Average the two percentages and calculate the percentage difference. given : Concentration of acetic acid in the 100 mL sample of titration 1 = 3.238 % Concentration of acetic acid in the 100 mL sample of titration 2 = 3.099 % 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

Average the two percentages and calculate the percentage difference. given : Concentration of acetic acid in the 100 mL sample of titration 1 = 3.238 % Concentration of acetic acid in the 100 mL sample of titration 2 = 3.099 % 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 = MV = 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 = MV = 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 = MV = 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 = MV = 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

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

Average the two percentages and calculate the percentage difference.

given :

Concentration of acetic acid in the 100 mL sample of titration 1 = 3.238 %

Concentration of acetic acid in the 100 mL sample of titration 2 = 3.099 %

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 H₂SO₄used = 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

Expert Solution

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.