What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl? ( Tris. At 25 ∘C, Tris has a pKb of 5.91 )

Question

a) The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0250 mol of hydrogen ions without changing the volume. What is the pH of the final solution?

I've solved it = 6.676mL

Im stuck with b)
The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0250 mol of hydrogen ions without changing the volume. What is the pH of the final solution?

from part a) result
mol TrisH+ = 0.1332
mol Tris = 0.06676

mol TrisH+ = 0.1332 +0.0250 = 0.1582
mol Tris = 0.06676 -0.0250 = 0.04176

Using HH equation
pH = pKa + log [A-] / [HA]
pH = 8.09 + log (0.04176 / 0.1582)
pH = 7.51

the answer is incorrect. Thanks

Answer 1

Thanks for showing your work.
At the end of part A, you're correct that you have 0.1333 mols TrisH+ and 0.0666 mols Tris. So when you remove half of it (500 mL of the 1L) you have 0.1333/2 mols TrisH+ and 0.0666/2 mols Tris. Those are the numbers you're starting with and not the values you have. See if that doesn't work. I raced through the calculations and I have approx 7.05