Question

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

Answer 1

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%`.