What is the Kb value for a 0.20M solution with a 3.0% ionization? NH3 + H2O <—> NH4^+ + OH^-

To find the K_b value for the given solution, we first need to calculate the concentration of OH^- produced using the initial concentration of NH_3 and the ionization percentage.
Given:
- Initial concentration [NH_3] = 0.20 M
- Ionization percentage = 3.0%

Concentration of OH^-:
\text{Concentration of } OH^- = 0.20 \, \text{M} \times \frac{3.0}{100} = 0.006 \, \text{M}

Since each molecule of NH_3 produces one NH_4^+ ion and one OH^- ion, we have:
K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{0.006 \times 0.006}{0.20} = 1.8 \times 10^{-4}

Therefore, the K_b value for the given 0.20 M solution of NH_3 with a 3.0% ionization is approximately 1.8 \times 10^{-4}.

\boxed{K_b \approx 1.8 \times 10^{-4}}