I need help filling the rest of the table Activity 1 Data Table 1: Titration of

ID: 1036188 • Letter: I

Question

I need help filling the rest of the table

Activity 1 Data Table 1: Titration of Vinegar with Sodium Hydroxide (Sample Provided) Vinegar Sample 1 Vinegar Sample 2 14.95 Vinegar Sample 3 Mass of Vinegar Sample 15.01 (Vinegar Density 1.005 g/mL) Initial NaOH Volume in 40.06 40.04 40.10 Final NaOH Volume in Total Volume of NaOH Moles of NaOH Moles of Acetic Acid in Sample Mass of Acetic Acid in Sample Calculated Molarity of Acetic Acid in the Sampled Vinegar Calculated Percent Mass of Acetic Acid in Sampled Vinegar Average Calculated Percent Mass of Acetic Acid in Sampled Vinegar If the accepted Percent Mass of Acetic acid in vinegar is 4.7%, what is your percent error?

Explanation / Answer

Vinegar Sample 1

Vinegar Sample 2

Vinegar Sample 3

Mass of vinegar sample (Vinegar density = 1.005 g/mL)

15.01 g

14.95 g

14.80 g

Initial NaOH volume in syringe

21.00 mL

21.12 mL

21.10 mL

Final NaOH volume in syringe

40.06 mL

40.04 mL

40.10 mL

Total volume of NaOH = (final NaOH) – (initial NaOH)

(40.06 – 21.00) mL = 19.06 mL

(40.04 – 21.12) mL = 18.92 mL

(40.10 – 21.10) mL = 19.00 mL

Moles of NaOH = (volume of NaOH in L)*(molarity of NaOH)

(Molarity of NaOH = 1 M)

(19.06 mL)*(1 L/1000 mL)*(1 M) = 0.01906 mole

(18.92 mL)*(1 L/1000 mL)*(1 M) = 0.01892 mole

(19.00 mL)*(1 L/1000 mL)*(1 M) = 0.01900 mole

Mass of acetic acid in sample (check sample calculation 1 below)

1.14 g

1.14

Calculated molarity of acetic acid in the sample vinegar (check sample calculation 2 below)

1.279 M

1.278 M

1.292 M

Calculated percent mass of acetic acid in sampled vinegar (%) = (mass of acetic acid in vinegar)/(mass of vinegar)*100

(1.14 g)/(15.01 g)*100 = 7.5949% ? 7.59%

(1.14 g)/(14.85 g)*100 = 7.6768% ? 7.68%

(1.14 g)/(14.80 g)*100 = 7.7027% ? 7.70%

Sample calculation 1

Moles of acetic acid = moles of NaOH at the equivalence point.

Therefore, moles of acetic acid = 0.01906 mole (take trail 1 as an example).

Molar mass of acetic acid, HC2H3O2 = (1*1.008 + 2*12.011 + 3*1.008 + 2*15.999) g/mol = 60.052 g/mol.

Mass of acetic acid = (moles of acetic acid)*(molar mass of acetic acid) = (0.01906 mole)*(60.052 g/mol) = 1.1446 g ? 1.14 g (correct to 2 decimal places).

Sample calculation 2

Molarity of acetic acid = (moles of acetic acid)/(volume of solution in L).

We are given the density of acetic acid solution; therefore, we can easily calculate the volume of acetic acid solution taken.

Trial 1

Trial 2

Trail 3

Mass of vinegar sample (g)

15.01

14.95

14.80

Density of vinegar sample (g/mL)

1.005

Volume of vinegar sample (mL) = (mass of vinegar sample)/(density of vinegar sample)

(15.01)/(1.005) = 14.9353

(14.95)/(1.005) = 14.8456

(14.80)/(1.005) = 14.7264

Volume of vinegar sample in L

(14.9353 mL)*(1 L/1000 mL) = 0.0149353 L ? 0.0149 L

(14.8456 mL)*(1 L/1000 mL) = 0.0148456 L ? 0.0148 L

(14.7264 mL)*(1 L/1000 mL) = 0.0147264 L ? 0.0147 L

Molarity of acetic acid (trial 1 as an example) = (0.01906 mole)/(0.0149 L) = 1.279 mol/L ? 1.28 M.