To find the pH of the resultant solution when 100 mL of 0.1 M H₃PO₄ is mixed with 50 mL of 0.1 M NaOH, we can follow these steps:
### Step 1: Calculate the moles of H₃PO₄ and NaOH
1. **Calculate moles of H₃PO₄:**
\[
\text{Moles of H₃PO₄} = \text{Volume (L)} \times \text{Concentration (M)} = 0.1 \, \text{L} \times 0.1 \, \text{M} = 0.01 \, \text{mol} \, (10 \, \text{mmol})
\]
2. **Calculate moles of NaOH:**
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.05 \, \text{L} \times 0.1 \, \text{M} = 0.005 \, \text{mol} \, (5 \, \text{mmol})
\]
### Step 2: Determine the reaction between H₃PO₄ and NaOH
H₃PO₄ is a weak acid and NaOH is a strong base. The reaction can be represented as:
\[
\text{H₃PO₄} + \text{NaOH} \rightarrow \text{NaH₂PO₄} + \text{H₂O}
\]
### Step 3: Calculate the final amounts after the reaction
- **Initial moles:**
- H₃PO₄: 0.01 mol (10 mmol)
- NaOH: 0.005 mol (5 mmol)
- NaH₂PO₄: 0 mol
- **After the reaction:**
- H₃PO₄ remaining: \(0.01 - 0.005 = 0.005 \, \text{mol}\) (5 mmol)
- NaOH remaining: \(0.005 - 0.005 = 0 \, \text{mol}\)
- NaH₂PO₄ formed: \(0 + 0.005 = 0.005 \, \text{mol}\) (5 mmol)
### Step 4: Calculate the pH of the resultant solution
Since we have a weak acid (H₃PO₄) and its conjugate base (NaH₂PO₄), we can use the Henderson-Hasselbalch equation:
\[
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)
\]
- **Given pKₐ values:**
- \(K_a1 = 10^{-3}\) → \(pK_a1 = 3\)
- **Concentrations:**
- \([\text{Salt}] = [\text{NaH₂PO₄}] = \frac{0.005 \, \text{mol}}{0.15 \, \text{L}} = \frac{0.005}{0.15} = 0.0333 \, \text{M}\)
- \([\text{Acid}] = [\text{H₃PO₄}] = \frac{0.005 \, \text{mol}}{0.15 \, \text{L}} = \frac{0.005}{0.15} = 0.0333 \, \text{M}\)
### Step 5: Substitute values into the Henderson-Hasselbalch equation
\[
\text{pH} = 3 + \log\left(\frac{0.0333}{0.0333}\right) = 3 + \log(1) = 3 + 0 = 3
\]
### Final Answer:
The pH of the resultant solution is **3**.
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