How do I find the #K_c#? (Question in description)

1.40mol of propanoic acid (#"CH"_3"CH"_2"COOH"#) and 5.40mol of ethanol are added together at #30^circ"C"#. After the system has come to equilibrium, a portion of the mixture is titrated against #0.4mol# #dm^-3# of #"NaOH"# solution. The titration shows that the whole of the equilibrium mixture requires #149cm^3# of standard alkali for neutralisation. Find the value of #K_c# for the esterification process.

1 Answer
Feb 21, 2018

#K_c~=0.20#

Explanation:

#"H"_3"CCH"_2"C(=O)OH" + "HOCH"_2"CH"_3 stackrel(Delta)rarr"H"_3"CCH"_2"C(=O)OCH"_2"CH"_3+"H"_2"O"#

We started with #0.140*mol# of the acid and excess ethanol....and we finished with ....

#0.40*mol*dm^-3xx149*cm^3xx10^-3*dm^3*cm^-3=0.0596*mol# with respect to the carboxylic acid....

And therefore....

#underbrace(0.140*mol)_"starting moles of acid"-0.0596*mol=underbrace(0.0804*mol)_"moles of acid that reacted"#

And these #"moles of acid that reacted"# are equal to the #"moles of ethanol that reacted"#, and the #"moles of ester that were produced"#

And since #K_c=({"moles of ester"}{"moles of water"}xx1/V^2)/({"moles of acid"}{"moles of alcohol"}xx1/V^2)#

.....we gots...#K_c-=(0.0804)^2/[(0.0596)(0.540-0.0804)]=0.24#

Do you follow?