Answer:
A.) [tex]K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
Explanation:
The general Kb expression is:
[tex]K_b = \frac{[HA][OH^-]}{[A^-]}[/tex]
In this equation
-----> Kb = equilibrium constant
-----> [HA] = acid
-----> [A⁻] = base
Since liquids are not included in equilibrium expressions, H₂O should not be present. The products are in the numerator while the reactant are in the denominator. In this reaction, CH₃NH₂ is acting as a base and CH₃NH₃⁺ is acting as an acid.
As such, the expression is:
[tex]K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
