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

##### 1 Answer

Since

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

#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

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

However, the true