Question

calculate the pH of an ammonia with the molarity of 0.33M

Answer 1

To calculate the pH of a 0.33 M ammonia (NH₃) solution, we first need to recognize that ammonia is a weak base that partially ionizes in water according to the following equilibrium:

\[
\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-
\]

The equilibrium constant for this reaction is known as the base dissociation constant (\(K_b\)). At 25°C, the \(K_b\) for ammonia is approximately \(1.8 \times 10^{-5}\).

1. **Set up the expression for \(K_b\):**
\[
K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}
\]

2. **Define the changes in concentration:**
Let's assume \(x\) is the concentration of \(\text{OH}^-\) (and therefore \(\text{NH}_4^+\)) formed at equilibrium. The initial concentration of \(\text{NH}_3\) is 0.33 M, and at equilibrium, the concentrations will be:
- \([\text{NH}_3] = 0.33 - x\) (approximately 0.33 M since \(x\) is small)
- \([\text{NH}_4^+] = x\)
- \([\text{OH}^-] = x\)

3. **Substituting into the \(K_b\) expression:**
\[
1.8 \times 10^{-5} = \frac{x \cdot x}{0.33 - x} \approx \frac{x^2}{0.33}
\]

4. **Solving for \(x\):**
\[
x^2 = (1.8 \times 10^{-5}) \times 0.33
\]
\[
x^2 = 5.94 \times 10^{-6}
\]
\[
x = \sqrt{5.94 \times 10^{-6}} \approx 2.44 \times 10^{-3} \text{ M}
\]

5. **Now, calculate \([\text{OH}^-]\):**
\[
[\text{OH}^-] = x \approx 2.44 \times 10^{-3} \text{ M}
\]

6. **Calculate the pOH:**
\[
\text{pOH} = -\log_{10}([\text{OH}^-]) = -\log_{10}(2.44 \times 10^{-3}) \approx 2.61
\]

7. **Finally, calculate the pH:**
\[
\text{pH} + \text{pOH} = 14
\]
\[
\text{pH} = 14 - \text{pOH} = 14 - 2.61 \approx 11.39
\]

Thus, the pH of a 0.33 M ammonia solution is approximately **11.39**.