The volume of 0.1 M `Ca (OH)_(2)` needed for the neutralization of 40 ml of 0.05 M oxalic acid is

Options

A. 10 ml
B. 20 ml
C. 30 ml
D. 40 ml

Answer

To solve the problem of how much volume of 0.1 M calcium hydroxide (Ca(OH)₂) is needed to neutralize 40 ml of 0.05 M oxalic acid (H₂C₂O₄), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Neutralization Reaction**:
The reaction between calcium hydroxide and oxalic acid can be represented as:
\[ \text{Ca(OH)}_2 + \text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{CaC}_2\text{O}_4 + 2\text{H}_2\text{O} \]
From this reaction, we can see that 1 mole of calcium hydroxide reacts with 1 mole of oxalic acid.

2. **Calculate the Number of Equivalents of Oxalic Acid**:
The normality (N) of a solution is defined as the number of equivalents of solute per liter of solution. For oxalic acid (H₂C₂O₄), which can donate 2 H⁺ ions, the normality is twice the molarity:
\[ N_2 = 2 \times M_2 = 2 \times 0.05 \, \text{M} = 0.1 \, \text{N} \]
The volume of oxalic acid (V₂) is given as 40 ml or 0.040 L. Therefore, the number of equivalents of oxalic acid is:
\[ \text{Equivalents of } H_2C_2O_4 = N_2 \times V_2 = 0.1 \, \text{N} \times 0.040 \, \text{L} = 0.004 \, \text{equivalents} \]

3. **Calculate the Number of Equivalents of Calcium Hydroxide Required**:
According to the law of equivalence, the number of equivalents of calcium hydroxide (N₁) must equal the number of equivalents of oxalic acid:
\[ N_1 \times V_1 = 0.004 \, \text{equivalents} \]

4. **Determine the Normality of Calcium Hydroxide**:
Calcium hydroxide (Ca(OH)₂) can donate 2 OH⁻ ions, so its normality is:
\[ N_1 = 2 \times M_1 = 2 \times 0.1 \, \text{M} = 0.2 \, \text{N} \]

5. **Set Up the Equation**:
Now we can set up the equation using the equivalents:
\[ 0.2 \, \text{N} \times V_1 = 0.004 \, \text{equivalents} \]

6. **Solve for V₁**:
Rearranging the equation to find the volume of calcium hydroxide:
\[ V_1 = \frac{0.004 \, \text{equivalents}}{0.2 \, \text{N}} = 0.02 \, \text{L} = 20 \, \text{ml} \]

### Final Answer:
The volume of 0.1 M calcium hydroxide needed for the neutralization of 40 ml of 0.05 M oxalic acid is **20 ml**.