156 8 Titrations of Neutralization Reactions 44 Titration of

156 8 Titrations of Neutralization Reactions 4.4: Titration of a vinegar sample using the phenolphthalein indicator Balanced reactionAORI Vinegar Unknown Sample Trial #1 Trial #2 Average molarity, NaOH (from Table 4.3) Volume of vinegar sample Mass, vinegar sample (with a density of 1.008 g/mL) Initial buret reading Buret reading at the end point 28L 023cL 8435 Volume titrant added at the end point 10 Moles, sodium hydroxide used at the equivalence point 11 Moles, acetic acid 12 Mass, acetic acid Mass percentage of acetic acid in the vinegar sample Molarity of acetic acid in vinegar sam 13 14 15 Average Mass Percent 16 Average Molarity calculations. were shaded.

Solution

10. Moles of NaOH

In both the trial, molarity of NaOH = 0.975mol/L = 0.975 M

Molar mass of NaOH = (23 + 16 + 1) g/mol= 40 g/mol

Volume of NaOH = 0.0208 L

Molarity is given by no. of moles divided by the volume of NaOH used till equivalence point in L.

M = n/V

Therefore, n = M x V

n = 0.975 mol/L x 0.0208 L

n = 0.0203 mol

Therefore, no. of moles of NaOH = 0.0203 mol.

11. Moles of acetic acid

Trial 1

Volume of vinegar sample = 50.2 mL = 0.0502 L

First calculate the molarity of vinegar sample using the equation

(MV)_sample = (MV)_NaOH

Therefore,

M_sample = (MV)_NaOH/V_sample

M_sample = (0.975M x 0.0208L)/ 0.0502 L

M_sample = 0.4040 M = 0.4040 mol/L

But, M = n/V

Therefore, no. of moles of acetic acid is

n = M x V

where M = molarity of vinegar

V = Volume of sample = 0.0502 L

n = 0.4040 mol/L x 0.0502 L

n = 0.0203 mol

Therefore, no.of moles of acetic acid = 0.0203 mol

Similarly, you calculate for trial 2.

12. Mass of acetic acid

No. of moles of acetic acid = 0.0203 mol

Molar mass of CH₃COOH = (12+3+12+16+16+1) g/mol = 60 g/mol

That is, 1 mol of acetic acid has mass of 60 g.

Therefore,

0.0203 mol of acetic acid will have mass

= 60 g x 0.0203

= 1.218 g

Therefore, mass of acetic acid = 1.218 g

Similarly calculate for trial 2.

13. Mass percentage of acetic acid

Density of vinegar sample = 1.008 g/mL

That is, 1 mL of the sample has mass of 1.008g

Volume of sample taken = 50.2 mL

Therefore, mass of the sample of 50.2 mL will be

= 1.008g x 50.2

= 50.602g

So, mass of the sample = 50.602g

Mass of acetic acid (calculated in point 12) = 1.218 g

Mass percentage of acetic acid will be

= (mass of acetic acid/ mass of sample) x 100

= (1.218g/50.602g) x 100

= 0.0241 x 100

= 2.41 %

Therefore, mass percentage of acetic acid in the sample is 2.41%

Similarly, calculate for trial 2.

14. Molarity of acetic acid in vinegar sample

Mass of acetic acid in the sample = 1.218 g

Molar mass of acetic acid = 60 g/mol

That is, 60 g of acetic has 1 mol

Therefore, 1.218g of acetic acid will have no. of moles

= (1mol/60g) x 1.218g

= 0.0203 mol

Volume of sample = 50.2 mL = 0.0502 L

Molarity is

M = n/V

M = 0.0203 mol/ 0.0502 L

M = 0.4044 mol/L

M = 0.4044 M

Therefore, molarity of acetic acid in vinegar sample is 0.4044 M.

Similarly, calculate for trial 2.