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

##### 1 Answer

Now, this is actually something that requires the full ICE table. The

#"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^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

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

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