To find the pH of the solution obtained by mixing 100 ml of 0.1 M H₃PO₄ and 150 ml of 0.1 M NaOH, we will follow these steps:
### Step 1: Calculate the moles of H₃PO₄ and NaOH
- **Moles of H₃PO₄**:
\[
\text{Volume} = 100 \, \text{ml} = 0.1 \, \text{L}
\]
\[
\text{Concentration} = 0.1 \, \text{M}
\]
\[
\text{Moles of H₃PO₄} = \text{Volume} \times \text{Concentration} = 0.1 \, \text{L} \times 0.1 \, \text{mol/L} = 0.01 \, \text{mol} = 10 \, \text{millimoles}
\]
- **Moles of NaOH**:
\[
\text{Volume} = 150 \, \text{ml} = 0.15 \, \text{L}
\]
\[
\text{Concentration} = 0.1 \, \text{M}
\]
\[
\text{Moles of NaOH} = \text{Volume} \times \text{Concentration} = 0.15 \, \text{L} \times 0.1 \, \text{mol/L} = 0.015 \, \text{mol} = 15 \, \text{millimoles}
\]
### Step 2: Determine the reaction between H₃PO₄ and NaOH
The reaction between the weak acid (H₃PO₄) and the strong base (NaOH) will occur as follows:
\[
\text{H₃PO₄} + \text{NaOH} \rightarrow \text{NaH₂PO₄} + \text{H₂O}
\]
### Step 3: Calculate the remaining moles after the reaction
- **Initial moles**:
- H₃PO₄: 10 mmoles
- NaOH: 15 mmoles
- **After reaction**:
- All 10 mmoles of H₃PO₄ will react with 10 mmoles of NaOH.
- Remaining NaOH: \( 15 - 10 = 5 \, \text{mmoles} \)
- Moles of NaH₂PO₄ formed: 10 mmoles
### Step 4: Determine the buffer solution composition
After the reaction, we have:
- NaH₂PO₄: 10 mmoles (acid)
- Na₂HPO₄: 5 mmoles (conjugate base)
### Step 5: Use the Henderson-Hasselbalch equation to find pH
The Henderson-Hasselbalch equation is given by:
\[
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
\]
Where:
- \(\text{Base} = \text{Na₂HPO₄}\)
- \(\text{Acid} = \text{NaH₂PO₄}\)
- **Calculate pKₐ**:
- For H₂PO₄⁻ ↔ HPO₄²⁻, \(K_2 = 10^{-8}\)
- \(\text{pK}_a = -\log(10^{-8}) = 8\)
### Step 6: Substitute values into the Henderson-Hasselbalch equation
\[
\text{pH} = 8 + \log\left(\frac{5 \, \text{mmoles}}{10 \, \text{mmoles}}\right)
\]
\[
\text{pH} = 8 + \log(0.5)
\]
\[
\text{pH} = 8 - 0.301 = 7.699 \approx 8
\]
### Final Answer
The pH of the resulting buffer solution is approximately **8**.
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