Answer
$K_a (HOCl) = 2.9 \times 10^{-8} $
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HOCl ]& [ ClO^- ]& [ H_3O^+ ]\\
Initial& 0.150 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.150 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ ClO^- ][ H^+ ]}{[ HOCl ]}$$
$$K_a = \frac{(x)(x)}{[ HOCl ]_{initial} - x}$$
3. Using the pH, find the $H_3O^+$ concentration:
$$[H_3O^+] = 10^{-pH} = 10^{-4.18} = 6.6 \times 10^{-5} \space M$$
$[H_3O^+] = x = 6.6 \times 10^{-5} $
4. Substitute the value of x and calculate the $K_a$:
$$K_a = \frac{( 6.6 \times 10^{-5} )^2}{ 0.150 - 6.6 \times 10^{-5} }$$
$K_a = 2.9 \times 10^{-8} $
