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)`