Calculate the ph for ch3co2h,3% dissociated at a concentration of 0.02 mol dm-3.

Use caps where needed. ch3co2h means nothing. I assume you meant to write CH3COOH. Let me call that HAc (H for the last H on the right, the acid H, and Ac for the rest of the molecule).

............3% = 0.03 dissociated
.............HAc ==> H^+ + Ac^-
initial.....0.02M....0.....0
change.....-6E-4....6E-4...6E-4
equil...0.0194.....6E-4....6E-4
(Note:6E-4 = 0.02*0.03
Then plug H^+ into pH = -log(H^+)

To calculate the pH for a weak acid like CH3CO2H (acetic acid) that is partially dissociated, we need to use the dissociation constant expression for the acid. The dissociation constant expression for acetic acid is:

Ka = [H+][CH3CO2-] / [CH3CO2H]

In this case, acetic acid is only 3% dissociated, which means that 97% remains undissociated. Thus, we can assume that the initial concentration of acetic acid ([CH3CO2H]) is equal to 0.02 mol dm-3, and the concentration of acetate ions ([CH3CO2-]) is 0.03% of that (0.0003 * 0.02 = 0.000006 mol dm-3). Since the concentration of H+ ions is the same as the concentration of acetate ions (according to the dissociation of acetic acid), it is also 0.000006 mol dm-3.

Now, we can substitute these values into the dissociation constant expression and solve for [H+].

Ka = [H+][CH3CO2-] / [CH3CO2H]
Ka = (0.000006)(0.000006) / 0.02
Ka = 0.000000000000036 / 0.02
Ka = 1.8 x 10^-12

The dissociation constant for acetic acid (Ka) is 1.8 x 10^-12. This is a small value, indicating that acetic acid is a weak acid.

To calculate the pH, we can use the equation:

pH = -log[H+]

Substituting the value of [H+] into the equation:

pH = -log(0.000006)
pH = -(-5.22) (using the calculator or logarithm tables)
pH ≈ 5.22

Therefore, the pH for a 3% dissociated acetic acid solution with a concentration of 0.02 mol dm-3 is approximately 5.22.