________ ml of 0.1 M `H_(2)SO_(4)` is required to neutralise 50 ml of 0.2 M NaOH solution

Options

A. 25
B. 100
C. 75
D. 50

Explanation

To solve the question of how many milliliters of 0.1 M \( H_2SO_4 \) are required to neutralize 50 mL of 0.2 M NaOH solution, we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation for the neutralization reaction between sulfuric acid and sodium hydroxide is:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]
From this equation, we can see that 1 mole of \( H_2SO_4 \) reacts with 2 moles of \( NaOH \).

### Step 2: Calculate the number of moles of NaOH.
Using the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
We can rearrange this to find the number of moles:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume in liters} \]
Given that the molarity of NaOH is 0.2 M and the volume is 50 mL (which is 0.050 L):
\[ \text{Number of moles of NaOH} = 0.2 \, \text{M} \times 0.050 \, \text{L} = 0.01 \, \text{moles} \]

### Step 3: Determine the number of moles of \( H_2SO_4 \) required.
From the balanced equation, we know that 1 mole of \( H_2SO_4 \) is required for every 2 moles of \( NaOH \). Therefore, the number of moles of \( H_2SO_4 \) needed is:
\[ \text{Number of moles of } H_2SO_4 = \frac{0.01 \, \text{moles of NaOH}}{2} = 0.005 \, \text{moles} \]

### Step 4: Calculate the volume of \( H_2SO_4 \) required.
Using the molarity of \( H_2SO_4 \) (0.1 M), we can find the volume needed using the rearranged molarity formula:
\[ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume of } H_2SO_4 = \frac{0.005 \, \text{moles}}{0.1 \, \text{M}} = 0.050 \, \text{L} \]

### Step 5: Convert the volume from liters to milliliters.
Since 1 L = 1000 mL:
\[ 0.050 \, \text{L} = 0.050 \times 1000 \, \text{mL} = 50 \, \text{mL} \]

### Final Answer:
50 mL of 0.1 M \( H_2SO_4 \) is required to neutralize 50 mL of 0.2 M NaOH solution.

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