Volume of `0.1M""H_(2)SO_(4)` required to neutralize 30 mL of `0.2 N""NaOH` is

Options

A. 30 mL
B. 15 mL
C. 40 mL
D. 60 mL

Answer

To determine the volume of `0.1 M H2SO4` required to neutralize `30 mL` of `0.2 N NaOH`, we can follow these steps:

### Step 1: Write the Neutralization Reaction
The neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) can be represented as:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]

### Step 2: Calculate the Milliequivalents of NaOH
We know that normality (N) is defined as the number of equivalents per liter of solution. For NaOH:
- Normality (N) = 0.2 N
- Volume (V) = 30 mL = 0.030 L

The milliequivalents (mEq) of NaOH can be calculated using the formula:
\[ \text{mEq} = N \times V \]
\[ \text{mEq of NaOH} = 0.2 \, \text{N} \times 0.030 \, \text{L} = 0.006 \, \text{equivalents} = 6 \, \text{mEq} \]

### Step 3: Determine the Normality of H2SO4
For sulfuric acid (H2SO4):
- Molarity (M) = 0.1 M
- The n-factor for H2SO4 is 2 (since it can donate 2 H+ ions).

Normality (N) can be calculated as:
\[ N = M \times n \]
\[ N_{H_2SO_4} = 0.1 \, \text{M} \times 2 = 0.2 \, \text{N} \]

### Step 4: Apply the Law of Equivalents
According to the law of equivalents, the milliequivalents of H2SO4 must equal the milliequivalents of NaOH for complete neutralization:
\[ \text{mEq of H2SO4} = \text{mEq of NaOH} \]
Let \( V \) be the volume of H2SO4 in liters. Then:
\[ N_{H_2SO_4} \times V = 6 \, \text{mEq} \]
Substituting the known values:
\[ 0.2 \, \text{N} \times V = 6 \, \text{mEq} \]

### Step 5: Solve for Volume (V)
Rearranging the equation to find \( V \):
\[ V = \frac{6 \, \text{mEq}}{0.2 \, \text{N}} \]
\[ V = 30 \, \text{mL} \]

### Conclusion
The volume of `0.1 M H2SO4` required to neutralize `30 mL` of `0.2 N NaOH` is **30 mL**.

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