Express the concentration of acetic acid in both samples as % by mass of acetic per 100mL of solution given : %= mass/ volumex100% Given: NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of H2SO4Volume of NaOH = 10.0 mL * 0.205 M31.7 mL10.0 mL * 0.205 M31.7 mL = 0.0647 M Estimation of NaOH NaOH vs H2SO4 Burette solution is NaOH and the pipette solution is 10.0 mL of 0.205 M H2SO4 # Calculation of NaOH concentration: Balanced equation for the reaction of NaOH and H2SO4 is: 2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l) Hence mole ratio is: 2 mol NaOH / 1 mol H2SO4 or 2/1 Given volume of 0.205 M H2SO4 = 10.0 mL or 0.0100 L => Moles of 0.205 M H2SO4 used = M*V = 0.205 mol/L * 0.0100 L = 0.00205 mol H2SO4 => Moles of NaOH in the burette = 0.00205 mol H2SO4 * (2 mol NaOH / 1 mol H2SO4) = 0.0041 mol Average volume of NaOH in the burette = 31.7 mL or 0.0317 L => Molarity of NaOH = 0.0041 mol / 0.0317 L = 0.129 M # Calculation of acetic acid concentration in 5.0 mL vinegar solution: Balanced equation for the reaction of NaOH and acetic acid (CH3COOH) is: NaOH(aq) + CH3COOH(aq) -----> CH3COONa(aq) + H2O(l) Hence mole ratio is: 1 mol NaOH / 1 mol CH3COOH or 1/1 Average volume of 0.129 M NaOH consumed = 20.3 mL or 0.0203 L => Moles of 0.129 M NaOH consumed = M*V = 0.129 mol/L * 0.0203 L = 0.00262 mol NaOH => Moles of acetic acid in the pipette = 0.00262 mol NaOH * (1 mol CH3COOH / 1 mol NaOH) = 0.00262 mol Volume of acetic acid = 5.0 mL or 0.0050 L => Molarity of acetic acid = 0.00262 mol / 0.0050 L = 0.524 M Calculation of moles of acetic acid in 100mL of both samples of vinegar : Titration-1: Volume of 0.129 M NaOH consumed = 20.9 mL or 0.0209 L => Moles of 0.129 M NaOH consumed = M*V = 0.129 mol/L * 0.0209 L = 0.0026961 mol NaOH => Moles of acetic acid in the pipette = 0.0026961 mol NaOH * (1 mol CH3COOH / 1 mol NaOH) = 0.0026961 mol => moles of acetic acid in 100 mL of the sample = 0.0026961 mol * (100 mL / 5.0 mL) = 0.053922 mol Titration-2: Volume of 0.129 M NaOH consumed = 20.0 mL or 0.0200 L => Moles of 0.129 M NaOH consumed = M*V = 0.129 mol/L * 0.0200 L = 0.00258 mol NaOH => Moles of acetic acid in the pipette = 0.00258 mol NaOH * (1 mol CH3COOH / 1 mol NaOH) = 0.00258 mol => moles of acetic acid in 100 mL of the sample = 0.00258 mol * (100 mL / 5.0 mL) = 0.0516 mol Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 31.8 mL 31.8 mL 31.7 mL Titration 1 0.0 mL 31.5 mL 31.5 mL Titration 2 0.0 mL 31.7 mL 31.7 mL
Express the concentration of acetic acid in both samples as % by mass of acetic per 100mL of solution given :
%= mass/ volumex100%
Given:
NaOH vs CH3COOH
Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar
Titration | Initial burette reading | Final burette reading | Volume of NaOH consumed | Average volume of NaOH |
Approximate | 0.0 mL | 20.1 mL | 20.1 mL | 20.3 mL |
Titration 1 | 0.0 mL | 20.9 mL | 20.9 mL | |
Titration 2 | 0.0 mL | 20.0 mL | 20.0 mL |
To find the average volume
Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3
= 20.3 mL
Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar
= 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL
= 0.2627 M
Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of H2SO4Volume of NaOH
= 10.0 mL * 0.205 M31.7 mL10.0 mL * 0.205 M31.7 mL
= 0.0647 M
Estimation of NaOH
NaOH vs H2SO4
Burette solution is NaOH and the pipette solution is 10.0 mL of 0.205 M H2SO4
# Calculation of NaOH concentration:
Balanced equation for the reaction of NaOH and H2SO4 is:
2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)
Hence mole ratio is: 2 mol NaOH / 1 mol H2SO4 or 2/1
Given volume of 0.205 M H2SO4 = 10.0 mL or 0.0100 L
=> Moles of 0.205 M H2SO4 used = M*V = 0.205 mol/L * 0.0100 L = 0.00205 mol H2SO4
=> Moles of NaOH in the burette = 0.00205 mol H2SO4 * (2 mol NaOH / 1 mol H2SO4) = 0.0041 mol
Average volume of NaOH in the burette = 31.7 mL or 0.0317 L
=> Molarity of NaOH = 0.0041 mol / 0.0317 L = 0.129 M
# Calculation of acetic acid concentration in 5.0 mL vinegar solution:
Balanced equation for the reaction of NaOH and acetic acid (CH3COOH) is:
NaOH(aq) + CH3COOH(aq) -----> CH3COONa(aq) + H2O(l)
Hence mole ratio is: 1 mol NaOH / 1 mol CH3COOH or 1/1
Average volume of 0.129 M NaOH consumed = 20.3 mL or 0.0203 L
=> Moles of 0.129 M NaOH consumed = M*V = 0.129 mol/L * 0.0203 L = 0.00262 mol NaOH
=> Moles of acetic acid in the pipette = 0.00262 mol NaOH * (1 mol CH3COOH / 1 mol NaOH) = 0.00262 mol
Volume of acetic acid = 5.0 mL or 0.0050 L
