What is the pH of the solution created by combining 1.00 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)?
NaOH + HCl ==> NaCl + H2O
So all the NaOH is neutralized leaving NaCl, which will not affect the pH either way, with excess HCl remaining.
THEN you have HC2H3O2, which is a weak acid with a Ka of 1.75 x 10^-5.
Calculate (HCl) = 0.7 mmols/17 mL (1 mL from NaOH + 8.00 mL from HCl + 8.00 mL from HC2H3O2) = 0.0412. (HAc) is done similarly, 8.00 x 0.1 = 0.8 millimols/17 mL = 0.0471.
If we call acetic acid, HC2H3O2 simply HAc, then HAc ==> H^+ + Ac^- and
Ka = (H^+)(Ac^-)/(HAc)
Then (H^+) = x from the HAc + 0.0412 from the HCl. (Ac^-) = x and (HAc) = 0.0471 - x. Plug all that into Ka and solve for x, then add that to (HCl) to obtain the total. Post your work if you get stuck. You will need to solve the quadratic or you may do it simpler by successive approximations. OR you can estimate the amount of H^+ coming from the HAc.