A 100 ml solution of 0.1N-HCl was titrated with 0.2 N-NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25N-KOH solution. The volume of KOH required for completing the titration is

Options

A. 16 ml
B. 32 ml
C. 35 ml
D. 70 ml

Explanation

To solve the problem, we need to determine the volume of KOH required to complete the titration after adding 30 ml of NaOH to the HCl solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the number of equivalents of HCl
Given:
- Volume of HCl solution = 100 ml
- Normality of HCl = 0.1 N

Number of equivalents of HCl can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (in L)} \]
Converting 100 ml to liters:
\[ 100 \, \text{ml} = 0.1 \, \text{L} \]
Now, substituting the values:
\[ \text{Number of equivalents of HCl} = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} \]

### Step 2: Calculate the number of equivalents of NaOH added
Given:
- Volume of NaOH solution added = 30 ml
- Normality of NaOH = 0.2 N

Converting 30 ml to liters:
\[ 30 \, \text{ml} = 0.03 \, \text{L} \]
Now, substituting the values:
\[ \text{Number of equivalents of NaOH} = 0.2 \, \text{N} \times 0.03 \, \text{L} = 0.006 \, \text{equivalents} \]

### Step 3: Calculate the remaining equivalents of HCl
The remaining equivalents of HCl after adding NaOH can be calculated as follows:
\[ \text{Remaining equivalents of HCl} = \text{Initial equivalents of HCl} - \text{Equivalents of NaOH added} \]
Substituting the values:
\[ \text{Remaining equivalents of HCl} = 0.01 - 0.006 = 0.004 \, \text{equivalents} \]

### Step 4: Calculate the volume of KOH required to neutralize the remaining HCl
Given:
- Normality of KOH = 0.25 N

Using the formula for the number of equivalents:
\[ \text{Volume of KOH (in L)} = \frac{\text{Remaining equivalents of HCl}}{\text{Normality of KOH}} \]
Substituting the values:
\[ \text{Volume of KOH} = \frac{0.004}{0.25} = 0.016 \, \text{L} \]
Converting this to ml:
\[ 0.016 \, \text{L} = 16 \, \text{ml} \]

### Final Answer
The volume of KOH required to complete the titration is **16 ml**.

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