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?

ACID =
with 8.00 mL of the 0.10 M HCl(aq)?
with 8.00 mL of the 0.10 M HC2H3O2(aq?

im not sure Katie I'll tell you when i know but anyway its a great question i hope someone answers for you

1. HCl (strong acid completely split into ions)

*Find the concentration after dilution using:
V1M1 = V2M2
(8.00mL)(0.10 M) = (100.0 mL)(M2)
*M2 = (8.00mL)(0.10 M) / (100 mL) = ?______M
*pH = -log(M2)

2. HC2H3O2 (weak acid only partially dissociated into ions):
*Find the diluted concentration, M2 for HC2H3O2 as you did for HCl, then ..... let [H+] = x
x^2 / ( M2 - x ) = Ka (exact formula)
*x^2 / ( M2 ) = Ka (approximate formula, easier to use)
*Look up the Ka for acetic acid, HC2H3O2
*Substitute the values for M2 and Ka in to the approximate formula and solve for x
* pH = -log(x)

To calculate the pH values after dilution, we need to consider the dilution formula:

C1V1 = C2V2

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

In this case, we have 8.00 mL of acid (HCl or HC2H3O2) with a concentration of 0.10 M. The acid is diluted with 100 mL of water.

For the first case (HCl):
C1 = 0.10 M
V1 = 8.00 mL = 0.008 L
C2 = ?
V2 = 8.00 mL + 100 mL = 108 mL = 0.108 L

Using the dilution formula, we can rearrange it to solve for C2:

C2 = (C1 × V1) / V2
C2 = (0.10 M × 0.008 L) / 0.108 L
C2 = 0.00741 M

Now, to calculate the pH value, we use the formula:

pH = -log[H+]

For hydrochloric acid (HCl), it is a strong acid and dissociates completely in water, so its concentration of H+ ions is the same as the concentration of the acid. Therefore:

pH = -log(0.00741)
pH ≈ 2.13

For the second case (HC2H3O2):
Follow the same steps as above, but with the new acid concentration.

C1 = 0.10 M
V1 = 8.00 mL = 0.008 L
C2 = ?
V2 = 8.00 mL + 100 mL = 108 mL = 0.108 L

C2 = (C1 × V1) / V2
C2 = (0.10 M × 0.008 L) / 0.108 L
C2 = 0.00741 M

Since acetic acid (HC2H3O2) is a weak acid, we can't assume complete dissociation. Therefore, we need to consider its acid dissociation constant, Ka, which is 1.8 × 10^-5.

[H+] = √(Ka × C2)
[H+] = √(1.8 × 10^-5 × 0.00741)
[H+] ≈ 0.00258 M

Therefore:

pH = -log(0.00258)
pH ≈ 2.59