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

Mar 4, 2017

Now, this is actually something that requires the full ICE table. The ${K}_{a 1}$ is not small enough to use the small x approximation, so...

${\text{H"_2"A"(aq) + "H"_2"O"(l) rightleftharpoons "HA"^(-) + "H"_3"O}}^{+} \left(a q\right)$

$\text{I"" ""0.530 M"" "" "" "" "" ""0.00 M"" ""0.00 M}$
$\text{C"" ""-x M"" "" "" "" "" "" ""+x M"" ""+x M}$
$\text{E"" ""0.530-x M"" "" "" "" ""x M"" "" ""x M}$

You still have to go through with the quadratic equation.

${K}_{a 1} = 5.90 \times {10}^{- 2} = \left({\left[\text{HA"^(-)]_(eq)["H"_3"O"^(+)]_(eq))/(["H"_2"A}\right]}_{e q}\right)$

$= {x}^{2} / \left(\left[\text{H"_2"A}\right] - x\right)$

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

${x}^{2} + {K}_{a 1} x - {K}_{a 1} \left[\text{H"_2"A}\right] = 0$

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

${x}^{2} + {\cancel{{K}_{a 1} x}}^{\text{small???") - K_(a1)["H"_2"A}} = 0$

=> 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 $\text{pH}$ 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:

$x = \frac{- {K}_{a 1} \pm \sqrt{{K}_{a 1}^{2} - 4 \left(1\right) \left(- {K}_{a 1} \left[\text{H"_2"A}\right]\right)}}{2 \left(1\right)}$

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

$\boldsymbol{x} = \frac{- {K}_{a 1} \pm \sqrt{{K}_{a 1}^{2} + 4 {K}_{a 1} \left[\text{H"_2"A}\right]}}{2}$

$= \frac{- {K}_{a 1} \pm 2 \sqrt{{\left({K}_{a 1} / 2\right)}^{2} + {K}_{a 1} \left[\text{H"_2"A}\right]}}{2}$

$= \boldsymbol{- {K}_{a 1} / 2 \pm \sqrt{{\left({K}_{a 1} / 2\right)}^{2} + {K}_{a 1} \left[\text{H"_2"A}\right]}}$

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

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