1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 mL. On titration, 25 mL of this solution neutralise 20 mL of a solution of sulphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly N/12 ?

Answer

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Calculate the Strength of Sodium Carbonate Solution
We know that 1.325 g of anhydrous sodium carbonate (Na2CO3) is dissolved in 250 mL of solution. To find the strength in grams per liter, we can use the formula:

\[ \text{Strength} = \left(\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}}\right) \times 1000 \]

Substituting the values:

\[ \text{Strength} = \left(\frac{1.325 \, \text{g}}{250 \, \text{mL}}\right) \times 1000 = 5.3 \, \text{g/L} \]

### Step 2: Calculate the Equivalent Mass of Sodium Carbonate
The molar mass of Na2CO3 is 106 g/mol. The n-factor for Na2CO3, which can donate 2 equivalents of Na+ or CO3^2-, is 2.

\[ \text{Equivalent Mass} = \frac{\text{Molar Mass}}{\text{n-factor}} = \frac{106}{2} = 53 \, \text{g/equiv} \]

### Step 3: Calculate the Normality of Sodium Carbonate Solution
Normality (N) can be calculated using the formula:

\[ \text{Normality} = \frac{\text{Strength (g/L)}}{\text{Equivalent Mass (g/equiv)}} \]

Substituting the values:

\[ \text{Normality of Na2CO3} = \frac{5.3}{53} = \frac{1}{10} \, \text{N} = 0.1 \, \text{N} \]

### Step 4: Use Titration Data to Find Normality of Sulfuric Acid
From the titration, we know that 25 mL of sodium carbonate solution neutralizes 20 mL of sulfuric acid solution. The number of equivalents of Na2CO3 equals the number of equivalents of H2SO4:

\[ N_{Na2CO3} \times V_{Na2CO3} = N_{H2SO4} \times V_{H2SO4} \]

Substituting the known values:

\[ 0.1 \times 25 = N_{H2SO4} \times 20 \]

Solving for \(N_{H2SO4}\):

\[ N_{H2SO4} = \frac{0.1 \times 25}{20} = \frac{2.5}{20} = 0.125 \, \text{N} = \frac{1}{8} \, \text{N} \]

### Step 5: Calculate Volume of Water to be Added for Dilution
We need to dilute 450 mL of sulfuric acid solution from \( \frac{1}{8} \, \text{N} \) to \( \frac{1}{12} \, \text{N} \). Using the dilution equation:

\[ N_1 V_1 = N_2 V_2 \]

Where:
- \(N_1 = \frac{1}{8}\)
- \(V_1 = 450 \, \text{mL}\)
- \(N_2 = \frac{1}{12}\)

Substituting the values:

\[ \frac{1}{8} \times 450 = \frac{1}{12} \times V_2 \]

Solving for \(V_2\):

\[ V_2 = \frac{450 \times 12}{8} = 675 \, \text{mL} \]

### Step 6: Calculate the Volume of Water to be Added
To find the volume of water to be added:

\[ \text{Volume of Water} = V_2 - V_1 = 675 \, \text{mL} - 450 \, \text{mL} = 225 \, \text{mL} \]

### Final Answer
The volume of water that should be added to 450 mL of the sulfuric acid solution to make it exactly \( \frac{1}{12} \, \text{N} \) is **225 mL**.

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