To find the pH of the resultant solution when 20 mL of 0.1 M H₃PO₄ is mixed with 20 mL of 0.1 M Na₃PO₄, we can follow these steps:
### Step 1: Determine the moles of each species
- For H₃PO₄:
\[
\text{Moles of H₃PO₄} = \text{Volume (L)} \times \text{Concentration (M)} = 0.020 \, \text{L} \times 0.1 \, \text{M} = 0.002 \, \text{mol}
\]
- For Na₃PO₄:
\[
\text{Moles of Na₃PO₄} = \text{Volume (L)} \times \text{Concentration (M)} = 0.020 \, \text{L} \times 0.1 \, \text{M} = 0.002 \, \text{mol}
\]
### Step 2: Identify the reaction
When H₃PO₄ (a weak acid) reacts with Na₃PO₄ (which provides PO₄³⁻ ions), the following reaction occurs:
\[
\text{H₃PO₄} + \text{PO₄}^{3-} \rightleftharpoons \text{H₂PO₄}^- + \text{HPO₄}^{2-}
\]
### Step 3: Calculate the concentrations after mixing
The total volume of the solution after mixing is:
\[
\text{Total Volume} = 20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL} = 0.040 \, \text{L}
\]
The concentrations of H₃PO₄ and Na₃PO₄ in the mixed solution are:
- Concentration of H₃PO₄:
\[
[\text{H₃PO₄}] = \frac{0.002 \, \text{mol}}{0.040 \, \text{L}} = 0.05 \, \text{M}
\]
- Concentration of Na₃PO₄ (which dissociates to give PO₄³⁻):
\[
[\text{PO₄}^{3-}] = \frac{0.002 \, \text{mol}}{0.040 \, \text{L}} = 0.05 \, \text{M}
\]
### Step 4: Use the Henderson-Hasselbalch equation
The pH of the solution can be calculated using the Henderson-Hasselbalch equation:
\[
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Where:
- \(\text{HA} = \text{H₂PO₄}^-\) (the conjugate acid)
- \(\text{A}^- = \text{HPO₄}^{2-}\) (the conjugate base)
### Step 5: Determine the pKₐ values
For phosphoric acid (H₃PO₄):
- \( \text{pK}_a1 \approx 2.15 \)
- \( \text{pK}_a2 \approx 7.21 \)
- \( \text{pK}_a3 \approx 12.37 \)
Since we are using H₂PO₄⁻ and HPO₄²⁻, we will use pK₂:
\[
\text{pK}_a = 7.21
\]
### Step 6: Apply concentrations to the equation
Since both the concentrations of H₂PO₄⁻ and HPO₄²⁻ are equal (0.05 M), we can substitute into the equation:
\[
\text{pH} = 7.21 + \log\left(\frac{0.05}{0.05}\right) = 7.21 + \log(1) = 7.21 + 0 = 7.21
\]
### Final Answer
The pH of the resultant solution is approximately **7.21**.
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