Question

Data Table 1: Titration of Vinegar with Sodium Hydroxide Vinegar Sample 1 Vinegar Sample 2 Vinegar Sample 3 Mass of Vinegar (6) Volume of Vinegar (ml) Density - 1.005 g/ml Initial NaOH volume in syringe (ml) Final NaOH volume in syringe (ml) Volume of NaOH delivered (mL) Volume of NaOH delivered (L) Molarity of NaOH Moles of NaOH delivered Reaction of NaOH with Acetic Acid Moles of Acetic Acid in vinegar sample Molar mass of Acetic Acid Mass of Acetic Acid in vinegar sample Calculated Molarity of Acetic Acid in vinegar sample Average Molarity Calculated Percent Mass of Acetic in vinegar sample Average % mass

i need all answers

Answer 1

1)

a) Molarity = Moles of solute/volume of solution in liter

Moles of solute = mass in grams / molar mass

b) Volume increase means concentration decrease

Volume decrease means concentration increase

c)

NaOH + CH3COOH ---> CH3COONa + H2O

Stoichiometry ratio of NaOH and CH3COOH = 1 : 1

Thus moles of NaOH = moles of CH3COOH

d)

Higher amount of NaOH means higher amount of acetic acid but mole ration will be same i.e. 1:1

2)

a) too much NaOH

b) Moles = molarity * volume in liters

c) moles of NaOH high

d) moles of acetic acid = moles of NaOH

Mass of acetic acid in grams = moles * molar mass

e) asprin will be low