Question

If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (\(K_b\)) of ethylamine is \(10^{-x}\).

The value of x is             (nearest integer).
The degree of ionization of ethylamine can be neglected with respect to unity.

Hint:

The key here is to recognize the ionization equation of ethylamine and use the given pH to find the pOH and, consequently, the concentration of OH\(^-\). The degree of ionization \( \alpha \) is assumed to be very small, so we approximate it as 1 for the calculation.

Answer 1

The reaction for ethylamine ionization in water is: \[ C_2H_5NH_2(aq) + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^- \] Given, \( C = 10^{-3} \, M \) and \( K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]} \). We are also given that the pH = 9, so: \[ \text{pOH} = 5 \quad \text{(since pH + pOH = 14)} \] Therefore, \[ [OH^-] = 10^{-5} \, M \] Now using the approximation \( 1 - \alpha \approx 1 \), where \( \alpha \) is the degree of ionization: \[ C_2H_5NH_2 \approx C = 10^{-3} \, M \] Substituting the values into the \( K_b \) expression: \[ K_b = \frac{(10^{-5})(10^{-5})}{10^{-3}} = 10^{-7} \] Thus, \( K_b = 10^{-7} \), and the value of \( x \) is 7.