Final answer:
To make a 450 mL of potassium phosphate buffer solution at pH = 6.97, you will only need 450 mL of the 1.00 M KH2PO4 stock solution. No K2HPO4 is required since the desired pH can be obtained solely with KH2PO4 based on the Henderson-Hasselbalch equation.
Step-by-step explanation:
To create a 450 mL buffer solution with a pH of 6.97 using 1.00 M stocks of KH2PO4 and K2HPO4, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where pH is the target pH of the buffer, pKa is the pH at which the acid is 50% dissociated, [A-] is the concentration of the base form of the buffer, and [HA] is the concentration of the acid form of the buffer.
By rearranging the equation, we can determine the ratio needed of base to acid to achieve the desired pH:
log([A-]/[HA]) = pH - pKa
log([A-]/[HA]) = 6.97 - 7.21 = -0.24
[A-]/[HA] = 10-0.24 ≈ 0.57
This implies for every 1 mole of KH2PO4 (HA), 0.57 moles of K2HPO4 (A-) are required to achieve the desired pH. To get the volumes needed, this ratio must be applied to the final volume of the buffer solution (450 mL).
Moles needed of KH2PO4 for 450 mL of 1.00 M solution = 0.45 moles
Therefore, moles of K2HPO4 needed = 0.45 moles * 0.57 = 0.2565 moles
Since we already have a 1.00 M stock of each, to find the volume (in liters), we'll divide moles by concentration:
Volume of KH2PO4 needed = 0.45 moles / 1.00 M = 0.45 L or 450 mL (which is the whole solution)
Because the volume of K2HPO4 would be calculated in a similar manner, we can see that it wouldn't be necessary to add any K2HPO4 since the desired pH is achieved solely with the KH2PO4. Thus, you will only need 450 mL of the 1.00 M KH2PO4 stock solution to make the buffer.