# Question d408a

Apr 17, 2015

The generic equilibrium reaction for a weak acid is

$H {A}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{-} + {A}_{\left(a q\right)}^{-}$

Start from the pH of the solution. You can use the pH to determine the concentration of ${H}^{+}$ in solution by

[H^(+)] = 10^(-pH_"sol") = 10^-3.22 = "0.0006026 M"

Calculate percent ionization by dividing the concentration of hydrogen ions produced in solution by the initial concentration of the weak acid, and multiplying by 100

$\text{% ionization} = \frac{\left[{H}^{+}\right]}{\left[H A\right]} \cdot 100$

"% ionization" = (0.0006026cancel("M"))/(0.2cancel("M")) * 100 = color(green)("0.3%")#

The acid dissociation constant is determined by using the equilibrium concentrations of all the species involved in the reaction.

Since you have $1 : 1$ mole ratios between all the species, you can say that the concentration of $H A$ decreased by the same amount as the concentrations of ${H}^{+}$ and ${A}^{-}$ increased.

This means that the equilibrium concentrations for all three species will be

$\left[{H}^{+}\right] = \left[{A}^{-}\right] = \text{0.0006020 M}$
$\left[H A\right] = {\left[H A\right]}_{0} - \left[{H}^{+}\right] = 0.2 - 0.0006026 = \text{0.19940 M}$

By definition, ${K}_{a}$ will be

${K}_{a} = \frac{\left[{H}^{+}\right] \cdot \left[{A}^{-}\right]}{\left[H A\right]}$

${K}_{a} = \frac{0.0006026 \cdot 0.0006026}{0.19940} = \textcolor{g r e e n}{1.82 \cdot {10}^{- 6}}$

Calculate $p {K}_{a}$ by using

$p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(1.82 \cdot {10}^{- 6}\right) = \textcolor{g r e e n}{5.74}$