Question #b2abc

1 Answer
Aug 14, 2016

Answer:

Here's what I got.

Explanation:

The idea here is that the number of moles of strong base added to the equilibrium mixture will give you the number of moles of acetic acid present at equilibrium.

The balanced chemical equation that describes this esterification reaction looks like this

#"CH"_ 3"CH"_ 2"OH"_ ((l)) + "CH"_ 3"COOH"_ ((l)) rightleftharpoons "CH"_ 3"COOCH"_ 2"CH"_ (3(l)) + "H"_ 2"O"_ ((l))#

Now, after the equilibrium is established, the solution will contain a number of moles of acetic acid. When you titrate this solution with sodium hydroxide, a strong base can be represented by #"OH"^(-)#, a neutralization reaction takes place

#"CH"_ 3"COOH"_ ((l)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

Notice that you have a #1:1# mole ratio between acetic acid and the hydroxide anions. Use the molarity and volume of the sodium hydroxide solution used for the titration to calculate how many moles of hydroxide anions were consumed by this reaction

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have -- kepp in mind that #"1 mL " = "1 cm"^3# !

#n_("OH"^(-)) = "1.0 mol" color(red)(cancel(color(black)("dm"^(-3)))) * overbrace(165 * 10^(-3)color(red)(cancel(color(black)("dm"^3))))^(color(blue)("the volume in dm"^3))#

#n_("OH"^(-)) = "0.165 moles OH"^(-)#

This means that the equilibrium mixture must have contained

#0.165 color(red)(cancel(color(black)("moles OH"^(-)))) * ("1 mole CH"_3"COOH")/(1color(red)(cancel(color(black)("mole OH"^(-))))) = "0.165 moles CH"_3"COOH"#

Now you're ready to set up an ICE table to find the value of the equilibrium constant, #K_c#, for the initial reaction. For the sake of simplicity, I'll use

#"CH"_ 3"CH"_ 2"OH"_ ((l)) -> "EtOH"_ ((l))#

#"CH"_ 3"COOH"_ ((l)) -> "HAc"_ ((l))#

#"CH"_ 3"COOCH"_ 2"CH"_ (3(l)) -> "EtOAc"_ ((l))#

Remember that

#" " "EtOH"_ ((l)) + "HAc"_ ((l)) rightleftharpoons "EtOAc"_ ((l)) + "H"_ 2"O"_ ((l))#

#color(purple)("I")color(white)(aaaaacolor(black)(0.2)aaaaaaacolor(black)(0.2)aaaaaaaaacolor(black)(0)aaaaaaaacolor(black)(0)#
#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaacolor(black)((-x))aaaaacolor(black)((+x))aaaaacolor(black)((+x))#
#color(purple)("E")color(white)(aaacolor(black)(0.2-x)aaaacolor(black)(0.2-x)aaaaaaacolor(black)(x)aaaaaaaacolor(black)(x)#

Now, you know that your equilibrium mixture contained #0.165# moles of acetic acid. This means that

#0.2 -x = 0.165 implies x= 0.035#

Therefore, the reaction consumed

#"0.035 moles EtOH "# and #" " "0.035 moles HAc"#

and produced

#"0.035 moles EtOAc "# and #" " "0.035 moles H"_2"O"#

You can now say thatbefore the addition of the strong base, the equilibrium mixture contained

#{("0.165 moles EtOH"), ("0.165 moles HAc"), ("0.035 moles EtOAc"), ("0.035 moles H"_2"O"):}#

By definition, the equilibrium constant for this reaction will be

#K_c = (["EtOAc"] * ["H"_2"O"])/(["EtOH"] * ["HAc"])#

It's important to realize that the equilibrium concentration of water must also be added to the expression of the equilibrium constant because this reaction is not taking place in aqueous solution.

As a result, water's concentration can no longer be considered constant and omitted from the expression of the equilibrium constant.

Since the volume is the same for all species involved in the reaction, you can treat moles and concentration interchangeably. You can thus say that at the particular temperature at which this reaction takes place, you have

#K_c = (0.035 color(red)(cancel(color(black)("mol dm"^(-3)))) * 0.035color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.165color(red)(cancel(color(black)("mol dm"^(-3)))) * 0.165color(red)(cancel(color(black)("mol dm"^(-3))))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.045)color(white)(a/a)|)))#

As you can see, the equilibrium constant is unitless. I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the moles of ethanol and acetic acid.