Question

(TITRATION) What would be the calculations based on the trial data gathered?

I obtained 50mL of NaOH solution and then pipetted 5.0mL of pickle juice into an Erlenmeyer flask. I then added 50mL of distilled water and 2 drops of indicator.

Molarity of NaOH: 0.1001M Trial 1 Vinitial: 24.50 mL Vfinal: 45.72 mL Trial 2 Vinitial: 2.42 mL Vfinal: 23.74 mL Trial 3 Vinitial: 23.74 mL Vfinal: 44.22 mL

Answer 1

	1	2	3
Molarity of NaOH	0.1001 M	0.1001 M	0.1001 M
Final volume of NaOH, mL	45.72	23.74	44.22
Initial volume of NaOH, mL	24.50	2.42	23.74
Net volume of NaOH, mL = (final volume) – (initial volume)	21.22	21.32	20.48
Volume, mL of vinegar or pickles	5.0	5.0	5.0
Molarity of CH3COOH (calculation 1 below)	0.0386 M	0.0388 M	0.0373 M
Average molarity of CH3COOH	1/3*(0.0386 + 0.0388 + 0.0373) M = 0.0382
% by weight of CH3COOH (calculation 2 below)	0.126 g
Average % by weight of CH3COOH (calculation 3 below)	2.52

Calculation 1:

Take trial 1 as an example.

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

= (21.22 mL)*(1 L)/(1000 mL)*(0.1001 M)

= 0.002124 mol.

Consider the reaction between NaOH and CH3COOH.

NaOH (aq) + CH3COOH (aq) -----------> NaCH3COO (aq) + H2O (l)

As per the stoichiometric equation above,

1 mol NaOH = 1 mol CH3COOH.

Therefore,

Mol(s) CH3COOH = mol(s) NaOH = 0.002124 mol.

Volume of the solution titrated = (5.0 + 50.0) mL = 55.0 mL = (55.0 mL)*(1 L)/(1000 mL)

= 0.055 L.

[CH3COOH] = 0.0386 M

Calculation 2:

Average molarity of CH3COOH = 0.0382 M.

Volume of solution titrated = 55.0 mL = (55.0 mL)*(1 L)/(1000 mL) = 0.055 L.

Mol(s) CH3COOH = (0.055 L)*(0.0382 M)

= 0.002101 mol.

Molar mass of CH3COOH = 60.052 g/mol

Mass of CH3COOH corresponding to 0.002101 mol = (0.002101 mol)*(60.052 g/mol)

= 0.126 g

Calculation 3:

Density of pickle = 1.05 g/mL.

Mass of pickle corresponding to 5.0 mL = (1.05 g/mL)*(5.0 mL) = 5.25 g.

Percent CH3COOH in pickle = (0.126 g CH3COOH)/(5.25 g pickle)*100

= 2.52