What volume of 0.1 M `Ba(OH)_(2)` will be required to neutralise a mixture of 50 " mL of " 0.1 M HCl and 100 " mL of " 0.2 M `H_(3)PO_(4)` using methyl red indicator?

Options

A. 25 mL
B. 50 mL
C. 100 mL
D. 125 mL

Explanation

To solve the problem of determining the volume of 0.1 M Ba(OH)₂ required to neutralize a mixture of 50 mL of 0.1 M HCl and 100 mL of 0.2 M H₃PO₄, we can follow these steps:

### Step 1: Calculate the moles of HCl
- **Volume of HCl** = 50 mL = 0.050 L
- **Molarity of HCl** = 0.1 M
- **Moles of HCl** = Molarity × Volume = 0.1 mol/L × 0.050 L = 0.005 moles

### Step 2: Determine the equivalent of HCl
- HCl is a strong acid that dissociates completely and provides 1 H⁺ ion per molecule.
- **Equivalents of HCl** = Moles of HCl = 0.005 equivalents

### Step 3: Calculate the moles of H₃PO₄
- **Volume of H₃PO₄** = 100 mL = 0.100 L
- **Molarity of H₃PO₄** = 0.2 M
- **Moles of H₃PO₄** = Molarity × Volume = 0.2 mol/L × 0.100 L = 0.020 moles

### Step 4: Determine the equivalent of H₃PO₄
- H₃PO₄ can donate 1 H⁺ ion in the first ionization step when using methyl red as an indicator.
- **Equivalents of H₃PO₄** = Moles of H₃PO₄ = 0.020 equivalents

### Step 5: Calculate the total equivalents of acid
- **Total equivalents of acid** = Equivalents of HCl + Equivalents of H₃PO₄
- Total = 0.005 + 0.020 = 0.025 equivalents

### Step 6: Determine the equivalents of Ba(OH)₂ required
- Ba(OH)₂ provides 2 OH⁻ ions per molecule.
- Therefore, the number of equivalents of Ba(OH)₂ needed to neutralize the acids will be equal to the total equivalents of the acids.
- **Equivalents of Ba(OH)₂** = 0.025 equivalents

### Step 7: Calculate the volume of Ba(OH)₂ required
- **Normality of Ba(OH)₂** = Molarity × Basicity = 0.1 M × 2 = 0.2 N
- Using the formula:
\[ \text{Equivalents} = \text{Normality} \times \text{Volume (L)} \]
\[ 0.025 = 0.2 \times V \]
\[ V = \frac{0.025}{0.2} = 0.125 \text{ L} = 125 \text{ mL} \]

### Final Answer
The volume of 0.1 M Ba(OH)₂ required to neutralize the mixture is **125 mL**.

---