one last question DrBob222 U wrote:

Question

I did not guess. You said a weak acid and NaOH so I picked acetic acid (which I will call HAc) with NaOH. I used 0.1M HAc and titrated with 0.1M NaOH. At the equivalence point we will have 0.05M NaAc which is hydrolyzed in the water solution.
...........Ac^= + HOH ==> HAc + OH^-
initial....0.05M...........x.....x
change.....-x..............x......x
equil....0.05-x..........x.......x
Kb for the Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05)
Ka is 1.8E-5, Kw is 1E-14 and I obtained 5.27E-6 = x = (OH^-). I converted that to pOH and stuck it into pH + pOH = pKw = 14 and pH = 8.72.

my question is how did u get the Ka ????
pleaseeeeee

Answer 1

In the appendix of most texts (and in tables specifically made for the job) is a listing of the Ka values for weak acids as well as a table of Kb constants for weak bases. I have done this so many times that I've memorized many of the Ka values. If you look in your text, usually at the end just before the index starts, you will find the Ka and Kb tables along with tables for other constants also. Your text may not list 1.8E-5 exactly, some have 1.75E-5 and some have 1.7E=5 but all will be right around 1.8E-5 for acetic acid.

Answer 2

youre the best!!!
thanx!!!!!!!!!!!!!!111