100 ml of 0.01 M `H_2 SO_4` is titrated against 0.2 M `Ca (OH)_2`, volume of `Ca (OH)_2` required to reach end point will be :

Options

A. 5 ml
B. 10 ml
C. 20 ml
D. 15 ml

Answer

To solve the problem of titrating 100 mL of 0.01 M H₂SO₄ against 0.2 M Ca(OH)₂, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and calcium hydroxide (Ca(OH)₂) is:
\[ H_2SO_4 + Ca(OH)_2 \rightarrow CaSO_4 + 2H_2O \]
This shows that 1 mole of H₂SO₄ reacts with 1 mole of Ca(OH)₂.

### Step 2: Determine the number of moles of H₂SO₄
To find the number of moles of H₂SO₄, we use the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (L)} \]
Given:
- Molarity of H₂SO₄ (M₁) = 0.01 M
- Volume of H₂SO₄ (V₁) = 100 mL = 0.1 L

Calculating the moles:
\[ \text{Moles of H₂SO₄} = 0.01 \, \text{mol/L} \times 0.1 \, \text{L} = 0.001 \, \text{mol} \]

### Step 3: Determine the number of moles of Ca(OH)₂ required
From the balanced equation, we see that 1 mole of H₂SO₄ reacts with 1 mole of Ca(OH)₂. Therefore, the moles of Ca(OH)₂ required will also be:
\[ \text{Moles of Ca(OH)₂} = 0.001 \, \text{mol} \]

### Step 4: Calculate the volume of Ca(OH)₂ needed
Now, we can use the molarity of Ca(OH)₂ to find the volume required:
Given:
- Molarity of Ca(OH)₂ (M₂) = 0.2 M

Using the formula for moles:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (L)} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{Number of moles}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume of Ca(OH)₂} = \frac{0.001 \, \text{mol}}{0.2 \, \text{mol/L}} = 0.005 \, \text{L} \]
Converting to mL:
\[ \text{Volume of Ca(OH)₂} = 0.005 \, \text{L} \times 1000 \, \text{mL/L} = 5 \, \text{mL} \]

### Final Answer
The volume of Ca(OH)₂ required to reach the endpoint is **5 mL**.

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