO4] / [KH2PO4]

10 ^ -0.24 = [K2HPO4] / [KH2PO4] = 0.5754

0.5754 = (0.450 - x) / x

0.5754x = 0.450 - x

1.5754x = 0.450 L

x = 0.28564 = volume of KH2PO4

0.450 - 0.28564 =  0.16436 L K2HPO4

You have to add 285.642 mL of KH2PO4