What volume of 0.2 M KOH will be requried to neutralise 100 " mL of " 0.1 M `H_(3)PO_(4)` using methyl red indicator (change of colour pink `rarr` yellow) and then bromothymol blue indicator is added.

Options

A. 50 mL
B. 100 mL
C. 150 mL
D. 200 mL

Explanation

To solve the problem of determining the volume of 0.2 M KOH required to neutralize 100 mL of 0.1 M H₃PO₄, we will follow these steps:

### Step 1: Understand the Neutralization Reaction
The neutralization of phosphoric acid (H₃PO₄) with potassium hydroxide (KOH) involves the release of hydrogen ions (H⁺) from the acid, which react with hydroxide ions (OH⁻) from the base to form water (H₂O).

### Step 2: Determine the Ionization of H₃PO₄
H₃PO₄ is a triprotic acid, meaning it can donate three protons (H⁺). However, in this case, we will consider two ionizations because we are using two indicators:
1. The first ionization (with methyl red) will neutralize one H⁺.
2. The second ionization (with bromothymol blue) will neutralize another H⁺.

### Step 3: Calculate the Number of Equivalents of H₃PO₄
The normality (N) of H₃PO₄ is given by the formula:
\[ N = M \times n \]
where:
- M = molarity of the acid (0.1 M)
- n = number of acidic protons available for reaction (1 for the first ionization and 1 for the second ionization)

For the first ionization:
\[ N_1 = 0.1 \, \text{M} \times 1 = 0.1 \, \text{N} \]

For the second ionization:
\[ N_2 = 0.1 \, \text{M} \times 1 = 0.1 \, \text{N} \]

### Step 4: Calculate the Total Milliequivalents of H₃PO₄
The total volume of H₃PO₄ is 100 mL, so:
\[ \text{Milliequivalents of H₃PO₄} = N \times V \]
\[ = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} \]

### Step 5: Use the Neutralization Equation
The neutralization reaction can be represented as:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 \) = Normality of KOH (0.2 N)
- \( V_1 \) = Volume of KOH (unknown)
- \( N_2 \) = Normality of H₃PO₄ (0.1 N)
- \( V_2 \) = Volume of H₃PO₄ (0.1 L or 100 mL)

### Step 6: Solve for the Volume of KOH
Rearranging the equation gives:
\[ V_1 = \frac{N_2 \times V_2}{N_1} \]
Substituting the values:
\[ V_1 = \frac{0.1 \, \text{N} \times 0.1 \, \text{L}}{0.2 \, \text{N}} \]
\[ V_1 = \frac{0.01}{0.2} = 0.05 \, \text{L} = 50 \, \text{mL} \]

### Step 7: Repeat for Second Ionization
Using the same approach for the second ionization:
\[ V_1 = \frac{0.1 \, \text{N} \times 0.1 \, \text{L}}{0.2 \, \text{N}} \]
This will also yield:
\[ V_1 = 50 \, \text{mL} \]

### Step 8: Total Volume of KOH Required
The total volume of KOH required for both ionizations is:
\[ V_{\text{total}} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} \]

Thus, the final answer is:
**100 mL of 0.2 M KOH is required to neutralize 100 mL of 0.1 M H₃PO₄.**

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