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

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

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.

ooh. That makes more sense. Thank you!

To find the pH values in table 2, we need to consider the dilution of the acid (HCl) with water. When an acid is diluted with water, the concentration of the acid decreases, which affects the pH of the resulting solution.

In this case, we have 8.00 mL of 0.10 M HCl, which was diluted with 100 mL of water. To calculate the new concentration of the diluted HCl, we can use the following equation:

C1V1 = C2V2

Where:
C1 = initial concentration of the acid (0.10 M)
V1 = initial volume of the acid (8.00 mL)
C2 = final concentration of the diluted acid (unknown)
V2 = final volume of the diluted acid (8.00 mL + 100 mL = 108.00 mL)

Using the equation, we can solve for C2:

(0.10 M)(8.00 mL) = C2(108.00 mL)
0.80 = C2(108.00 mL)
C2 = 0.80 / 108.00
C2 ≈ 0.00741 M

Now, to find the pH of the solution created by combining 1.00 mL of the 0.10 M NaOH with the diluted HCl, we can use the same approach as before:

For the HCl solution:
pH = -log[H+]
[H+] = concentration of HCl

Using the new concentration of the diluted HCl (0.00741 M), we can find the pH:

pH wHCl = -log(0.00741)
pH wHCl ≈ 2.13

Therefore, the pH value in table 2 for the solution created by combining 1.00 mL of the 0.10 M NaOH with the diluted HCl is approximately 2.13