4.4: Titration of a vinegar sample using the phenolphthalein indicator Balanced

Question

4.4: Titration of a vinegar sample using the phenolphthalein indicator Balanced reaction 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) ,05 mu Initial buret reading Buret reading at the end point Volume titrant added at 9the end point 23.9 ml23.9 Moles, sodium hydroxide used at the eq 10 uivalence point 1 Moles, acetic acid 12 Mass, acetic acid Mass percentage of acetic acid in the vinegar sample Molarity of acetic acid in vinegar 13 14 15 Average Mass Percent 6 Average Molarity Calculations were performed for rows that As always, attach your sample calculations. were shaded.

Explanation / Answer

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,

Msample = (MV)NaOH/Vsample

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

Msample = 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.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.