I got part 1:
What is the pH of the solution created by combining 1.10 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
pH w/HCl =1.12 w/HC2H3O2(aq)= 3.94
But how do you do part two?
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
thanks
How did you do part 1. You had moles NaOH and moles HCl (HCl in excess), yo subtracted moles HCl-moles NaOH to arrive at moles HCl in excess. I think that is 0.69 millimoles and that divided by the volume (8 mL + 1.1 mL = 9.1 mL) = 0.0758 M and pH = 1.12. So for part 2, you have the same of everything except the 0.69 millimoles is now in 100 + 1.1 = 101.1 mL. Convert that to pH.
The 3.94 you obtained for the initial part was obtained by using the Henderson-Hasselbalch equation. That is
pH = pKa + log (base)/(acid)
Notice, however, in this one that the pH is determined by a ratio (not like HCl/NaOH). When you dilute the base by what ever factor (8 mL up to 100) that it dilutes BOTH the base AND the acid the same amount and since mole/L is concn, the volume cancels and its just moles/moles and the volume doesn't matter. The end result is that dilution has no effect on the buffer solution.