Find the pH of `0.001 M NH_3` `(K_b =1.8×10^(−5))`

Explanation

To find the pH of a 0.001 M solution of NH₃ (ammonia), we will follow these steps:

### Step 1: Write the equilibrium reaction
When ammonia (NH₃) dissolves in water, it establishes an equilibrium with water:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

### Step 2: Set up the expression for \( K_b \)
The base dissociation constant \( K_b \) for ammonia is given as \( 1.8 \times 10^{-5} \). The expression for \( K_b \) is:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]

### Step 3: Define the concentrations at equilibrium
Let \( x \) be the concentration of \( \text{OH}^- \) produced at equilibrium. Since \( \text{NH}_4^+ \) is produced in the same amount as \( \text{OH}^- \), we have:
- \([\text{NH}_4^+] = x\)
- \([\text{OH}^-] = x\)
- \([\text{NH}_3] = 0.001 - x \approx 0.001\) (since \( x \) will be very small compared to 0.001)

### Step 4: Substitute into the \( K_b \) expression
Substituting these values into the \( K_b \) expression gives:
\[ 1.8 \times 10^{-5} = \frac{x \cdot x}{0.001} \]
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.001} \]

### Step 5: Solve for \( x^2 \)
Rearranging gives:
\[ x^2 = 1.8 \times 10^{-5} \times 0.001 \]
\[ x^2 = 1.8 \times 10^{-8} \]

### Step 6: Calculate \( x \)
Taking the square root of both sides:
\[ x = \sqrt{1.8 \times 10^{-8}} \]
\[ x \approx 1.34 \times 10^{-4} \text{ M} \]
This value represents the concentration of \( \text{OH}^- \).

### Step 7: Calculate \( pOH \)
Using the concentration of hydroxide ions, we can find \( pOH \):
\[ pOH = -\log[\text{OH}^-] \]
\[ pOH = -\log(1.34 \times 10^{-4}) \]
\[ pOH \approx 3.87 \]

### Step 8: Calculate \( pH \)
Finally, we can find \( pH \) using the relationship:
\[ pH + pOH = 14 \]
\[ pH = 14 - pOH \]
\[ pH = 14 - 3.87 \]
\[ pH \approx 10.13 \]

### Final Answer
The pH of the 0.001 M NH₃ solution is approximately **10.13**.