30 ml of N/10 HC1 is required to neutralize 50 ml of a sodium carbonate solution. What volume of water (in ml) must be added to 30 ml of `Na_2 CO_3` solution so that the final solution has concentration N/50?

Options

A. 20 ml
B. 30 ml
C. 50 ml
D. 60 ml

Answer

To solve the problem step by step, we will follow the given information and apply the concepts of normality and dilution.

### Step 1: Determine the normality of sodium carbonate solution

We know that 30 ml of N/10 HCl is required to neutralize 50 ml of sodium carbonate solution.

Using the formula for neutralization:
\[ n_1 \times V_1 = n_2 \times V_2 \]

Where:
- \( n_1 \) = Normality of HCl = \( \frac{1}{10} \) N
- \( V_1 \) = Volume of HCl = 30 ml
- \( n_2 \) = Normality of Na2CO3 (unknown)
- \( V_2 \) = Volume of Na2CO3 = 50 ml

Substituting the values into the equation:
\[ \left(\frac{1}{10}\right) \times 30 = n_2 \times 50 \]

### Step 2: Calculate the normality of Na2CO3

Now, we can solve for \( n_2 \):
\[ \frac{30}{10} = n_2 \times 50 \]
\[ 3 = n_2 \times 50 \]
\[ n_2 = \frac{3}{50} \]

So, the normality of the sodium carbonate solution is \( \frac{3}{50} \) N.

### Step 3: Set up the equation for dilution

We need to find out how much water must be added to the 30 ml of Na2CO3 solution to achieve a final concentration of \( \frac{1}{50} \) N.

Using the dilution formula:
\[ N_{initial} \times V_{initial} = N_{final} \times V_{final} \]

Where:
- \( N_{initial} = \frac{3}{50} \) N
- \( V_{initial} = 30 \) ml
- \( N_{final} = \frac{1}{50} \) N
- \( V_{final} = V_{initial} + V_{water} \)

Substituting the known values:
\[ \left(\frac{3}{50}\right) \times 30 = \left(\frac{1}{50}\right) \times V_{final} \]

### Step 4: Calculate the final volume

Calculating the left side:
\[ \frac{3 \times 30}{50} = \frac{90}{50} = 1.8 \]

Now, substituting back into the equation:
\[ 1.8 = \left(\frac{1}{50}\right) \times V_{final} \]

To find \( V_{final} \):
\[ V_{final} = 1.8 \times 50 = 90 \text{ ml} \]

### Step 5: Calculate the volume of water to be added

The volume of water to be added is given by:
\[ V_{water} = V_{final} - V_{initial} \]
\[ V_{water} = 90 \text{ ml} - 30 \text{ ml} = 60 \text{ ml} \]

### Final Answer

The volume of water that must be added is **60 ml**.

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