# The pH of a 0.100 M solution of an aqueous weak acid (HA) is 4. What is the Ka for the weak acid?

Feb 13, 2016

$1.0 \cdot {10}^{- 7}$

#### Explanation:

Start by writing a balanced chemical equation for the partial ionization of the acid

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

Notice that you have $1 : 1$ mole ratios across the board. For every mole of acid that ionizes in aqueous solution, you get one mole of its conjugate base and one mole of hydronium ions, ${\text{H"_3"O}}^{+}$.

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 $4$, which means that you'll have

["H"_3"O"^(+)] = 10^(-4)"M"

By definition, the acid dissociation constant, ${K}_{a}$, will be equal to

${K}_{a} = \left(\left[\text{A"^(-)] * ["H"_3"O"^(+)])/(["HA}\right]\right)$

The expression for the acid dissociation constant is written using equilibrium concentrations. So, if the reaction produced a concentration of hydronium ions equal to ${10}^{- 4} \text{M}$, it follows that it also produced a concentration of conjugate base equal to ${10}^{- 4} \text{M}$.

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

${K}_{a} = \frac{{10}^{- 4} \cdot {10}^{- 4}}{0.100} = \textcolor{g r e e n}{1.0 \cdot {10}^{- 7}}$

This is the underlying concept behind an ICE table

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

color(purple)("I")" " " "0.100" " " " " " " " " " " " " " " "0" " " " " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " "(+x)
color(purple)("E")" "0.100-x" " " " " " " " " " " " " "x" " " " " " " " " " "x

Here $x$ represents the equilibrium concentration for the conjugate acid and hydronium ions. Since you know that $x = {10}^{- 4}$, you will have

${K}_{a} = \frac{{10}^{- 4} \cdot {10}^{- 4}}{0.100 - {10}^{- 4}}$

Once again, you can use

$0.100 - {10}^{- 4} = 0.0999 \approx 0.100$

to get

${K}_{a} = {10}^{- 8} / 0.100 = \textcolor{g r e e n}{1.0 \cdot {10}^{- 7}}$