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 need to determine how much water must be added to dilute a sodium carbonate solution to a desired normality.

### Step 1: Determine the normality of Na2CO3 solution
We know that 30 ml of N/10 HCl is required to neutralize 50 ml of Na2CO3 solution.

- Given:
- Volume of HCl (V1) = 30 ml
- Normality of HCl (N2) = N/10 = 0.1 N
- Volume of Na2CO3 solution (V2) = 50 ml

Using the formula for neutralization:
\[ N_1 \times V_1 = N_2 \times V_2 \]
where \(N_1\) is the normality of Na2CO3.

Substituting the known values:
\[ N_1 \times 50 = 0.1 \times 30 \]

Calculating \(N_1\):
\[ N_1 = \frac{0.1 \times 30}{50} = \frac{3}{50} = 0.06 \, \text{N} \]

### Step 2: Determine the final normality after dilution
We want to dilute this Na2CO3 solution to a final normality of N/50.

- Final Normality (N_final) = N/50 = 0.02 N

### Step 3: Use the dilution formula to find the final volume
We can use the dilution equation:
\[ N_1 \times V_1 = N_2 \times V_2 \]
where:
- \(N_1 = 0.06 \, \text{N}\)
- \(V_1 = 30 \, \text{ml}\) (the volume of Na2CO3 solution)
- \(N_2 = 0.02 \, \text{N}\) (the desired final normality)
- \(V_2\) is the total volume after adding water.

Rearranging the equation to solve for \(V_2\):
\[ V_2 = \frac{N_1 \times V_1}{N_2} \]
Substituting the known values:
\[ V_2 = \frac{0.06 \times 30}{0.02} = \frac{1.8}{0.02} = 90 \, \text{ml} \]

### Step 4: Calculate the volume of water to be added
The total volume after dilution \(V_2\) is 90 ml. Since we started with 30 ml of the Na2CO3 solution, the volume of water to be added is:
\[ \text{Volume of water} = V_2 - V_1 = 90 \, \text{ml} - 30 \, \text{ml} = 60 \, \text{ml} \]

### Final Answer
The volume of water that must be added to 30 ml of Na2CO3 solution to achieve a final concentration of N/50 is **60 ml**.

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