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^+)
Calculate the ph for ch3co2h,3% dissociated at a concentration of 0.02 mol dm-3.
1 answer
1 answer