Answer
$K_a = 2.2 \times 10^{-7} $
Work Step by Step
$$[HX] = \frac{0.250 \space mol}{655 \space mL} \times \frac{1000 \space mL}{1 \space L} = 0.382 \space M$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HX ]& [ X^- ]& [ H_3O^+ ]\\
Initial& 0.382 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.382 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ X^- ][ H^+ ]}{[ HX ]}$$
$$K_a = \frac{(x)(x)}{[ HX ]_{initial} - x}$$
3. Using the pH, find the $H_3O^+$ concentration:
$$[H_3O^+] = 10^{-pH} = 10^{-3.54} = 2.9 \times 10^{-4} \space M$$
$[H_3O^+] = x = 2.9 \times 10^{-4} $
4. Substitute the value of x and calculate the $K_a$:
$$K_a = \frac{( 2.9 \times 10^{-4} )^2}{ 0.382 - 2.9 \times 10^{-4} }$$
$K_a = 2.2 \times 10^{-7} $
