Question

Given `K_a = 5.90 xx 10^(-2)` for `"0.530 M"` lactic acid, what is the `["H"^+]` at equilibrium?

Answer 1

Now, this is actually something that requires the full ICE table. The `K_(a1)` is not small enough to use the small x approximation, so...

`"H"_2"A"(aq) + "H"_2"O"(l) rightleftharpoons "HA"^(-) + "H"_3"O"^(+)(aq)`

`"I"" ""0.530 M"" "" "" "" "" ""0.00 M"" ""0.00 M"`
`"C"" ""-x M"" "" "" "" "" "" ""+x M"" ""+x M"`
`"E"" ""0.530-x M"" "" "" "" ""x M"" "" ""x M"`

You still have to go through with the quadratic equation.

`K_(a1) = 5.90 xx 10^(-2) = (["HA"^(-)]_(eq)["H"_3"O"^(+)]_(eq))/(["H"_2"A"]_(eq))`

`= x^2/(["H"_2"A"] - x)`

Solving for the correct equation form, we obtain a general quadratic equation for one-proton dissociations:

`x^2 + K_(a1)x - K_(a1)["H"_2"A"] = 0`

You can see that had we assumed `x` was small, this still reduces to the correct expression for `x`:

`x^2 + cancel(K_(a1)x)^("small???") - K_(a1)["H"_2"A"] = 0`

`=> color(red)(x) ~~ sqrt(K_(a1)["H"_2"A"]) ~~ color(red)"0.177 M" => 33.4%` dissociation
(upon rejecting the nonphysical `x`)

This is a great example of how NOT to use the small `x` approximation, because it gives too large of an error (the `"pH"` would have been `0.752` under this failed approximation). This acid dissociates to one third of its original concentration...

In full then, the true answer is:

`x = (-K_(a1) pm sqrt(K_(a1)^2 - 4(1)(-K_(a1)["H"_2"A"])))/(2(1))`

You could plug stuff in right now... but here's an easier formula to use on many weak-acid one-proton dissociation equilibria:

`bbx = (-K_(a1) pm sqrt(K_(a1)^2 + 4K_(a1)["H"_2"A"]))/(2)`

`= (-K_(a1) pm 2sqrt((K_(a1)/2)^2 + K_(a1)["H"_2"A"]))/(2)`

`= bb(-K_(a1)/2 pm sqrt((K_(a1)/2)^2 + K_(a1)["H"_2"A"]))`

So, your true `x` is, upon rejecting the nonphysical `x`:

`color(blue)(x = ["H"^(+)] = "0.150 M")`