=> Molarity of acetic acid = 0.00262 mol / 0.0050 L = 0.524 M
Calculation of moles of acetic acid in 100mL of both samples of vinegar :
Titration-1:
Volume of 0.129 M NaOH consumed = 20.9 mL or 0.0209 L
=> Moles of 0.129 M NaOH consumed = M*V = 0.129 mol/L * 0.0209 L = 0.0026961 mol NaOH
=> Moles of acetic acid in the pipette = 0.0026961 mol NaOH * (1 mol CH3COOH / 1 mol NaOH) = 0.0026961 mol
=> moles of acetic acid in 100 mL of the sample = 0.0026961 mol * (100 mL / 5.0 mL) = 0.053922 mol
Titration-2:
Volume of 0.129 M NaOH consumed = 20.0 mL or 0.0200 L
=> Moles of 0.129 M NaOH consumed = M*V = 0.129 mol/L * 0.0200 L = 0.00258 mol NaOH
=> Moles of acetic acid in the pipette = 0.00258 mol NaOH * (1 mol CH3COOH / 1 mol NaOH) = 0.00258 mol
=> moles of acetic acid in 100 mL of the sample = 0.00258 mol * (100 mL / 5.0 mL) = 0.0516 mol
Titration | Initial burette reading | Final burette reading | Volume of NaOH consumed | Average volume of NaOH |
Approximate | 0.0 mL | 31.8 mL | 31.8 mL | 31.7 mL |
Titration 1 | 0.0 mL | 31.5 mL | 31.5 mL | |
Titration 2 | 0.0 mL | 31.7 mL |
31.7 mL
|
![Experimental Procedure:
Part A. Preparation of the NAOH solution
Weigh approximately 2 g of sodium hydroxide pellets on a watch glass and transfer
the solid to your large beaker.
1.
Add about 400 mL of deionized water and stir to dissolve the solid. This solution has
a concentration of approximately 0.1 M NAOH. Its molarity must be determined
exactly by titrating it with sulfuric acid according to equation (1) above.
2.
Part B. Titration of NaOH with sulfuric acid
1. Rinse a clean buret with about 10 mL of the NaOH solution. Add the solution, briefly
drain a small volume to remove bubbles from the buret tip, then fill the buret with the
solution to about the 0 mL line. Record the actual volume.
2.
Collect about 40 mL of standard sulfuric acid solution in a small beaker. Record the
molarity of the acid.
3.
Rinse a 10.00 mL pipet with a little of the sulfuric acid (2-3 mL) and discard the acid.
4.
Using the pipet, transfer 10.00 mL of sulfuric acid solution to a clean flask (the flask
does not have to be dry and can be wet with water, but do not rinse the flask with
acid).
5.
Add 20 mL deionized water and 3 drops ofphenolphthalein indicator. Place the flask
on a sheet of white paper so the color change will be more clearly visible.
6.
Now add the sodium hydroxide solution from the buret until a faint pink color
persists for at least 30 seconds. During the addition swirl the flask gently. During this
titration, the sodium hydroxide can be added fairly rapidly to give an approximate
idea of how much solution is needed to neutralize the acid in the flask.
7.
Record the volume of sodium hydroxide used, but do not use it in your calculations.
8.
Refill the buret with sodium hydroxide.
9.
Rinse out the flask with water and again pipet 10.00 mL of the sulfuric acid solution
into the flask, followed by about 20 mL of water and 3 drops.of indicator.
0.
Repeat the titration more carefully using the volume of sodium hydroxide solution
from step 7 to estimate the approximate volume required for the final titration. Ade
20 - 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5656bf79-0e9e-499f-b5be-5c0fd163f502%2F32d1aca4-d839-486c-9a31-bd447e17cedc%2Flzfzirb_processed.jpeg&w=3840&q=75)
![of one drop of sodium hydroxide to
faint pink color.
11.
Record the volume of sodium hydroxide used, and use this volume to calculate the
molarity.
12.
Feriorm a second, careful titration by repeating steps 8 - 11 with another 10 mL
sample of the standard sulfuric acid solution.
Part C. Analysis of vinegar
1.
Collect about 40 mL of the vinegar solution in a small beaker.
Pipet 5.00 mL of the vinegar solution into a flask and add 20 mL of deionized water,
3 drops of phenolphthalein and swirl to mix.
2.
3.
Refill the buret with sodium hydroxide solution.
Titrate with sodium hydroxide solution in the buret until the faint pink color persists
for at least 30 seconds. As in the previous analysis, the first titration gives an estimate
of the volume of sodium hydroxide required.
4.
Repeat the titration more carefully with two additional samples of vinegar, and use
these volumes of sodium hydroxide in the calculations below.
5.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5656bf79-0e9e-499f-b5be-5c0fd163f502%2F32d1aca4-d839-486c-9a31-bd447e17cedc%2F7pmfas9_processed.jpeg&w=3840&q=75)
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