Question

How many grams each of solid KH2PO4 (MW= 136g/mol) and solid K2HPO4 (MW=174g/mol) would you use to prepare 2L of 0.3 M phosphate buffer (phosphoric acid) at pH 6.8?

Answer 1

To prepare a 2L 0.3 M phosphate buffer at pH 6.8, you can use the Henderson-Hasselbalch equation to determine the ratio of HPO42- and H2PO4-. Then, you can calculate the amounts in grams of KH2PO4 and K2HPO4 required.

Step-by-step explanation:

To prepare a 2L 0.3 M phosphate buffer at pH 6.8, you would need to determine the amounts of KH2PO4 and K2HPO4 required. The pH of a phosphate buffer is determined by the ratio of the concentrations of H2PO4- and HPO42-. To calculate the required amounts, you can use the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation for a phosphate buffer is:

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

In this case, the pKa of phosphoric acid is around 2.1. By substituting the given pH of 6.8 and rearranging the equation, you can solve for the ratio of [HPO42-]/[H2PO4-]. Once you have the ratio, you can use it to calculate the amounts in grams of KH2PO4 and K2HPO4 required in order to make a 2L solution.