equilibrium constant data lab questions Hopefully, someone can help me solve the

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equilibrium constant data lab questions

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The amount of acetic acid remaining after equilibrium is reached will be determined by titrating the reaction mixture and the reaction mixture at equilibrium with standardized sodium hydroxide Acetic acid reacts with sodium hydroxide on a 1:1 mole ratio as shown below. roxide solution. the initial solution CH3COOH + NaOH CH3COONa + H2O in the two titrations Since the molarity of the NaOH is known, the difference in the volume of NaOH used can be converted to moles of NaOH. The difference in moles of NaOH is equal to the moles converted to product in the reaction. It is also equal to the moles of propyl acetate and equilibrium mixture. With this information the equilibrium constant can be calculat of acetic acid water formed in the

Explanation / Answer

The amount of NaOH used = 6M 8.4 mL

it contains, (6/1000) * 8.4 = 0.05 moles

since molarity is the number of moles of solute in 1000 mL

Initial amount of acetic acid = 6 g (5.7 mL)

molar mass of acetic acid = 60 g/mol

number of moles of acetic acids = 6/60 = 0.1 moles

At equilibrium, 0.05 moles of acetic acid exists

ie 0.05 moles got consumed.

this will yield an equal number of moles of propyl acetate ie 0.05 moles (similarly for water also)

amount of propyl alcohol taken = 6 g

molar mass of propyl alcohol = 60

number of moles used = 6/60 = 0.1

amount consumed = 0.05 moles or at equilibria 0.05 moles exists

Equilibrium constant = [propyl acetate] [water]/ [acetic acid] [propyl alcohol]

Here we can substitute the number of moles instead of concentration

Equilibrium constant = 0.05 x 0.05/ 0.05 x 0.05 = 1

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