Activity Coefficient for Phosphate Buffers: How would I calculate the predicted pH for Na2HPO4 and NaH2PO4? For hypothetical pH 6- we added 94mL of H2PO4 and 6mL of HPO4 (for a total of 100mL). But how would I calculate "actual pH"?

We measured at two different strengths, 0.1M and 0.01M. For HPO4 the correction factor for 0.1M-0.45 and 0.01-0.75 and for H2PO4 0.1-0.74 and 0.01-0.93.

I know it's a similar equation to the HH equation with the added log(A coeff/HA coeff) but I'm unsure how to proceed from there.

How many moles of conjugate base (Na2HPO4) are present in 40mL of .010M buffer at pH 7.5 vs at pH 6?

Buffer was made up of Na2HPO4 and NaH2PO4.
pH=pKa +log[A/HA]; pKa=7.2.... When I did the calculations I got the same amount as if it were in a 0.1M buffer at the same pH's so I think I'm doing something wrong.

Ignore the second post at 10:32, I'm going to post it as it's own question!

To calculate the actual pH of a phosphate buffer solution, you can use the Henderson-Hasselbalch equation, which is:

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

In this case, [A-] represents the concentration of the conjugate base (HPO4-) and [HA] represents the concentration of the weak acid (H2PO4-).

To apply the Henderson-Hasselbalch equation, you need to know the pKa value of the acid-base pair. In the case of the phosphate buffer, the pKa value is approximately 7.2.

Now, let's calculate the actual pH for both the 0.1M and 0.01M phosphate buffer solutions.

For the 0.1M phosphate buffer:
1. Calculate the concentrations of HPO4- and H2PO4-. Since you added 94mL of H2PO4 and 6mL of HPO4 to make a total volume of 100mL, the concentration of H2PO4- is (94/100) * 0.1M = 0.094M, and the concentration of HPO4- is (6/100) * 0.1M = 0.006M.
2. Use the Henderson-Hasselbalch equation:

pH = 7.2 + log(0.006/0.094) = 7.2 - 0.934 = 6.266

So, the actual pH for the 0.1M phosphate buffer is approximately 6.266.

For the 0.01M phosphate buffer:
1. Calculate the concentrations of HPO4- and H2PO4-. Using the same method as before, the concentration of H2PO4- is (94/100) * 0.01M = 0.0094M, and the concentration of HPO4- is (6/100) * 0.01M = 6e-4M.
2. Apply the Henderson-Hasselbalch equation:

pH = 7.2 + log(6e-4/0.0094) = 7.2 - 0.75 = 6.45

So, the actual pH for the 0.01M phosphate buffer is 6.45.

To consider the activity coefficient correction, you need to multiply the concentration of each species by its respective activity coefficient (A coefficient). In this case, you mentioned correction factors for HPO4 and H2PO4 at different strengths.

For example, let's use the correction factor of 0.75 for the 0.01M HPO4. The corrected concentration is:

Corrected [HPO4-] = 0.75 * 6e-4M = 4.5e-4M

You can then substitute this corrected concentration into the Henderson-Hasselbalch equation to calculate the actual pH.

Repeat the same procedure for the other concentrations and correction factors to calculate the actual pH considering the activity coefficients.