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

1 Answer
Apr 24, 2017

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)