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.

To find the pH of the solution created by combining 1.00 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq), we need to consider the reactions that occur.

The reaction between NaOH and HCl produces NaCl and water. Since NaCl is a salt, it does not affect the pH of the solution. Therefore, the remaining component that can affect the pH is the excess HCl.

Next, we need to consider the reaction between HC2H3O2 (acetic acid) and water. Acetic acid is a weak acid with a Ka value of 1.75 x 10^-5.

To calculate the concentration of HCl in the solution, we take into account the 1.00 mL of NaOH and the 8.00 mL of HCl, giving us a total volume of 17 mL.

(HCl) = (0.10 M) * (8.00 mL / 17 mL) = 0.0471 M

To calculate the concentration of HC2H3O2, we multiply the concentration (0.10 M) by the volume (8.00 mL) and divide by the total volume (17 mL).

(HC2H3O2) = (0.10 M) * (8.00 mL / 17 mL) = 0.0471 M

Now, let's assume that the concentration of H^+ ions from HC2H3O2 is x, and the concentration of Ac^- ions is also x. The concentration of HC2H3O2 is 0.0471 M - x.

Using the equilibrium equation for the dissociation of HC2H3O2:

Ka = (H^+)(Ac^-) / (HC2H3O2)

Substituting the values:

1.75 x 10^-5 = (x)(x) / (0.0471 - x)

Solving this equation will give us the concentration of H^+ ions (x) from the HC2H3O2. We then add this to the concentration of H^+ ions from HCl (0.0412) to obtain the total concentration of H^+ ions in the solution.

Finally, to find the pH, we take the negative logarithm of the H^+ concentration:

pH = -log10([H^+])

Remember to keep track of the units throughout the calculations in order to obtain the correct answer.