The pH of the `10^(-3)M NH_(4)OH(Kb = 10^(-5))` is

Options

A. 3
B. 11
C. 10
D. 4

Explanation

To find the pH of a `10^(-3) M NH4OH` solution, we can follow these steps:

### Step 1: Understand the dissociation of NH4OH
NH4OH (ammonium hydroxide) is a weak base that dissociates in water as follows:
\[ \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

### Step 2: Set up the equilibrium expression
Let the initial concentration of NH4OH be \( C = 10^{-3} \, M \) and let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will be:
- \([\text{NH}_4^+] = C\alpha\)
- \([\text{OH}^-] = C\alpha\)
- \([\text{NH}_4\text{OH}] = C - C\alpha \approx C\) (since \( \alpha \) is small)

### Step 3: Write the expression for Kb
The base dissociation constant \( K_b \) for NH4OH is given by:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4\text{OH}]} \]
Substituting the equilibrium concentrations:
\[ K_b = \frac{(C\alpha)(C\alpha)}{C - C\alpha} \approx \frac{C\alpha^2}{C} = C\alpha^2 \]

### Step 4: Solve for alpha
Rearranging gives:
\[ \alpha^2 = \frac{K_b}{C} \]
Thus,
\[ \alpha = \sqrt{\frac{K_b}{C}} \]

### Step 5: Substitute the values
Given \( K_b = 10^{-5} \) and \( C = 10^{-3} \):
\[ \alpha = \sqrt{\frac{10^{-5}}{10^{-3}}} = \sqrt{10^{-2}} = 10^{-1} = 0.1 \]

### Step 6: Calculate the concentration of OH⁻
The concentration of hydroxide ions \([\text{OH}^-]\) is:
\[ [\text{OH}^-] = C\alpha = (10^{-3})(0.1) = 10^{-4} \, M \]

### Step 7: Calculate pOH
Using the concentration of hydroxide ions, we can find pOH:
\[ \text{pOH} = -\log [\text{OH}^-] = -\log (10^{-4}) = 4 \]

### Step 8: Calculate pH
Finally, we can find pH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} = 14 - 4 = 10 \]

### Final Answer
The pH of the `10^(-3) M NH4OH` solution is **10**.

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