Assuming the final solution will be diluted to 1.00 , how much more should you add to achieve the desired pH?

Question

Information:
Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.80.
You have in front of you:
100 ml of 7.00×10−2 M HCl,
100 ml of 5.00×10−2 M NaOH and
plenty of distilled water.

You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 85.0 mL of HCl and 89.0 ml of NaOH left in their original containers.

Answer 1

You added 15 mL x 0.07M HCl = 1.05 millimoles initially.
Then you added 11 mL x 0.05M NaOH = 0.55 mmoles NaOH. Those react in 1:1 ratio to form 0.55 mmoles NaCl and this leaves 1.05-0.55 = 0.50 mmoles HCl remaining.
A pH of 2.80 = -log(H^+) and solve for (H^+); I obtained 0.00158M is what you want to produce.
0.00158 moles desired for the 1 L.
-0.0005 on hand with the 26 ml.
-------------
0.00108 moles extra needed.

M = moles/L and
L = moles/M = 0.00108/0.07 = ? L HCl needed, the make to 1L final volume.
Check my work.

Answer 2

Thank you!