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 follow these steps:

### Step 1: Understand the neutralization reaction
When an acid reacts with a base, they neutralize each other. The number of equivalents of the acid will equal the number of equivalents of the base at neutralization.

### Step 2: Calculate the number of equivalents of phosphoric acid
The number of equivalents of an acid can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (L)} \]
For phosphoric acid (H₃PO₄):
- Normality (N) = 0.1N
- Volume = 10 mL = 10/1000 L = 0.01 L

Now, substituting the values:
\[ \text{Number of equivalents of H₃PO₄} = 0.1 \times 0.01 = 0.001 \text{ equivalents} \]

### Step 3: Set up the equation for NaOH
Since the number of equivalents of NaOH will equal the number of equivalents of H₃PO₄, we have:
\[ \text{Number of equivalents of NaOH} = \text{Number of equivalents of H₃PO₄} = 0.001 \text{ equivalents} \]

### Step 4: Calculate the volume of NaOH required
Using the same formula for NaOH:
\[ \text{Number of equivalents of NaOH} = \text{Normality} \times \text{Volume (L)} \]
Let \( V \) be the volume of NaOH in liters. We know:
- Normality of NaOH = 0.1N

Thus:
\[ 0.001 = 0.1 \times V \]
Now, solving for \( V \):
\[ V = \frac{0.001}{0.1} = 0.01 \text{ L} \]

### Step 5: Convert volume from liters to milliliters
To convert liters to milliliters, multiply by 1000:
\[ V = 0.01 \text{ L} \times 1000 = 10 \text{ mL} \]

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

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