You are titrating 500 mL of 01 M acetic acid 01M with a 01 M

You are titrating 50.0 mL of 0.1 M acetic acid (0.1M) with a 0.1 M sodium hydroxide.

a) What is the volume of titrant delivered at equivalence

b) Calculate the pH at the following points corresponding to the volume of titrant delivered: V= 0; V=Veq/4; V = Veq/2; V=0.75Veq; V = 0.9*Veq; V = Veq; V = 1.05*Veq; V = 1.5*Veq; V=2*Veq

Solution

CH3COOH is a weak acid - calculate [H+] from the Ka equation for question a)
When you add NaOH to the acid you produce a buffer solution Use the Henderson Hasselbalch equation
At the half equivalence point , pH = pKa . pKa for the acid is 4.77 . The half equivalence point is c) So the pH here will be 4.77

a) You have a weak acid. In order to calculate pH you first have to determine the [H+] in the solution. You do this using the Ka equation:

Ka = [H+] [CH3COO-] / [CH3COOH]
You know that [H+] = [CH3COO-] so for product we write [H+] ²
Because the acid is weak , we say [CH3COOH] = 0.10
Substitute:
1.7*10^-5 = [H+] ² / 0.1
[H+] ² = (1.7*10^-5) *0.1
[H+] ² = 1.7*10^-6
[H+] = √(1.7*10^-6)
[H+] = 1.30*10^-3
pH = -log [H+]
pH = -log (1.3*10^-3)
pH = 2.88

Now we go to b) but first some theory:
When you add NaOH to the CH3COOH solution you form CH3COONa. You then have a solution of a weak acid and a salt of this acid. This is a buffer solution . The pH of the buffer solution is calculated using the Henderson - Hasselbalch equation.
pH = pKa + log ([salt] /[acid])
Some preliminary calculations :
pKa = -log Ka = - log (1.7*10^-5) = 4.77

Mol CH3COOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol
Mol NaOH in 25mL of 0.1M solution = 25/1000*0.1 = 0.0025 mol
The NaOH reacts with the CH3COOH in 1:1 ratio
Therefore you produce 0.0025 mol CH3COONa
and remaining unreacted is 0.0075 mol CH3COOH
In total volume 125mL = 0.125 L
Molarity of CH3COONa = 0.0025/0.125 = 0.02M
Molarity of CH3COOH = 0.0075 / 0.125 = 0.06M
Now use the H-H equation:
pH = pKa + log ([salt] /[acid])
pH = 4.77 + log ( (0.02/0.06)
pH = 4.77 + log 0.333
pH = 4.77 + (-0.48)
pH = 4.29

c) This is the really interesting situation : remember I said that here pH = pKa. Here\'s how:
Mol CH3COOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol
Mol NaOH in 50mL of 0.1M solution = 25/1000*0.1 = 0.005 mol
The NaOH reacts with the CH3COOH in 1:1 ratio
Therefore you produce 0.005 mol CH3COONa
and remaining unreacted is 0.005 mol CH3COOH
In total volume 150mL = 0.150 L
Molarity of CH3COONa = 0.005/0.150 = 0.0333M
Molarity of CH3COOH = 0.005 / 0.150 = 0.0333M
Substitute into H-H equation:
pH = 4.77 + log (0.0333*0.0333)
pH = 4.77 + log 1
pH = 4.77 + 0
pH = 4.77 which is the same as pKa

d) I am sure that you can do this by yourself: Follow the above system

e) When you have added 100mL of the NaOH solution you have neutralised all the acid. There is then only a solution of salt CH3COONa in solution . This is called the equivalence point - and you must calculate the pH at the equivalence point.

You have reacted 0.01mol CH3COOH with 0.01mol NaOH to produce 0.01 mol CH3COONa dissolved in 200mL = 0.20L solution:
Molarity of CH3COONa solution = 0.01/0.2 = 0.05M solution.
Calculate pH as follows:
The aim is to find the [OH-] of the solution You will know that a salt of a strong base and weak acid is basic.
The CH3COONa dissociates in water :
CH3COONa ↔ CH3COO- + Na+ The CH3COO- reacts with water:
CH3COO- + H2O ↔ CH3COOH + OH-

Ka for CH3COOH = 1.7*10^-5
Kw = Ka * Kb - we want Kb
Kb = Kw* Ka
Kb = 10^-14 / (1.7*10^-5)
Kb = 5.88*10^-10

Kb = [CH3COO-] *[[OH] / [CH3COONa]
We know that [CH3COO-] = [OH-] so product is [OH-] ²
[CH3COONa] = 0.05M

Kb = [OH] ²/0.05
(5.88*10^-10 ) * 0.05 = [OH[ ²
[OH] ² = 2.94*10^-11
[OH-] = 5.42*10^-6
To calculate pH you require [H+]
[OH-] *[H+] = 10^-14
[H+] = 10^-14/ [OH-]
[H+] = 10^-14 / (5.42*10^-6)
[H+] = 1.84*10^-9

pH = -log ( 1.84*10^-9)
pH = 8.73

And finally to f) You have added a very large excess of NaOH . All the acid has been neutralised. With this large excess of a strong base , the salt does not play any role in the pH of the final solution.
You have added 300mL of NaOH = 0.03 mol NaOH
This has reacted with 100mL = 0.01mol acid.
You have 0.02 mol NaOH remaining unreacted in a final volume of 400mL
Molarity of NaOH solution: 0.02/0.4 = 0.05 M NaOH

In 0.05M NaOH solution , [OH-] = 0.05M
[H+] = 10^-14 / 0.05
[H+] = 2.0*10^-13

pH = -log 2.0*10^-13
pH = 12.7