The volume of 0.1M `Ca(OH)_(2)` required to neutralize 10 mL of 0.1 N HCl

Options

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

Explanation

To find the volume of 0.1 M \( \text{Ca(OH)}_2 \) required to neutralize 10 mL of 0.1 N HCl, we can follow these steps:

### Step 1: Understand the Neutralization Reaction
In a neutralization reaction, the number of equivalents of acid must equal the number of equivalents of base. For \( \text{Ca(OH)}_2 \), it dissociates into \( \text{Ca}^{2+} \) and \( 2 \text{OH}^- \). Therefore, the n-factor (number of hydroxide ions produced) for \( \text{Ca(OH)}_2 \) is 2.

### Step 2: Calculate the Equivalents of HCl
Given:
- Volume of HCl = 10 mL = 0.01 L
- Normality of HCl = 0.1 N

The number of equivalents of HCl can be calculated using the formula:
\[ \text{Equivalents of HCl} = \text{Normality} \times \text{Volume (L)} \]
\[ \text{Equivalents of HCl} = 0.1 \, \text{N} \times 0.01 \, \text{L} = 0.001 \, \text{equivalents} \]

### Step 3: Calculate the Volume of \( \text{Ca(OH)}_2 \)
Now we need to find out how much volume of \( \text{Ca(OH)}_2 \) is required to provide the same number of equivalents.

Given:
- Molarity of \( \text{Ca(OH)}_2 \) = 0.1 M
- n-factor of \( \text{Ca(OH)}_2 \) = 2

First, we calculate the normality of \( \text{Ca(OH)}_2 \):
\[ \text{Normality of } \text{Ca(OH)}_2 = \text{Molarity} \times \text{n-factor} = 0.1 \, \text{M} \times 2 = 0.2 \, \text{N} \]

Now, we can use the equivalence relation:
\[ n_1 v_1 = n_2 v_2 \]
Where:
- \( n_1 \) = Normality of \( \text{Ca(OH)}_2 = 0.2 \, \text{N} \)
- \( v_1 \) = Volume of \( \text{Ca(OH)}_2 \) (in liters)
- \( n_2 \) = Normality of HCl = 0.1 N
- \( v_2 \) = Volume of HCl = 0.01 L

Substituting the values:
\[ 0.2 \, v_1 = 0.1 \times 0.01 \]
\[ 0.2 \, v_1 = 0.001 \]
\[ v_1 = \frac{0.001}{0.2} = 0.005 \, \text{L} \]

### Step 4: Convert Volume to mL
Since 1 L = 1000 mL:
\[ v_1 = 0.005 \, \text{L} \times 1000 \, \text{mL/L} = 5 \, \text{mL} \]

### Final Answer
The volume of 0.1 M \( \text{Ca(OH)}_2 \) required to neutralize 10 mL of 0.1 N HCl is **5 mL**.

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