one last question DrBob222 U wrote:
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
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.
youre the best!!!
thanx!!!!!!!!!!!!!!111
To find the value of Ka (acid dissociation constant), you can refer to a table of Ka values for various acids. In this case, you mentioned that you used acetic acid (HAc). The Ka for acetic acid is a known value, which you can look up in a table of acid dissociation constants. The Ka for acetic acid is approximately 1.8 x 10^-5.
The Ka value is a measure of the strength of the acid. Strong acids have larger Ka values, indicating that they fully dissociate in water, while weak acids have smaller Ka values, indicating that they only partially dissociate. In this case, acetic acid is a weak acid.