50 mL of 0.1 M NaOH is added to 60 mL of 0.15 M `H_(3)PO_(4)` solution `(K_(1), K_(2) " and " K_(3) " for " H_(3)PO_(4) " are " 10^(-3), 10^(-8) " and " 10^(-13)` respectively). The pH of the mixture would be about (log 2=0.3) :

Options

A. `3.1`
B. `5.5`
C. `4.1`
D. `6.5`

Answer

To solve the problem, we need to determine the pH of the solution after mixing 50 mL of 0.1 M NaOH with 60 mL of 0.15 M H₃PO₄. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of NaOH and H₃PO₄

1. **Moles of NaOH**:
\[ \text{Volume of NaOH} = 50 \, \text{mL} = 0.050 \, \text{L} \]
\[ \text{Concentration of NaOH} = 0.1 \, \text{M} \]
\[ \text{Moles of NaOH} = \text{Volume} \times \text{Concentration} = 0.050 \, \text{L} \times 0.1 \, \text{mol/L} = 0.005 \, \text{mol} \]

2. **Moles of H₃PO₄**:
\[ \text{Volume of H₃PO₄} = 60 \, \text{mL} = 0.060 \, \text{L} \]
\[ \text{Concentration of H₃PO₄} = 0.15 \, \text{M} \]
\[ \text{Moles of H₃PO₄} = \text{Volume} \times \text{Concentration} = 0.060 \, \text{L} \times 0.15 \, \text{mol/L} = 0.009 \, \text{mol} \]

### Step 2: Determine the neutralization reaction

H₃PO₄ is a triprotic acid, which means it can donate three protons (H⁺). The reaction with NaOH will proceed as follows:
- The first reaction is:
\[ \text{H}_3\text{PO}_4 + \text{NaOH} \rightarrow \text{NaH}_2\text{PO}_4 + \text{H}_2\text{O} \]

### Step 3: Calculate the moles after neutralization

- Since we have 0.005 moles of NaOH, it will neutralize 0.005 moles of H₃PO₄:
\[ \text{Remaining moles of H₃PO₄} = 0.009 - 0.005 = 0.004 \, \text{mol} \]
- Moles of NaH₂PO₄ formed (salt) = 0.005 moles.

### Step 4: Calculate the concentrations of the remaining acid and salt

- Total volume of the solution after mixing:
\[ \text{Total Volume} = 50 \, \text{mL} + 60 \, \text{mL} = 110 \, \text{mL} = 0.110 \, \text{L} \]

- Concentration of remaining H₂PO₄⁻ (acid):
\[ [\text{H}_2\text{PO}_4^-] = \frac{0.004 \, \text{mol}}{0.110 \, \text{L}} \approx 0.03636 \, \text{M} \]

- Concentration of NaH₂PO₄ (salt):
\[ [\text{NaH}_2\text{PO}_4] = \frac{0.005 \, \text{mol}}{0.110 \, \text{L}} \approx 0.04545 \, \text{M} \]

### Step 5: Calculate pKa and pH using the Henderson-Hasselbalch equation

1. **Calculate pKa**:
\[ K_1 = 10^{-3} \implies pK_1 = -\log(10^{-3}) = 3 \]

2. **Use the Henderson-Hasselbalch equation**:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
\[ \text{pH} = 3 + \log\left(\frac{0.04545}{0.03636}\right) \]
\[ \text{pH} = 3 + \log(1.25) \approx 3 + 0.09691 \approx 3.1 \]

### Final Answer

The pH of the mixture would be approximately **3.1**.

---