Help with Chemistry?

Question

given a 1M solution of HAc, and that [H_3_0+] =.001 M, how would you calculate [Ac-] and [HAc] at equalibrium?

Answer 1

First, you need to recall that the expression for Ka is:

Ka=[H3O+][Ac-]/[HAc]

First, you need to recognize that IF you started with pure HAc and water, that [H3O+] =[Ac-] because the whenever you form a hydronium ion, you'll also form an acetate ion. If you look at the initial concentration of HAc and the final concentration of Ac-, you'll see that the [HAc] hasn't changed much. In most instances, for weak acids, you can usually assume that the initial and final concentrations of the weak acid are the same. For some acids, that's not a safe assumption, but in most instances it is. So, at this point, you've got everything you need, I think...