Calculate the #"pH"# for #"0.270 M"# benzoic acid? #K_a = 6.5 xx 10^(-5)#

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"" ""0.270 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M"#
#"C"" "-x" "" "" "" "-" "" "" "+x" "" "" "+x#
#"E"" ""(0.270 - x) M"" "-" "" "" "x" ""M"" "" "x" ""M"#

where #"HA"# is benzoic acid and #"A"^(-)# is therefore benzoate. So, its equilibrium expression (its mass action expression) would be:

#K_a = (["H"_3"O"^(+)]["A"^(-)])/(["HA"]) = (x^2)/(0.270 - x)#

In the small #x# approximation, we say that #x# #"<<"# #0.270#, i.e. that #0.270 - x ~~ 0.270#. Therefore:

#0.270K_a ~~ x^2#

#=> x ~~ sqrt(0.270stackrel(6.5 xx 10^(-5))overbrace(K_a)) = 4.19 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

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

#= (4.19 xx 10^(-3) "M")/("0.270 M") xx 100%#

#~~ color(blue)(1.55%)#

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 #4.16 xx 10^(-3) "M"#... close enough. The true percent dissociation would then be #1.53_96% ~~ 1.54%#.

Therefore, the #"pH"# to a good approximation is

#color(blue)("pH") = -log(4.19 xx 10^(-3) "M") ~~ 2.377_9 ~~ color(blue)(2.38)#