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?
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.
To find the initial concentration of acetic acid, we can use the concept of stoichiometry from the balanced equation of the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
Given that 22.5 ml of 0.21 M NaOH is required to reach equivalence, we can calculate the number of moles of NaOH delivered:
Number of moles of NaOH = volume (L) x concentration (mol/L)
= 0.0225 L x 0.21 mol/L
= 0.004725 mol
Since the stoichiometric ratio between acetic acid and NaOH is 1:1, the number of moles of acetic acid present in the 50.0 ml solution is also 0.004725 mol.
To find the initial concentration of acetic acid, we need to calculate its molarity:
Initial concentration of acetic acid (M) = moles/volume (L)
= 0.004725 mol/0.050 L
= 0.0945 M
So, the initial concentration of acetic acid is 0.0945 M.
Next, let's move on to determining the pH of the solution.
The pH of a solution of acetic acid can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In the case of acetic acid, it dissociates into acetate (A-) and a hydronium ion (H3O+). The pKa value for acetic acid is 4.76.
At the half-equivalence point, the concentrations of acetic acid and acetate are equal, so [A-] = [HA].
Using this information, we can calculate the pH at the half-equivalence point:
pH = pKa + log(1)
= 4.76 + log(1)
= 4.76
Therefore, the pH at the half-equivalence point is 4.76.
To calculate the pH after the addition of 20.0 ml of NaOH, we need to determine the remaining concentration of acetic acid.
The reaction between acetic acid and NaOH can be used to calculate the moles of NaOH required to reach the equivalence point:
Number of moles of NaOH = volume (L) x concentration (mol/L)
= 0.0200 L x 0.21 mol/L
= 0.0042 mol
Since the stoichiometric ratio between acetic acid and NaOH is 1:1, the number of moles of acetic acid remaining after the addition of NaOH is:
Remaining moles of acetic acid = Initial moles - Moles of NaOH
= 0.004725 mol - 0.0042 mol
= 0.000525 mol
Now we can calculate the remaining concentration of acetic acid:
Remaining concentration of acetic acid (M) = moles/volume (L)
= 0.000525 mol/0.050 L
= 0.0105 M
With this information, we can calculate the pH after the addition of NaOH:
pH = pKa + log([A-]/[HA])
= 4.76 + log(0.0105/0.0105)
= 4.76 + log(1)
= 4.76
Therefore, the pH after the addition of NaOH remains the same at 4.76.
To find the volume of titrant (NaOH) required to reach a pH of 4.74, we can use the Henderson-Hasselbalch equation again. Rearranging the equation, we get:
pH = pKa + log([A-]/[HA])
log([A-]/[HA]) = pH - pKa
[A-]/[HA] = 10^(pH - pKa)
Since we want to reach a pH of 4.74, plug in the values:
[A-]/[HA] = 10^(4.74 - 4.76)
[A-]/[HA] = 10^(-0.02)
We can then solve for [A-]/[HA]:
[A-]/[HA] = 0.6309
At equivalence, [A-] and [HA] cancel each other out, so we can assume that the volume of NaOH to reach a pH of 4.74 is equal to the volume of acetic acid initially present, which is 50.0 ml.
Therefore, the volume of titrant (NaOH) to reach a pH of 4.74 is 50.0 ml.