Question

buffer solution contains 0.2 mole of NH OH and 0.25 mole of NH CI per litre. Calculate the pH of this solution if dissociation constant of NH OH at this temperature is 1.81 x 10.5

Answer 1

To calculate the pH of the buffer solution containing \( \text{NH}_4\text{OH} \) (ammonium hydroxide) and \( \text{NH}_4\text{Cl} \) (ammonium chloride), we will use the Henderson-Hasselbalch equation given by:

\[
\text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)
\]

Where:
- \([\text{base}]\) is the concentration of the weak base (in this case, \( \text{NH}_4\text{OH} \)),
- \([\text{acid}]\) is the concentration of the conjugate acid (in this case, \( \text{NH}_4^+ \) from \( \text{NH}_4\text{Cl} \)),
- \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant.

1. **Identify the concentrations**:
- Concentration of \( \text{NH}_4\text{OH} \) (base) = 0.2 M
- Concentration of \( \text{NH}_4\text{Cl} \) (acid) = 0.25 M

2. **Find the dissociation constant \( K_b \)**:
The dissociation constant of \( \text{NH}_4\text{OH} \) is given as \( K_b = 1.81 \times 10^{-5} \).

3. **Find \( K_a \)**:
The relation between \( K_a \) and \( K_b \) for a conjugate acid-base pair can be given by the water dissociation constant:

\[
K_w = K_a \cdot K_b
\]

At \( 25^\circ \text{C} \), \( K_w = 1.0 \times 10^{-14} \).

Thus:
\[
K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.81 \times 10^{-5}} \approx 5.52 \times 10^{-10}
\]

4. **Calculate \( \text{pK}_a \)**:
\[
\text{pK}_a = -\log(K_a) \approx -\log(5.52 \times 10^{-10}) \approx 9.26
\]

5. **Plug into Henderson-Hasselbalch equation**:
\[
\text{pH} = 9.26 + \log \left( \frac{0.2}{0.25} \right)
\]
\[
= 9.26 + \log (0.8)
\]
\[
\log (0.8) \approx -0.0969
\]

6. **Calculate pH**:
\[
\text{pH} = 9.26 - 0.0969 \approx 9.16
\]

Therefore, the pH of the solution is approximately **9.16**.