Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.

Question

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)?

table 1:
mL NaOH=1.00
pH wHCl=1.12
pH wHC2H3O2=3.87


Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
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)?

table 2:
mL NaOH=1.00
pH wHCl=?
pH wHC2H3O2=3.87

As you can see, I got all of the answers correct except for the last HCL one. I thought it was 1.12 like the other table, but I was wrong. I need help

Answer 1

It's a little difficult to read but as I understand it, I think you have failed to take into account the dilution by the water.
1.00 mL x 0.1M NaOH = 0.1 millimols NaOH
8.00 mL x 0.1 M HCl = 0.8 mmols HCl
Excess HCl = 0.7 mmol and
M HCl = 0.7mmol/101 mL = 0.00693 and pH = approximately 2.

Don't mistake the impact of this experiment. Note that adding 100 mL H2O to the HCl/NaOH (an un-buffered solution) changes the pH from 1.12 to a little over 2 which is a change of 10 for H^+ just by diluting it. BUT, note that adding 100 mL H2O to the acetic acid/sodium acetate (a buffered solution) did not change the pH at all. Buffered solutions resist a change in pH. Un-buffered solutions do not.

Answer 2

ooh. That makes more sense. Thank you!