Chemistry help needed! (calculating the Ka or Kb value of salt in solution) *help?*?

Question

I drew the question in paint.

here it is:

http://img256.imageshack.us/img256/8877/ddml6.jpg

Answer 1

Sodium acetate dissolves totally in water. Some of the acetate ion reacts with water:
OAc- + H2O = HOAc + OH-
Since the pH + pOH = 14, pOH= 4.75.
Then OH- = 10^-4.75 = 10-5 * 10^(.25) =
1.8x10-5
Since [OH]- is of the order of 10-5 and equals [HOAc], compared to an initial conc of 1 M of acetate ion, the equilibrium concentration is just about 1 M.
Define Kb= [HOAC][OH-]/[OAc]-, then,
We get Kb = [1.8x10-5]^2 = 3.2 x 10-10.

BTW, Kb*Ka = Kw, as noted by initial answerer. Since you asked for either Kb or Ka, this approach was simpler.

Answer 2

convert the PH to H+. you can do this by anti logging it.

that that and square it. then divide it by (the molarity minus the H+)

*note your H+ will be a small number. (REALLY SMALL)
*note: when solving for PH, and you are given a Ka or Kb:
take your unknown (your H+, that you will convert to PH once you have a value for it) and divide it by the molarity. (skip the step of minusing the H+, since it is such a small number,... or else you get an unfactoralable quadratic, and that's too much work.

Answer 3

Make use of these equations

pH = pKb + log ([B]/[BH+]), where [B] is the concentration of the base (salt), and [BH+] is the concentration of your acid (conjugate acid).

pH = -log[H+]
[H+] = [BH+] = 10^-pH
pKw = 14 = pKa + pKb

You're given all the variables besides pKb in the first equation

pKb = pH - log ([B]/[BH+])
pKa = 14 - pKb