# Question 92bdd

Nov 29, 2015

${K}_{a} = 2.4 \cdot {10}^{- 2}$

#### Explanation:

Your strategy here will be to use the measure pH of the solution to determine the concentration of hydronium ions.

Once you know this value, use the weak acid's ionization equilibrium to determine the acid dissociation constant ,${K}_{a}$.

So, to get the concentration of hydronium ions from the solution's pH you need to use the equation

$\textcolor{b l u e}{\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)}$

In your case, a pH of $0.96$ corresponds to a concentration of hydronium ions equal to

["H"_3"O"^(+)] = 10^(-0.96) = "0.1096 M"

Now, use the number of moles of acid and the volume of the solution to determine its concentration

$\textcolor{b l u e}{c = \frac{n}{V}}$

$c = \text{0.25 moles"/(500 * 10^(-3)"L") = "0.50 M}$

When placed in aqueous solution, the weak acid will dissociate according to the following equilibrium reaction

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

Notice that you have a $1 : 1$ mole ratio between ${\text{H"_3"O}}^{+}$ and ${\text{A}}^{-}$. This means that the ionization of the acid produced equal numbers of moles of both ions.

Since all the chemical species share the same volume of solution, you can say that

["H"_3"O"^(+)] = ["A"^(-)] = "0.1096 M"#

By definition, the acid dissociation constant for this reaction will be

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

Plug in your values to get

${K}_{a} = \frac{0.1096 \cdot 0.0196}{0.50} = \textcolor{g r e e n}{2.4 \cdot {10}^{- 2}}$

I'll leave the answer rounded to two sig figs.