## Chemistry: Molecular Approach (4th Edition)

$K_a = 3.0 \times 10^{-6}$
1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HA ]& [ A^- ]& [ H_3O^+ ]\\ Initial& 0.085 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.085 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ A^- ][ H^+ ]}{[ HA ]}$$ $$K_a = \frac{(x)(x)}{[ HA ]_{initial} - x}$$ 3. Use the percent ionization to find x: $$Percent \space ionization = \frac{x}{[ HA ]_{initial}} \times 100\%$$ $$x = \frac{ 0.59 \% \times 0.085 }{100\%}$$ $x = 5.0 \times 10^{-4}$ 4. Substitute the value of x and calculate the $K_a$: $$K_a = \frac{( 5.0 \times 10^{-4} )^2}{ 0.085 - 5.0 \times 10^{-4} }$$ $K_a = 3.0 \times 10^{-6}$