If 1 mM solution of ethylamine produces pH=9, then the ionization constant $$(K_{b})$$ of ethylamine is $$10^{-x}$$. The value of is ______ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]

Given a 1 mM solution of ethylamine produces pH = 9.

Since ethylamine is a base, we find pOH first: $$ pOH = 14 - pH = 14 - 9 = 5 $$ and $$ [OH^-] = 10^{-5} \text{ M} $$.

The ionization of ethylamine in water is represented by the equilibrium $$ C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^- $$.

The base‐ionization constant is given by $$ K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]} $$.

Assuming the degree of ionization is negligible compared to unity, we have $$ K_b = \frac{[OH^-]^2}{C} $$ where $$ C = 1 \text{ mM} = 10^{-3} \text{ M} $$.

Substituting the concentration of hydroxide ions, $$ K_b = \frac{(10^{-5})^2}{10^{-3}} = \frac{10^{-10}}{10^{-3}} = 10^{-7} $$.

Thus $$ K_b = 10^{-x} $$ where $$ x = 7 $$.

The answer is 7.