Question

calculate the pH of a buffer solution obtained by dissolving 18.0g of KH2PO4 (s) and 34.0g of Na2HPO4 (s) in water and then diluting to 1.00L

Answer 1

To calculate the pH of a buffer solution made from \( KH_2PO_4 \) (potassium dihydrogen phosphate) and \( Na_2HPO_4 \) (disodium hydrogen phosphate), we can use the Henderson-Hasselbalch equation.

### Step 1: Calculate the number of moles of each component

1. **Moles of \( KH_2PO_4 \)**:
- Molar mass of \( KH_2PO_4 \):
\[
K = 39.10 \, \text{g/mol}, \, H = 1.01 \, \text{g/mol}, \, P = 30.97 \, \text{g/mol}, \, O = 16.00 \, \text{g/mol}
\]
\[
\text{Molar mass} = 39.10 + 2(1.01) + 30.97 + 4(16.00) = 136.09 \, \text{g/mol}
\]
- Moles of \( KH_2PO_4 \):
\[
\text{Moles} = \frac{18.0 \, \text{g}}{136.09 \, \text{g/mol}} \approx 0.132 \, \text{mol}
\]

2. **Moles of \( Na_2HPO_4 \)**:
- Molar mass of \( Na_2HPO_4 \):
\[
Na = 22.99 \, \text{g/mol}, \, H = 1.01 \, \text{g/mol}, \, P = 30.97 \, \text{g/mol}, \, O = 16.00 \, \text{g/mol}
\]
\[
\text{Molar mass} = 2(22.99) + 1.01 + 30.97 + 4(16.00) = 141.96 \, \text{g/mol}
\]
- Moles of \( Na_2HPO_4 \):
\[
\text{Moles} = \frac{34.0 \, \text{g}}{141.96 \, \text{g/mol}} \approx 0.239 \, \text{mol}
\]

### Step 2: Find the concentrations in a 1.00 L solution

- **Concentration of \( KH_2PO_4 \)**:
\[
[KH_2PO_4] = \frac{0.132 \, \text{mol}}{1.00 \, \text{L}} = 0.132 \, \text{M}
\]

- **Concentration of \( Na_2HPO_4 \)**:
\[
[Na_2HPO_4] = \frac{0.239 \, \text{mol}}{1.00 \, \text{L}} = 0.239 \, \text{M}
\]

### Step 3: Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by:
\[
pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)
\]
Where:
- \( [A^-] \) is the concentration of the base form \( Na_2HPO_4 \).
- \( [HA] \) is the concentration of the acid form \( KH_2PO_4 \).
- \( pK_a \) of \( H_2PO_4^- \) (the conjugate acid of \( HPO_4^{2-} \)) is approximately 7.2 (you may use a more precise value based on specific literature).

### Step 4: Substitute values into the equation

\[
pH = 7.2 + \log\left(\frac{0.239}{0.132}\right)
\]
Calculating the ratio:
\[
\frac{0.239}{0.132} \approx 1.808
\]
Now calculating the logarithm:
\[
\log(1.808) \approx 0.257
\]
Now, substituting back into the equation:
\[
pH = 7.2 + 0.257 \approx 7.457
\]

### Final Result

The pH of the buffer solution is approximately \( 7.46 \).