The ionization constant \( K_a \) for acetic acid is defined as:
\[
K_a = \frac{[ \text{H}^+ ] [ \text{CH}_3\text{COO}^- ]}{[ \text{CH}_3\text{COOH} ]}
\]
At equilibrium, the concentration of \( \text{CH}_3\text{COOH} \) is approximately \( 0.1 - x \), where \( x \) represents the concentration of \( \text{H}^+ \) ions. Given that \( [ \text{H}^+ ] = 1.0 \times 10^{-3} \) M, we can approximate \( [ \text{CH}_3\text{COO}^- ] = [ \text{H}^+ ] = 1.0 \times 10^{-3} \) M and \( [ \text{CH}_3\text{COOH} ] \approx 0.1 \) M.
Substituting these values into the equation yields:
\[
K_a = \frac{(1.0 \times 10^{-3}) (1.0 \times 10^{-3})}{0.1} = 1.0 \times 10^{-5}
\]
Therefore, the ionization constant \( K_a \) for acetic acid is \( 1.0 \times 10^{-5} \). The correct option is (1).