Complete the table below:

What is the pH of the solution created by combining 0.70 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)?
mL NaOH: 0.70
pH w/ HCl: 1.08
pH wHC2H3O2: 3.23

Complete the table below:
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)?
mL NaOH: 0.70
pH w/ HCl: 2.14
pH w/ HC2H3O2: ???

I got everything except for the 2nd pH w/ HC2H3O2. Please help!!

On the first part I agree with your answer of 1.08 but I obtained 3.73 if acetic acid is substituted for HCl.

For the second part, I don't agree with 2.14. I found 2.17. My guess is you did 0.73/100 mL = pH 2.14 but it should be 0.73/108.7 = pH 2.17.
If acetic acid is substituted for HCl in the second part, the addition of 100 mL water will not change the pH from what it was in the first part. Here is how I did the first part. HAc = acetic acid.
............HAc + NaOH ==> NaAc + H2O
initial....0.800...0........0......0
add................0.07.............
change.....-0.07...-0.07..+0.07..+.07
equil......0.73......0.....0.07...0.07

pH = pKa + log(Ac^-)/(HAc)
pH = 4.76+log(0.07/0.73)
pH = 3.74

To find the pH value when combining 0.70 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HC2H3O2(aq) that has been diluted with 100 mL of water, we need to use the concept of dilution and the equation for the pH of a weak acid.

The first step is to calculate the concentration of the acid after dilution.

The initial concentration of HC2H3O2 is 0.10 M. However, when we dilute it with 100 mL of water, the volume increases to 8.00 mL + 100 mL = 108 mL.

Using the equation for dilution, we can calculate the final concentration of HC2H3O2:

C1V1 = C2V2

where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume

0.10 M × 8.00 mL = C2 × 108 mL

C2 = (0.10 M × 8.00 mL) / 108 mL

C2 ≈ 0.0074 M

So, after dilution, the concentration of HC2H3O2 is approximately 0.0074 M.

Next, we can calculate the pH using the equation for the pH of a weak acid:

pH = pKa + log ([A-] / [HA])

In this equation, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

In this case, HC2H3O2 is a weak acid, and its conjugate base is the acetate ion (C2H3O2-). The pKa value of acetic acid (HC2H3O2) is approximately 4.75.

Therefore, the equation becomes:

pH = 4.75 + log ([C2H3O2-] / [HC2H3O2])

Substituting the concentrations:

pH = 4.75 + log (0.0074 M / 0.0074 M)

pH = 4.75 + log (1)

pH = 4.75 + 0

pH = 4.75

So, the pH value of the solution created by combining 0.70 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HC2H3O2(aq) that has been diluted with 100 mL of water is approximately 4.75.