The pH of a 0.100 M solution of an aqueous weak acid (#HA#) is 4. What is the #Ka# for the weak acid?
1 Answer
Explanation:
Start by writing a balanced chemical equation for the partial ionization of the acid
#"HA"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "A"_text((aq])^(-) + "H"_3"O"_text((aq])^(+)#
Notice that you have
In other words, the equation produces equal concentrations of conjugate base and hydronium ions.
Now, you can use the pH of the solution to calculate the equilibrium concentration of the hydronium ions.
#color(blue)("pH" = - log(["H"_3"O"^(+)]) implies ["H"_3"O"^(+)] = 10^(-"pH"))#
In your case, the pH of the solution is equal to
#["H"_3"O"^(+)] = 10^(-4)"M"#
By definition, the acid dissociation constant,
#K_a = (["A"^(-)] * ["H"_3"O"^(+)])/(["HA"])#
The expression for the acid dissociation constant is written using equilibrium concentrations. So, if the reaction produced a concentration of hydronium ions equal to
Because the initial concentration of the acid is considerably higher than the concentrations of the conjugate base and hydronium ions, you can approximate it to be constant.
This means that the acid dissociation constant for this acid will be
#K_a = (10^(-4) * 10^(-4))/0.100 = color(green)(1.0 * 10^(-7))#
This is the underlying concept behind an ICE table
#" ""HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "A"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#
Here
#K_a = (10^(-4) * 10^(-4))/(0.100 - 10^(-4))#
Once again, you can use
#0.100 - 10^(-4) = 0.0999 ~~ 0.100#
to get
#K_a = 10^(-8)/0.100 = color(green)(1.0 * 10^(-7))#