Consider a different titrant for this exercise. Suppose Ca(OH) 2 were used as th

Question

Consider a different titrant for this exercise. Suppose Ca(OH)2 were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. The analyte will still be HC2H3O2.

(a) What is the stoichiometry of HC2H3O2 to Ca(OH)2? (2:1)

(b) Complete the following table for this titration.

Data Table P1: Titration of acetic acid with calcium hydroxide

please explain how to do this as well so I can do the rest of the problems just like this in the assignment.

concentration of Ca(OH)2 0.351 M volume vinegar solution 11.85 mL mass vinegar solution 11.91 g volume of Ca(OH)2 solution 12.15 mL mmol of Ca(OH)2 _____mmol mmol of HC2H3O2 ______mmol mass of HC2H3O2 _____g mass % of HC2H3O2 in the original sample _____% molarity of HC2H3O2 in the original sample _____M

Explanation / Answer

(a) stoichiometry of HC2H3O2 (CH3COOH) to Ca(OH)2 = 2 : 1

2CH3COOH + Ca(OH)2 ---> 2(CH3COO)2Ca + 2H2O

(b) mmol of Ca(OH)2 = molarity (M) x volume (ml) = 0.351 x 12.15 = 4.265 mmol

mmol of HC2H3O2 = 2 x mmol of Ca(OH)2 = 2 x 4.265 = 8.530 mmol

mass of HC2H3O2 = mmol x molar mass = 8.530 x 60.05 = 512.23 mg

mass % of HC2H3O2 in the original sample = (0.512 g/11.91 g) x 100 = 4.301%

molarity of HC2H3O2 in the original sample = moles/L = 8.530/11.85 = 0.72 M

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