Answer
The $K_a$ for that weak acid is equal to $1.4 \times 10^{-4}$.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HA(aq) + H_2O(l) \lt -- \gt A^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $A^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [A^-] = 0 + x = x$
-$[HA] = [HA]_{initial} - x$
2. The percent dissociation formula is:
$\% dissociation = \frac{x}{[HA]_{initial}} \times 100\%$
$3.0\% = \frac{x}{0.15M} \times 100\%$
- Solve for "x:
$\frac{3.0\% \times 0.15M}{100\%} = x$
$x = 4.5 \times 10^{-3}M$
3. Now, use the Ka value and equation to find the 'x' value.
$K_a = \frac{[H_3O^+][A^-]}{ [HA]}$
$K_a = \frac{(x)(x)}{[HA]_{initial}-x}$
$K_a = \frac{(4.5 \times 10^{-3})^2}{0.15-4.5 \times 10^{-3}} = 1.4 \times 10^{-4}$
