How do I calculate the amount of grams of Na2HPO4 and NaH2PO4 to prepare 200 mL of a buffer with an 8.25 pH so the sum of the concentrations of HPO4 and H2PO4 ions is 0.5 M? Does this involve writing a Ka expression and equilibrium? I am confused and Thanks for your help!

It means using the Henderson-Hasselbalch equation.

How?

To calculate the amount of grams of Na2HPO4 and NaH2PO4 needed to prepare the buffer solution, we need to consider the Henderson-Hasselbalch equation and the equilibrium of the HPO4 and H2PO4 ions. Let's break down the steps:

Step 1: Write the equilibrium equation and the Ka expression:
The equilibrium equation for the dissociation of Na2HPO4 is:
Na2HPO4 ↔ 2Na+ + HPO4^2-

The equilibrium equation for the dissociation of NaH2PO4 is:
NaH2PO4 ↔ Na+ + H2PO4^-

The Ka expression for the dissociation of Na2HPO4 is:
Ka = [Na+][HPO4^2-] / [Na2HPO4]

The Ka expression for the dissociation of NaH2PO4 is:
Ka = [Na+][H2PO4^-] / [NaH2PO4]

Step 2: Determine the desired pH and concentrations:
Given that the pH of the buffer solution is 8.25, we can determine the concentration of HPO4^2- and H2PO4^- using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Here, [A-] represents the concentration of the conjugate base (HPO4^2-), and [HA] represents the concentration of the acid (H2PO4^-).

In our case, assuming a buffer solution with equal concentrations of HPO4^2- and H2PO4^- (0.5 M each), the pH is equal to the pKa:

8.25 = pKa + log(0.5/0.5)
8.25 = pKa + log(1)
8.25 = pKa

Therefore, the pKa required for the buffer solution is 8.25.

Step 3: Determine the pKa values:
To find the pKa values, you can use reference material or search for the acid dissociation constants (Ka) of HPO4^2- and H2PO4^- in a reliable source.

For HPO4^2-, the pKa is approximately 7.21.
For H2PO4^-, the pKa is approximately 6.86.

Step 4: Calculate the moles of HPO4^2- and H2PO4^-:
To determine the amount of moles of HPO4^2- and H2PO4^- needed, we use the formula:

moles = concentration (in M) x volume (in L)
However, in this case, we have the same concentration (0.5 M) for both HPO4^2- and H2PO4^-.

moles of HPO4^2- = 0.5 M x 0.2 L (200 mL in liters) = 0.1 mol
moles of H2PO4^- = 0.5 M x 0.2 L (200 mL in liters) = 0.1 mol

Step 5: Convert moles to grams:
Finally, using the molar mass of the respective compounds, we can calculate the weight in grams:

For Na2HPO4:
Molar mass = 22.99 g/mol (Na) + 2 * 1.01 g/mol (H) + 31.99 g/mol (P) + 4 * 16.00 g/mol (O) = 141.96 g/mol
grams of Na2HPO4 = moles of Na2HPO4 x molar mass = 0.1 mol x 141.96 g/mol = 14.196 g

For NaH2PO4:
Molar mass = 22.99 g/mol (Na) + 1.01 g/mol (H) + 31.99 g/mol (P) + 4 * 16.00 g/mol (O) = 119.98 g/mol
grams of NaH2PO4 = moles of NaH2PO4 x molar mass = 0.1 mol x 119.98 g/mol = 11.998 g

Therefore, to prepare 200 mL of a buffer solution with an 8.25 pH and a total concentration of 0.5 M for both HPO4^2- and H2PO4^-, you will need approximately 14.196 grams of Na2HPO4 and 11.998 grams of NaH2PO4.

Note: The pKa values used in this explanation are approximations and may vary slightly depending on the source.