Question

NCSUGenChem102LabV1 9. POST.01. RMy Notes 1/6 points 1. Previous Answers Ask Your Teacher Consider a different titrant for this exercise. Suppose Ca(OH), were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. The analyte will still be HC2H302. (a) What is the stoichiometry of HC2H302 to Ca(OH)2? 8:3 5:2 2:1 1:1 1:2 2:5 3:8 (b) Complete the following table for this titration. Data Table P1: Titration of acetic acid with calcium hydroxide concentration of Ca(OH)2 0.346 M volume vinegar solution 11.84 mL mass vinegar solution volume of Ca(OH)2 solution 11.90 g 11.19 mL mmol of Ca(0H)2 4.0 4.47 mmol X mmol of HC2H302 4.0 8.94 mmol X 4.0 0.536 mass of HC2H302 X 9 mass % of HC2H302 in the original sample 404.51 X % molarity of HC2H302 in the original sample 400.790 XM

Answer 1

concentration of Ca(OH)2	0.346 M
volume vinegar solution	11.84 mL
mass vinegar solution	11.90 g
volume of Ca(OH)2 solution	11.19 mL
mmol of Ca(OH)2	3.87 mmol
mmol of HC2H3O2	7.74 mmol
mass of HC2H3O2	0.465 g
mass % of HC2H3O2 in the original sample	3.91 %
molarity of HC2H3O2 in the original sample	0.692 M

moles Ca(OH)₂ = (concentration Ca(OH)₂) * (volume Ca(OH)₂)

moles Ca(OH)₂ = (0.346 M) * (11.19 mL)

moles Ca(OH)₂ = 3.87174 mmol

moles acetic acid = (moles Ca(OH)₂) * 2

moles acetic acid = (3.87174 mmol) * 2

moles acetic acid = 7.74348 mmol

mass acetic acid = (moles acetic acid) * (molar mass acetic acid)

mass acetic acid = (7.74348 mmol) * (60.05 g/mol)

mass acetic acid = 465.0 mg

mass acetic acid = 0.465 g

mass % acetic acid = (mass acetic acid / mass sample) * 100

mass % acetic acid = (0.465 g / 11.90 g) * 100

mass % acetic acid = 3.91 %

molarity acetic acid = (moles acetic acid) / (volume of sample)

molarity acetic acid = (7.74348 mmol) / (11.19 mL)

molarity acetic acid = 0.692 M