A solution of #"acetic acid"# that is formally #0.01*mol*L^-1# is #1.43%# dissociated exhibits what value of #K_a#?

1 Answer
Dec 10, 2017

We assess the equilibrium....

#HOAc(aq) + H_2O(l) rightleftharpoonsH_3O^+ + ""^(-)OAc#

Explanation:

And as usual we write the equilibrium expression....

#K_a=([AcO^(-)][H_3O^+])/([HOAc])#...

...and here we have the data to measure #K_a# directly, without all that tedious mucking about with approximations....

#[HOAc]=0.01*mol*L^-1xx98.57%=9.857xx10^-3*mol*L^-1#

#[H_3O^+]=0.01*mol*L^-1xx1.43%=1.43xx10^-4*mol*L^-1#

#[AcO^-]=0.01*mol*L^-1xx1.43%=1.43xx10^-4*mol*L^-1#

And so #K_a=(1.43xx10^-4*mol*L^-1)^2/(9.857xx10^-3*mol*L^-1)=2.074xx10^-6#....

#pK_a=5.68#...this is a little different from the quoted #pK_a=4.76#...but we are still in the right order of magnitude....