Titration of 50.0 ml of acetic acid reaches equivalence after delivery of 22.5 ml of standardized NaOH 0.21 M.
What is the initial concentration of acetic acid and what is the pH of the solution? What is the pH at equivalence? What is the pH after addition of 20.0 ml of NaOH? What volume of titrant (NaOH) must be delivered to reach a pH of 4.74?
3 answers
I responded earlier with this. How much do you know how to do?
i don't even know where to start, I'm really confused on this topic in chemistry, I just want to see at step by step way to solvethis problem.
HAc + NaOH ==> NaAc + H2O
First determine the M of the HAc.
mL x M = mL x M
22.5 x .21 = 50x (I worked this part for you at your original post yesterday. Apparently you never went back to look at it.)
M HAc = approx 0.9 but you need to do it more accurately.
Now divide the titration into four parts.
a. beginning (just "pure" 0.9M HAc).
b. between beginning and eq. pt.
c. eq pt.
d. after eq pt.
a. 0.9M HAc. I know you've seen this done.
..........HAc --> H^+ + Ac^-
I........0.9M.....0......0
C........-x.......x......x
E.........0.9-x...x.......x
Substitute the E line into Ka for HAc and solve for x,then convert to pH.
pH at eq pt is due to the hydrolysis of the acetate ion. (Ac^-) = (mmols Ac^-/mL solution) = [(0.21*22.5/(22.5+50)]
........Ac^- + HOH ==> HAc + OH^-
I...approx 0.6..........0.....0
C.........-x............x.....x
E......0.6-x............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.6-x) and solve for x = (OH^-) then convert to pH.
Anything between zero mL and 22.5 mL NaOH is that portion between the beginning and the eq pt; use the Henderson-Hasselbalch equation for that.
Finally, how much NaOH is need to produce a pH of 4.74 (which just happens to bre pKa for HAc)?
Since pH = pKa + log (base/acid) and you want pH = pKa, then you know log base/acid must be zero and that mans base = acid. Therefore, halfway to the eq pt will be that spot so 22.5 mL to the eq pt and 1/2 that will give you that spot Much of this I typed at the first post yesterday and I've just doubled my efforts today. Please go back and look at your earlier posts. We get to them eventually.
First determine the M of the HAc.
mL x M = mL x M
22.5 x .21 = 50x (I worked this part for you at your original post yesterday. Apparently you never went back to look at it.)
M HAc = approx 0.9 but you need to do it more accurately.
Now divide the titration into four parts.
a. beginning (just "pure" 0.9M HAc).
b. between beginning and eq. pt.
c. eq pt.
d. after eq pt.
a. 0.9M HAc. I know you've seen this done.
..........HAc --> H^+ + Ac^-
I........0.9M.....0......0
C........-x.......x......x
E.........0.9-x...x.......x
Substitute the E line into Ka for HAc and solve for x,then convert to pH.
pH at eq pt is due to the hydrolysis of the acetate ion. (Ac^-) = (mmols Ac^-/mL solution) = [(0.21*22.5/(22.5+50)]
........Ac^- + HOH ==> HAc + OH^-
I...approx 0.6..........0.....0
C.........-x............x.....x
E......0.6-x............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.6-x) and solve for x = (OH^-) then convert to pH.
Anything between zero mL and 22.5 mL NaOH is that portion between the beginning and the eq pt; use the Henderson-Hasselbalch equation for that.
Finally, how much NaOH is need to produce a pH of 4.74 (which just happens to bre pKa for HAc)?
Since pH = pKa + log (base/acid) and you want pH = pKa, then you know log base/acid must be zero and that mans base = acid. Therefore, halfway to the eq pt will be that spot so 22.5 mL to the eq pt and 1/2 that will give you that spot Much of this I typed at the first post yesterday and I've just doubled my efforts today. Please go back and look at your earlier posts. We get to them eventually.
2 answers