Final answer:
To prepare 40 liters of a 0.02 M phosphate buffer with 1 M K₂HPO₄, you will need 40 milliliters of 1 M K₂HPO₄.
Step-by-step explanation:
To prepare 40 liters of a 0.02 M phosphate buffer with 1 M K₂HPO₄, you will need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is pH = pKa + log([base]/[acid]). In this case, K₂HPO₄ is the base and H₂PO₄- is the acid.
Given that the pKa of H₂PO₄- is 7.21, let's calculate the moles of K₂HPO₄ needed for the buffer. Since pH = 7.21 + log([base]/[acid]) = 7.21 + log(1/0.02), [base]/[acid] = 0.2. This means the moles of K₂HPO₄ should be 0.2 times the moles of H₂PO₄-. Because the concentration of HPO₄²- is twice that of H₂PO₄-, you will need 0.04 moles of K₂HPO₄ to prepare the buffer solution.
To determine the volume of K₂HPO₄ required, divide the moles by the concentration (0.04 moles / 1 M), which gives a volume of 0.04 liters or 40 milliliters. Therefore, you will need 40 milliliters of 1 M K₂HPO₄ to prepare 40 liters of a 0.02 M phosphate buffer.