Question

If 16.0 mL of a 1.50M solution of butyric acid in diethyl ether is mixed with 16.0 mL of a 1.50M solution of ethanol also in diethyl ether, what will be the equilibrium concentration of ethyl butrate?

Answer 1

we know that

moles = molarity x volume (ml)

so

moles of butyric acid = 1.5 x 16 x 10-3

moles of butyric acid = 24 x 10-3

now

moles of ethanol taken = 1.5 x 16 x 10-3

so

moles of ethanol taken = 24 x 10-3

now

the reaction is

butyric acid + ethanol ----> ethyl butyrate + water

we can see that

moles of ethanol reacted = moles of butyric acid taken

so

moles of ethanol reacted = moles of butyric acid taken = 24 x 10-3

now

butyric acid + ethanol ----> ethyl butyrate + water

we can see that

moles of ethyl butyrate formed = moles of ethanol reacted

so

moles of ethyl butyrate formed = 24 x 10-3

now

final volume = 16 + 16 = 32 ml

now

we know that

concentration = moles x 1000 / volume (ml)

so

conc of ethyl butyrate = 24 x 10-3 x 1000 / 32

conc of ethyl butyrate = 0.75 M

so

the concentration of ethyl butyrate is 0.75 M