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

Let ethylamine be denoted as EtNH₂. It reacts with water according to the following equilibrium: \[EtNH_2(aq) + H_2O(l) \rightleftharpoons EtNH_3^+(aq) + OH^-(aq)\]

1. Given Information:
The pH of a 1 mM solution of ethylamine is 9. This means the pOH is 14-9=5. 

Therefore, \([OH^-] = 10^{-5} M\).

Also, the initial concentration of ethylamine is 1 mM = \(10^{-3} M\).

2. Equilibrium Expression:
The equilibrium expression for the base ionization constant is: \[K_b = \frac{[EtNH_3^+][OH^-]}{[EtNH_2]}\]

3. Approximations:
Since the degree of ionization is small, we can approximate that \([EtNH_2] \approx\) initial concentration of \(EtNH_2\) which is \(10^{-3} M\).

Also, since one mole of \(EtNH_3^+\) and \(OH^-\) are formed for every mole of \(EtNH_2\) that ionizes, we have \([EtNH_3^+] = [OH^-] = 10^{-5} M\).

4. Calculating K_b:
So, \[K_b = \frac{(10^{-5})(10^{-5})}{10^{-3}} = \frac{10^{-10}}{10^{-3}} = 10^{-7}\]

5. Final Value:
Given that \(K_b = 10^{-x}\), we have \(x = 7\).

Final Answer:
The final answer is $7$.