156 8 Titrations of Neutralization Reactions 4.4: Titration of a vinegar sample

Question

156 8 Titrations of Neutralization Reactions 4.4: Titration of a vinegar sample using the phenolphthalein indicator Balanced reaction YCOHrG Vinegar Unknown Sample Trial #1 Trial #2 Average molarity, NaOH (from Table 4.3) Volume of vinegar sample 4 S v Mass, vinegar sample (with a density of 1.008 g/ml.) So , 028L Initial buret reading 0230L Buret reading at the end point Volume titrant added at the end point Moles, sodium hydroxide used at the equivalence point L12021 11 Moles, acetic acid 12 Mass, acetic acid Mass percentage of acetic acid in the vinegar sample Molarity of acetic acid in vinegar sample 14 15 Average Mass Percent 16 Average Molarity As always, attach your sample calculations. Calculations were performed for rows that were shaded.

Explanation / Answer

Vinegar sample unknown

Trial 1

Trail 2

Average Molarity, NaOH

0.975 mol/L

Volume of vinegar sample

50.2 mL

50.4 mL

Mass vinegar sample

50.6 g

50.8 g

Initial buret reading

0.0028 L

0.0236 L

Buret reading at the end point

0.0236 L

0.0438 L

Volume of titrant added at the end point

0.0208 L

0.0202 L

Moles, sodium hydroxide at the equivalence point

(volume of titrant at the end point)*(average molarity of NaOH) = (0.0208 L)*(0.975 mol/L) = 0.02028 mole.

(volume of titrant at the end point)*(average molarity of NaOH) = (0.0202 L)*(0.975 mol/L) = 0.019695 mole.

Moles acetic acid

Acetic acid (CH3COOH) reacts with NaOH on 1:1 molar ratio as

CH3COOH + NaOH -------> CH3COONa + H2O

Moles acetic acid = moles sodium hydroxide at the equivalence point = 0.02028 mole.

Acetic acid (CH3COOH) reacts with NaOH on 1:1 molar ratio as

CH3COOH + NaOH -------> CH3COONa + H2O

Moles acetic acid = moles sodium hydroxide at the equivalence point = 0. 19695 mole.

Mass acetic acid

Molar mass of acetic acid, CH3COOH = (2*12.011 + 4*1.008 + 2*15.999) g/mol = 60.052 g/mol.

Mass acetic acid = (moles acetic acid)*(molar mass of acetic acid) = (0.02028 mole)*(60.052 g/mol) = 1.2178 g

Molar mass of acetic acid, CH3COOH = (2*12.011 + 4*1.008 + 2*15.999) g/mol = 60.052 g/mol.

Mass acetic acid = (moles acetic acid)*(molar mass of acetic acid) = (0.019695 mole)*(60.052 g/mol) = 1.1827 g

Mass percentage acetic acid in the vinegar sample

(mass acetic acid)/(mass vinegar sample)*100 = (1.2178 g)/(50.6 g)*100 = 2.4067 2.41%

(mass acetic acid)/(mass vinegar sample)*100 = (1.1827 g)/(50.8 g)*100 = 2.3281 2.33%

Molarity of acetic acid in vinegar sample

(moles of acetic acid)/(volume of vinegar sample) = (0.02028 mole)/[(50.2 mL)*(1 L/1000 mL)] = 0.40398 mol/L 0.404 mol/L

(moles of acetic acid)/(volume of vinegar sample) = (0.019695 mole)/[(50.4 mL)*(1 L/1000 mL)] = 0.39077 mol/L 0.391 mol/L

Average mass percent

½*(2.41 + 2.33)% = 2.37%

Average molarity

½*(0.404 + 0.391) mol/L = 0.3975 mol/L

Vinegar sample unknown

Trial 1

Trail 2

Average Molarity, NaOH

0.975 mol/L

Volume of vinegar sample

50.2 mL

50.4 mL

Mass vinegar sample

50.6 g

50.8 g

Initial buret reading

0.0028 L

0.0236 L

Buret reading at the end point

0.0236 L

0.0438 L

Volume of titrant added at the end point

0.0208 L

0.0202 L

Moles, sodium hydroxide at the equivalence point

(volume of titrant at the end point)*(average molarity of NaOH) = (0.0208 L)*(0.975 mol/L) = 0.02028 mole.

(volume of titrant at the end point)*(average molarity of NaOH) = (0.0202 L)*(0.975 mol/L) = 0.019695 mole.

Moles acetic acid

Acetic acid (CH3COOH) reacts with NaOH on 1:1 molar ratio as

CH3COOH + NaOH -------> CH3COONa + H2O

Moles acetic acid = moles sodium hydroxide at the equivalence point = 0.02028 mole.

Acetic acid (CH3COOH) reacts with NaOH on 1:1 molar ratio as

CH3COOH + NaOH -------> CH3COONa + H2O

Moles acetic acid = moles sodium hydroxide at the equivalence point = 0. 19695 mole.

Mass acetic acid

Molar mass of acetic acid, CH3COOH = (2*12.011 + 4*1.008 + 2*15.999) g/mol = 60.052 g/mol.

Mass acetic acid = (moles acetic acid)*(molar mass of acetic acid) = (0.02028 mole)*(60.052 g/mol) = 1.2178 g

Molar mass of acetic acid, CH3COOH = (2*12.011 + 4*1.008 + 2*15.999) g/mol = 60.052 g/mol.

Mass acetic acid = (moles acetic acid)*(molar mass of acetic acid) = (0.019695 mole)*(60.052 g/mol) = 1.1827 g

Mass percentage acetic acid in the vinegar sample

(mass acetic acid)/(mass vinegar sample)*100 = (1.2178 g)/(50.6 g)*100 = 2.4067 2.41%

(mass acetic acid)/(mass vinegar sample)*100 = (1.1827 g)/(50.8 g)*100 = 2.3281 2.33%

Molarity of acetic acid in vinegar sample

(moles of acetic acid)/(volume of vinegar sample) = (0.02028 mole)/[(50.2 mL)*(1 L/1000 mL)] = 0.40398 mol/L 0.404 mol/L

(moles of acetic acid)/(volume of vinegar sample) = (0.019695 mole)/[(50.4 mL)*(1 L/1000 mL)] = 0.39077 mol/L 0.391 mol/L

Average mass percent

½*(2.41 + 2.33)% = 2.37%

Average molarity

½*(0.404 + 0.391) mol/L = 0.3975 mol/L

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