Question

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!!

Answer 1

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