If the dissociation constant of `5 xx 10^(-4)` M aqueous solution of diethylamine is `2.5 xx 10^(-5)`, its pH value is

Options

A. 8
B. 3.95
C. 10.05
D. 2

Answer

To find the pH value of a 5 x 10^(-4) M aqueous solution of diethylamine with a dissociation constant (Kb) of 2.5 x 10^(-5), we can follow these steps:

### Step 1: Write the dissociation equation
Diethylamine (C2H5)2NH dissociates in water as follows:
\[ (C2H5)2NH + H2O \rightleftharpoons (C2H5)2NH2^+ + OH^- \]

### Step 2: Set up the equilibrium expression
The base dissociation constant (Kb) is given by the expression:
\[ K_b = \frac{[(C2H5)2NH2^+][OH^-]}{[(C2H5)2NH]} \]

### Step 3: Define initial concentrations and changes
Let the initial concentration of diethylamine be \( C = 5 \times 10^{-4} \) M. At equilibrium, if \( \alpha \) is the degree of dissociation, then:
- Concentration of \((C2H5)2NH\) at equilibrium = \( C(1 - \alpha) \)
- Concentration of \((C2H5)2NH2^+\) at equilibrium = \( C\alpha \)
- Concentration of \(OH^-\) at equilibrium = \( C\alpha \)

### Step 4: Substitute into the Kb expression
Substituting the equilibrium concentrations into the Kb expression gives:
\[ K_b = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]
Given \( K_b = 2.5 \times 10^{-5} \) and \( C = 5 \times 10^{-4} \), we can simplify this expression.

### Step 5: Approximate \(1 - \alpha\)
Since \( K_b \) is small, we can assume \( \alpha \) is small compared to 1, thus \( 1 - \alpha \approx 1 \):
\[ K_b \approx C\alpha^2 \]
Substituting the values:
\[ 2.5 \times 10^{-5} = (5 \times 10^{-4})\alpha^2 \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha^2 = \frac{2.5 \times 10^{-5}}{5 \times 10^{-4}} = 0.05 \]
Taking the square root:
\[ \alpha = \sqrt{0.05} \approx 0.2236 \]

### Step 7: Calculate the concentration of \(OH^-\)
Now, we can find the concentration of \(OH^-\):
\[ [OH^-] = C\alpha = (5 \times 10^{-4})(0.2236) \approx 1.118 \times 10^{-4} \text{ M} \]

### Step 8: Calculate pOH
Now we can calculate pOH:
\[ pOH = -\log[OH^-] = -\log(1.118 \times 10^{-4}) \approx 3.95 \]

### Step 9: Calculate pH
Finally, we can find pH using the relation \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 3.95 \approx 10.05 \]

### Final Answer
The pH value of the solution is approximately **10.05**.

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