Answer
$K_a = 9.0 \times 10^{-5} $
Work Step by Step
$$[HY] = \frac{4.85 \times 10^{-3}}{0.095 \space L} = 0.0511 \space M$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HY ]& [ Y^- ]& [ H_3O^+ ]\\
Initial& 0.0511 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.0511 -x& x& x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ Y^- ][ H^+ ]}{[ HY ]}$$
$$K_a = \frac{(x)(x)}{[ HY ]_{initial} - x}$$
3. Using the pH, find the $H_3O^+$ concentration:
$$[H_3O^+] = 10^{-pH} = 10^{-2.68} = 2.1 \times 10^{-3} \space M$$
$[H_3O^+] = x = 2.1 \times 10^{-3} $
4. Substitute the value of x and calculate the $K_a$:
$$K_a = \frac{( 2.1 \times 10^{-3} )^2}{ 0.0511 - 2.1 \times 10^{-3} }$$
$K_a = 9.0 \times 10^{-5} $
