25 ml of the given HCl solution requires 30 mL of 0.1M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2M aqueous NaOH solution ?

Options

A. 25 mL
B. 12.5 mL
C. 72 mL
D. 50 mL

Explanation

To solve the problem step by step, we will use the concept of normality and equivalence in titration.

### Step 1: Calculate the milliequivalents of HCl using Na2CO3
Given that 25 mL of HCl solution requires 30 mL of 0.1 M sodium carbonate (Na2CO3), we can first find the milliequivalents of Na2CO3 used.

1. **Calculate the number of equivalents of Na2CO3:**
\[ \text{Normality of Na2CO3} = 0.1 \, \text{M} \times 2 \, (\text{N factor of Na2CO3}) = 0.2 \, \text{N} \]
\[ \text{Volume of Na2CO3} = 30 \, \text{mL} = 0.030 \, \text{L} \]
\[ \text{Milliequivalents of Na2CO3} = \text{Normality} \times \text{Volume (L)} = 0.2 \times 0.030 = 0.006 \, \text{equivalents} \]

2. **Milliequivalents of HCl:**
Since HCl reacts with Na2CO3 in a 1:1 ratio, the milliequivalents of HCl will also be 0.006 equivalents.

### Step 2: Calculate the molarity of HCl
Using the milliequivalents of HCl, we can find its molarity.

1. **Convert equivalents to milliequivalents:**
\[ \text{Milliequivalents of HCl} = 0.006 \, \text{equivalents} \times 1000 = 6 \, \text{milliequivalents} \]

2. **Calculate the molarity of HCl:**
\[ \text{Molarity (M)} = \frac{\text{Milliequivalents}}{\text{Volume (L)}} = \frac{6}{0.025} = 240 \, \text{mM} = 0.240 \, \text{M} \]

### Step 3: Calculate the volume of HCl required to titrate NaOH
Now we need to find the volume of this HCl solution required to titrate 30 mL of 0.2 M NaOH.

1. **Calculate the equivalents of NaOH:**
\[ \text{Normality of NaOH} = 0.2 \, \text{M} \times 1 \, (\text{N factor of NaOH}) = 0.2 \, \text{N} \]
\[ \text{Volume of NaOH} = 30 \, \text{mL} = 0.030 \, \text{L} \]
\[ \text{Milliequivalents of NaOH} = 0.2 \times 0.030 = 0.006 \, \text{equivalents} \]

### Step 4: Set up the equation for HCl and NaOH
Using the equivalence principle:
\[ \text{Milliequivalents of HCl} = \text{Milliequivalents of NaOH} \]
Let \( V \) be the volume of HCl required in liters.

1. **Set up the equation:**
\[ \text{Normality of HCl} \times V = \text{Normality of NaOH} \times \text{Volume of NaOH} \]
\[ 0.240 \times V = 0.2 \times 0.030 \]

### Step 5: Solve for \( V \)
1. **Substituting the values:**
\[ 0.240 \times V = 0.006 \]
\[ V = \frac{0.006}{0.240} = 0.025 \, \text{L} = 25 \, \text{mL} \]

### Conclusion
The volume of the HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solution is **25 mL**.

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