This was a 2 part problem, and I believe I did the first part correctly; I just need help with the second.

The first part of the problem said this: A 0.25 M pH 8.5 phosphate buffer needs to be made. Give the amounts in grams of the acid and of the base you will use to make 1.5 L of this buffer.

I calculated that 22.67 g of K_{3}PO_{4} and 46.72 g of K_{2}HPO_{4} were needed. If someone finds this to be incorrect, I can give you my calculations to see where I went wrong. Otherwise, I'll assume these numbers are right.

The second part is where I am having trouble.

I need to calculate the pH of the resultant solution after adding 1.0 mL of a 1.0 M solution of NaOH to 100 mL of the buffer above.

I really had no clue where to begin, except that I remembered pH = -log [H+]

So (I don't know if this is right) I tried to calculate the [H+] for the solution before anything was added. The pH is 8.5, so 8.5 = -log [H+]

-8.5 = log [H+]. [H+] = 3.16 x 10^-9

NaOH I believe is [H+] = 10^-14

So then (here is where I will have a hard time explaining what I was trying to do, because I really don't know), for NaOH:

1.0 M / 101 mL = .0099 M

For the buffer:

0.25M / 101 mL = .002475 M

So for the [H+] of the solution:

(3.162 x 10^-9 / .002475) + (10^-14 / .0099) = 1.2776 x 10^-6

pH = -log(1.2776 x 10^-6)

pH = 5.894

Obviously, that’s wrong. The pH would not have decreased on adding a base. But I have no clue where I went wrong, or if I’m even using the right equation. Any help would be much appreciated!