Chemistry Tutor Site Help
- How many grams of CH3COOH in the 40 mL of vinegar if the density of vinegar is 0.96 g/mL ?
- How many grams of CH3COOH in the 5 M of 40 mL of CH3COOH solution ?
Chemistry Tutor Site Help
This is very easy to solve because the density of the vinegar has been known. To calculate the mass of the acetic acid we just multiply the volume of the vinegar with its density, and we get:
Mass of acetic acid = volume x density
Mass of acetic acid = 40 mL x 0.96 g/mL
Mass of acetic acid = 38.4 grams
In the second problem we have the volume and the molarity of the acetic acid solution. Since 1 Molar is one mole in the 1 liter of the solution, so we have 5 moles of acetic acid in the 1 liter of the solution. But in this problem we only have 40 mL of the solution, thus the moles are:
moles of acetic acid = Molarity x Volume
moles of acetic acid = 5 M x 0.04 L
moles of acetic acid = 0.2 moles
and the mass of acetic acid in the 40 mL of 5 M acetic acid solution,
mass = moles x molar mass
mass = 0.2 mol x 60 g/mol
mass = 12 grams
May 26th, 2008
Titration of acetic acid and sodium hydroxide is an example of weak acid versus strong base titration. We begin to analyze the titration of 50 mL of 0.1 M acetic acid with 0.1 M of sodium
CH3COOH + NaOH -> CH3COONa + H2O
Before the titration is started, we only have 50 mL of 0.1 acetic acid, and the pH is calculated for the weak acid formula:
[H+] = (Ka.[acid])1/2
[H+] = (1.75 x 10-5) x (0.1)
[H+] = 1.323 x 10-3
pH = -log(1.75 x 10-6)
pH = 2.878
or we can solve the problem as,
|
CH3COOH |
+ |
-> |
CH3COO- |
+ |
H+ |
| Initial |
0.1 M |
|
|
0 |
|
0 |
| reacted |
x |
|
|
x |
|
x |
| equilibrium |
0.1 M |
|
|
x |
|
x |
Ka = (x)(x)/0.1
x2 = Ka x 0.1
x2 = (1.75 x 10-5) x (0.1)
x2 = 1.75 x 10-6
x = 1.323 x 10-3
x = [H+] = 1.323 x 10-3
pH = 2.878
The pH of the solution when 10 mL of NaOH has been added to the solution,
mmol CH3COOH start = 0.1 M x 50 mL = 5 mmol
mmol OH- = 0.1 M x 10 mL = 1 mmol
|
CH3COOH |
+ |
NaOH |
-> |
CH3COONa |
+ |
H2O |
| initial |
5 |
|
1 |
|
0 |
|
0 |
| reacted |
1 |
|
1 |
|
1 |
|
1 |
| equilibrium |
4 |
|
0 |
|
1 |
|
1 |
Look from the equation above we have CH3COOH and CH3COONa in the final solution this means that we have buffer solution. We can use the Henderson-hasselbalch equation to calculate the pH of the solution. Since the volumes cancel, use the millimoles:
pH = pKa + log [CH3COO-]/[CH3COOH}
pH = 4.757 + log (1/4)
pH = 4.155
With the same way we can calculate the pH of the solution when the 20, 30, 40 of the NaOH added and the result are:
20 mL---pH 4.581
30mL---pH 4.933
40mL---pH 5.359
The very special one is reached when the 25 of 0.1 NaOH is added, because this circumtance is in the midle of the titration process so the concentration of CH3COOH and CH3COONa are equal so the pH will be equal to the pKa
pH = pKa = 4.757
At the equivalent point is reached when the 50 mL of NaOH is added, we only have CH3COONa. This salt is form from the weak acid and the strong base thus this salt will be partial hydrolized, and we calculate the pH as:
[OH-] = {(1.0×10-14/1.75×10-5)}1/2
[OH-] = 5.345×10-6
pOH = 5.272
pH = 14 -5.272
pH = 8.728
At 60 mL of NaOH we have a solution that containts of CH3COONa and excess added NaOH. Thus the pH is determined by the concentration of excess OH-.
|
CH3COOH |
+ |
NaOH |
-> |
CH3COONa |
+ |
H2O |
| initial |
5 |
|
6 |
|
0 |
|
0 |
| reacted |
5 |
|
5 |
|
5 |
|
5 |
| equilibrium |
0 |
|
1 |
|
5 |
|
5 |
[OH-] = 1/110 mL = 0.009 M
pOH = 2.041
pH = 14 - 2.041
pH = 11.959
To make the titration curve of acetic acid versus NaOH just plot the pH with the volume of NaOH added, and we have as follow:
May 2nd, 2008