Let's call acetic acid (CH3COOH) just HAc and call acetate ion (CH3COO^-) just Ac^-. That makes easier typing. Also, I don't know what you are using a molarity of 1.1009. That is far more accuracy than the pKa value. Those certainly are not that accurate. Anyway,
...............HAc ==> H^+ + Ac^-
I...............0.100......0.........0
C..............-x............x.........x
E.............0.100-x.....x.........x
Convert pKa to Ka with pKa = -log Ka, then
Substitute the E line above into the Ka expression of
Ka = (H^+)(Ac^-)/(HAc) and solve for x = (H^+), then convert that to pH.
For the acetate, the pH is determined by the hydrolysis of the acetate ion.
...............Ac^- + HOH ==> HAc + OH^-
I..............1.01......................0...........0
C..............-x.........................x...........x
E.............1.01-x....................x..........x
Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(HAc)
Plug the E line into the Kb expression and solve for x = (OH^-) and convert that to pH.
Post your work if you get stuck.
how can i calculate the ph of a stock acetic acid solution given the following: pKa of acetic acid=4.75, concentration of acetic acid = 1.1009M.
then, how can i calculate the ph of a stock sodium acetate solution given the following: concentration of sodium acetate = 1.0113M.
4 answers
im still not sure how to get Kb.
so for ka i did 10^-4.75 and got 1.78x10^-5. then using the ice table and the 1.1009 for HA, i got x=4.4x10^-3. then to get the ph, i did -log(x) = 2.35.
do i use the same ka of 1.78x10^-5 for kb?
so for ka i did 10^-4.75 and got 1.78x10^-5. then using the ice table and the 1.1009 for HA, i got x=4.4x10^-3. then to get the ph, i did -log(x) = 2.35.
do i use the same ka of 1.78x10^-5 for kb?
To find Kb the formula is Kb=Kw/Ka, and Kw=1x10^-14. Just replace the values, using the Ka you found.
got it, thank you!
4 answers