For #"1.25 M"# of acetic acid (#K_a = 1.8 xx 10^(-5)#), determine the percent dissociation?

Since #K_a# #~# #10^(-5)#, we expect to be able to use the small x approximation to simplify calculations. However, this is somewhat borderline, unless you already know what average concentration gives you a small percent dissociation.

Write out the dissociation reaction of a weak acid in water and construct its ICE table:

#"HA"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "A"^(-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" ""1.25 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M"#
#"C"" "-x" "" "" "" "-" "" "" "+x" "" "" "+x#
#"E"" ""(1.25 - x) M"" "-" "" "" "x" ""M"" "" "x" ""M"#

where #"HA"# is acetic acid and #"A"^(-)# is therefore acetate.

So, its equilibrium expression (its mass action expression) would be:

#K_a = (["H"_3"O"^(+)]["A"^(-)])/(["HA"]) = (x^2)/(1.25 - x)#

In the small #x# approximation, we say that #x# #"<<"# #1.25#, i.e. that #1.25 - x ~~ 1.25#. Therefore:

#1.25K_a ~~ x^2#

#=> x ~~ sqrt(1.25stackrel(1.8 xx 10^(-5))overbrace(K_a)) = 4.74 xx 10^(-3)# #"M" = ["H"^(+)] = ["H"_3"O"^(+)]#

(In general, under the small #x# approximation, #color(green)(x ~~ sqrt(["HA"]K_a))#.)

The percent dissociation is then

#color(blue)(%" dissoc") = (["HA"]_(lost))/(["HA"]_i) = x/(["HA"]_i)#

#= (4.74 xx 10^(-3) "M")/("1.25 M") xx 100%#

#~~ color(blue)(0.38%)#

Another measure to determine whether the small #x# approximation is valid is if the percent dissociation is under #5%#... and since it obviously is, we don't have to check the true answer (where we don't say #x# #"<<"# #["HA"]#).

However, the true #x# via the quadratic formula would have been #4.73 xx 10^(-3) "M"#... close enough. The true percent dissociation would then be #0.37_9% ~~ 0.38%#.