To hydrolyze ethyl acetate, a student measured the mass of 4.440 g of ethyl acet
ID: 492720 • Letter: T
Question
To hydrolyze ethyl acetate, a student measured the mass of 4.440 g of ethyl acetate and transferred the compound into a 125 mL flask labeled, flask 1. To the same flask, the student 5.0 mL of 3M HCI. The student then heated this flask in a water bath for 20 minutes. The density of ethyl acetate is 0.900 g/mL In another flask. labeled flask 2, the student pipet 5.0 mL of 3M HCI and 5.0 mL of water then titrated the solution with 1.4886 M NaOH solution. It required 13.4 mL of NaOH to reach the endpoint. After heating the first flask, and cooling it to room temperature, this solution was titrated with same NaOH solution. The first flask required 24.7 mL of NaOH to reach the endpoint The number of mols of OH- used to titrate flask 1_____ The number of mols of H_3O^+ in flask 1 ____ The number of mols of OH- used in flask 2 ____ The number of mols H3O^+ in flask 2 ____ The number of mols of acetic acid formed _____ The number of mols of ethyl alcohol formed _____ The initial number of mols of ethyl acetate ____ The number of mols of ethyl acetate at equilibrium ____ The concentration of acetic acid and ethyl acetate at equilibrium The concentration of ethyl acetate remaining at equilibrium The equilibrium constant for the reaction is Kc =Explanation / Answer
The HCl must be present to catalyze the reaction. Since this reaction is acid-catalyzed, the rate at which the reaction reaches equilibrium is greatly increased by the presence of this acid. Since the
catalyst is neither a reactant nor a product,the concentration of the catalyst DOES NOT appear in theequilibrium expression
given molarity of NaOH =1.4886M and volume of NaOH used to titrate with flask 1=24.7 mL=0.0247L
1) no. of mole of OH- used to titrate flask1 = M x V1= 1.4886 x 0.0247=0.03676 mole
2) no. of moles of H3O+ in flask 1:
since no. of moles of OH- in flask 1= no. of moles of H3O+ in flask 1= 0.0376 mole
3) no. of moles of OH- used to titrate flask 2:
molarity of NaOH = 1.4886M and volume of NaOH used to titrate flask 2 =13.4mL= 0.0134L
no. of moles of OH- used to titrate flask 2= M x V2= 1.4886x 0.0134=0.0199 mole
4) no. of moles of H3O+ in flask 2:
since moles of OH- in flask 2= moles of H3O+ in flask 2= 0.0199 mole
5) no. of moles of acetic acid formed
moles of NaOH required to titrate flask 1 =0.03676mole
moles of NaOH = moles of acid in flask 1= 0.03676 mole
Remove moles of HCl from total moles acid:
moles of HCl=VHCl x M HCl= 3 x 0.005=0.015 mole
no. of moles of acetic acid= total moles of acid - moles of HCl
=0.03676-0.015= 0.02176 moles
6) no. of moles of ethyl alcohol formed :
the no. of moles ethyl alcohol formed = no. of moles of acetic acid formed =0.02176 moles
7) initial no. of moles of ethyl acetate:
mass of ethyl acetate used= 4.44g and molar mass of ethyl acetate= 88.11g/mL
therefore initial no. of moles of ethyl acetate= 4.44/88.11=0.05039 moles
8) no. of moles of ethyl acetate at equlibrium:
at equlibrium no. of moles of ethyl acetate = initial no. of moles of ethyl acetate-0.02176
=0.05039- 0.02176=0.02863 moles
9) concentration of acetic acid and ethyl acetate at equilibrium:
concentration of acetic acid at equilibrium= no. of moles x molar mass= 0.02176 x 60.05=1.30 M
concentration of ethyl acetate at equilibrium =no. of moles x molar mass= 0.02863 x 88.11=2.522 M
11) concentration of ethyl acetate remaining at equilibrium:
moles of ethyl acetate remaining at equilibrium = initial no. of moles- no. of moles at equilibrium
=0.05039-0.02863=0.02176
therefore concentration of ethyl acetate remaining at equilibrium= no. of moles remaining x molar mass
= 0.02176 x 88.11=1.917 M
12) the equilibrium constant for the reaction:
the equilibrium constant is calculated using the equilibrium moles of acetic acid, Ethyl alcohol, Ethyl acetate, and water
to calculate the moles of water:
mass of water in HCl solution
density of 3M HCl solution =1.05 g/mL
mass of 5mL of 3M HCl: g of HCl soln =density x volume= 1.05 x 5=5.25g
mass of HCl in 3M HCl: M(HCl) x V (HCl) x molar mass (HCl)= 3 x 0.005 x 36.46=0.5469g of HCl in HCl solution
Mass of water in 5 mL of 3M of HCl soln:
mass of water in solution=mass total -mass HCL=5.25-0.5469=4.7g of water in 3M HCl soln
initial moles of water in 5mL soln of 3M HCl= mass of water/ molar mass=4.7/18=0.261moles
no. of moles of water at equilibrium=0.261-0.02176=0.239
Now, equilibrium constant Kc= [ethyl alcohol][acetic acid]/[ethyl acetate][water]
=(0.02176x0.02176)/(0.02863x0.239)=4.734x 10-4/6.842x10-3=6.92x 10-2
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