How much volume of 0.1N NaOH will neutralize 10mL of 0.1N phosphoric acid

Answer

To determine how much volume of 0.1N NaOH will neutralize 10 mL of 0.1N phosphoric acid (H₃PO₄), we can use the concept of normality and the neutralization reaction between an acid and a base.

### Step-by-Step Solution:

1. **Identify the Reaction**:
Phosphoric acid (H₃PO₄) is a triprotic acid, meaning it can donate three protons (H⁺ ions). Sodium hydroxide (NaOH) is a strong base that can react with these protons.

2. **Write the Neutralization Equation**:
The neutralization reaction can be represented as:
\[ H₃PO₄ + 3NaOH \rightarrow Na₃PO₄ + 3H₂O \]
From this equation, we see that 1 mole of H₃PO₄ reacts with 3 moles of NaOH.

3. **Calculate the Equivalent of Phosphoric Acid**:
The normality (N) of a solution is defined as the number of equivalents of solute per liter of solution. Since we have 0.1N phosphoric acid:
\[ \text{Equivalents of H₃PO₄} = \text{Normality} \times \text{Volume (L)} \]
Here, the volume of H₃PO₄ is 10 mL, which is 0.01 L.
\[ \text{Equivalents of H₃PO₄} = 0.1 \, \text{N} \times 0.01 \, \text{L} = 0.001 \, \text{equivalents} \]

4. **Determine the Required Equivalents of NaOH**:
From the balanced equation, we see that 1 equivalent of H₃PO₄ requires 3 equivalents of NaOH. Therefore:
\[ \text{Equivalents of NaOH} = 3 \times \text{Equivalents of H₃PO₄} = 3 \times 0.001 = 0.003 \, \text{equivalents} \]

5. **Calculate the Volume of NaOH Required**:
We know the normality of NaOH is 0.1N. Using the formula for equivalents:
\[ \text{Volume (L)} = \frac{\text{Equivalents}}{\text{Normality}} = \frac{0.003 \, \text{equivalents}}{0.1 \, \text{N}} = 0.03 \, \text{L} \]
Converting this to mL:
\[ 0.03 \, \text{L} = 30 \, \text{mL} \]

### Final Answer:
The volume of 0.1N NaOH required to neutralize 10 mL of 0.1N phosphoric acid is **30 mL**.