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 $7$.

Answer 2

Step 1: Write the ionization reaction

\[ CH_3CH_2NH_2 + H_2O \rightleftharpoons CH_3CH_2NH_3^+ + OH^- \]

Step 2: Given data

\[ C = 1\, \text{mM} = 1 \times 10^{-3}\, \text{M} \] \[ \text{pH} = 9 \Rightarrow \text{pOH} = 14 - 9 = 5 \] \[ [OH^-] = 10^{-\text{pOH}} = 10^{-5}\, \text{M} \]

Step 3: Apply the formula for weak base dissociation constant

For a weak base, \[ K_b = \frac{[BH^+][OH^-]}{[B]} \] Assuming ionization is small compared to initial concentration (\( [B] \approx C \)): \[ K_b = \frac{(10^{-5})^2}{10^{-3}} = \frac{10^{-10}}{10^{-3}} = 10^{-7} \]

Final Answer:

\[ K_b = 10^{-7} \] \[ \boxed{x = 7} \]