The pH of 0.05 M aqueous solution of diethyl amine is 12.0. The `K_(b) = x xx 10^(-3)` then what is x value?

Answer

To find the value of \( x \) in the expression \( K_b = x \times 10^{-3} \) for diethyl amine, we will follow these steps:

### Step 1: Calculate the pOH
Given that the pH of the solution is 12.0, we can find the pOH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
Substituting the known pH:
\[ \text{pOH} = 14 - 12 = 2 \]

### Step 2: Calculate the concentration of hydroxide ions \([OH^-]\)
Using the pOH, we can find the concentration of hydroxide ions:
\[ [OH^-] = 10^{-\text{pOH}} = 10^{-2} = 0.01 \, \text{M} \]

### Step 3: Set up the equilibrium expression
For the dissociation of diethyl amine (\( C_2H_5)_2NH \)):
\[ (C_2H_5)_2NH + H_2O \rightleftharpoons (C_2H_5)_2NH_2^+ + OH^- \]
Let \( x \) be the concentration of \( (C_2H_5)_2NH_2^+ \) and \( OH^- \) at equilibrium. Initially, we have:
- \([C_2H_5)_2NH] = 0.05 \, \text{M}\)
- \([OH^-] = 0\)

At equilibrium:
- \([C_2H_5)_2NH] = 0.05 - x\)
- \([C_2H_5)_2NH_2^+] = x\)
- \([OH^-] = x\)

From our previous calculation, we know \( x = 0.01 \, \text{M} \).

### Step 4: Substitute values into the \( K_b \) expression
The expression for \( K_b \) is:
\[ K_b = \frac{[(C_2H_5)_2NH_2^+][OH^-]}{[(C_2H_5)_2NH]} \]
Substituting the equilibrium concentrations:
\[ K_b = \frac{(0.01)(0.01)}{0.05 - 0.01} = \frac{0.0001}{0.04} = 0.0025 \]

### Step 5: Convert \( K_b \) to the required format
To express \( K_b \) in the form \( x \times 10^{-3} \):
\[ K_b = 0.0025 = 2.5 \times 10^{-3} \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{2.5} \]