Yes and no.
Percent dissociation has little to do with 5%. That's 5% by mass CH3COOH in water BUT that doesn't tell you that it is or is not dissociated by any amount.
Let's call vinegar HAc to avoid writing CH3COOH for it. H stands for H (of the COOH part) and Ac stands for the remainder of the moelcule (CH3COO).
HAc ==> H^+ + Ac^-
%diss = [(H^+)/(HAc)]*100
pH = 2.41. You can calculate (H^+) from that. You have (HAc) (which you calculate from the 5% and the density but I don't know that it is 1M). Is there something else in the problem?
Vinegar is a 5.0% solution by weight of acetic acid in water. Given that the pH for vinegar is 2.41 and that Ka= 1.8E-5 and the density s 1.00g/mL, what is the percent dissociation of acetic acid in vinegar?
I know the % dissociation equals the amount which did dissociate 5.0%divided by the amount which could have dissociated in this case 1.00 M because it says in the question that this is a 1.00 M solution of CH3COOH. Am I on the right track?
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