Question

How to calculate molarity of acetic acid in the picture?

als: Name Experiment 6 DATA SHEET Titration CM1004 Section Average molarity of NaOH, from Part I. Fall 2019 Instructor or TA Initials: 0.083 M II. Molarity of acetic acid in white vinegar Determination Determination Determination 3 (if needed) Step 1 Vinegar Initial burette reading, ml. 10.2 21.6 23.8 23.8 Step 2 Vinegar Final burette reading, ml 12.3 Volume of vinegar, mL (Step 2-Step 1) 2.2 18.2 Step 3 NaOH Initial burette reading, mł_10 18.5 39.2 Step 4 NaOH Final burette reading, mL 29 Volume of NaOH, ml. (Step 4 - Step 3) 19.3 20.7. Molarity of acetic acid in vinegar (M or moVL) = Show the calculation of molarity for each determination: M - (01) Average molarity: Show calculations:

Answer 1

The balanced neutralization reaction is :

NaOH (aq) + CH₃COOH (aq) $\rightarrow$ CH₃COONa (aq) + H₂O (l)

Determination 1

volume of vinegar = 2.1 mL = 2.1 x 10^-3 L

volume NaOH = 19.3 mL = 19.3 x 10^-3 L

moles NaOH = (concentration NaOH) * (volume NaOH is Liter)

moles NaOH = (0.083 M) * (19.3 x 10^-3 L)

moles NaOH = 1.6019 x 10^-3 mol

moles acetic acid in vinegar = moles NaOH

moles acetic acid in vinegar = 1.6019 x 10^-3 mol

Molarity acetic acid in vinegar = (moles acetic acid in vinegar) / (volume of vinegar in Liter)

Molarity acetic acid in vinegar = (1.6019 x 10^-3 mol) / (2.1 x 10^-3 L)

Molarity acetic acid in vinegar = 0.763 M

Determination 2

volume of vinegar = 2.2 mL = 2.2 x 10^-3 L

volume NaOH = 20.7 mL = 20.7 x 10^-3 L

moles NaOH = (concentration NaOH) * (volume NaOH is Liter)

moles NaOH = (0.083 M) * (20.7 x 10^-3 L)

moles NaOH = 1.7181 x 10^-3 mol

moles acetic acid in vinegar = moles NaOH

moles acetic acid in vinegar = 1.7181 x 10^-3 mol

Molarity acetic acid in vinegar = (moles acetic acid in vinegar) / (volume of vinegar in Liter)

Molarity acetic acid in vinegar = (1.7181 x 10^-3 mol) / (2.2 x 10^-3 L)

Molarity acetic acid in vinegar = 0.781 M

Average molarity = (molarity from determination 1 + molarity from determination 2) / 2

Average molarity = (0.763 M + 0.781 M) / 2

Average molarity = (1.544 M) / 2

Average molarity = 0.772 M