A mixture containing `Na_(2)CO_(3)`, NaOH and inert matter weighs 0.75 g. When the aqueous solution is titrated with 0.50 N HCl, the colour of the phenolphthalein disappears when 21.00 mL of the acid has been added. Methyl orange is then added and 7.00 mL more of the acid is required to give a red colour to the solution. The % of `Na_(2)CO_(3)` is:

Options

A. 49.5
B. 24.5
C. 37.1
D. None of these

Explanation

To solve the problem, we need to determine the percentage of sodium carbonate (Na₂CO₃) in a mixture containing Na₂CO₃, NaOH, and inert matter based on the titration results provided. Here’s a step-by-step solution:

### Step 1: Calculate the millimoles of HCl used at the phenolphthalein endpoint
The normality (N) of the HCl solution is 0.50 N, and the volume of HCl used is 21.00 mL.

\[ \text{Millimoles of HCl} = \text{Normality} \times \text{Volume (mL)} = 0.50 \, \text{N} \times 21.00 \, \text{mL} = 10.5 \, \text{mmol} \]

### Step 2: Set up the equation for the first endpoint
At the phenolphthalein endpoint, the HCl reacts with both NaOH and Na₂CO₃. Let \( x \) be the millimoles of NaOH and \( y \) be the millimoles of Na₂CO₃.

\[ \text{Millimoles of HCl} = \text{Millimoles of NaOH} + \text{Millimoles of Na₂CO₃} \]
\[ 10.5 = x + y \quad \text{(Equation 1)} \]

### Step 3: Calculate the millimoles of HCl used at the methyl orange endpoint
After adding methyl orange, an additional 7.00 mL of HCl is required.

\[ \text{Millimoles of HCl} = 0.50 \, \text{N} \times 7.00 \, \text{mL} = 3.5 \, \text{mmol} \]

At this point, the HCl reacts only with Na₂CO₃ (which has converted to NaHCO₃).

\[ \text{Millimoles of HCl} = \text{Millimoles of Na₂CO₃} \]
\[ 3.5 = y \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 2, we have \( y = 3.5 \). Substitute \( y \) into Equation 1:

\[ 10.5 = x + 3.5 \]
\[ x = 10.5 - 3.5 = 7.0 \]

### Step 5: Calculate the molar mass of Na₂CO₃
The molar mass of Na₂CO₃ is calculated as follows:

\[ \text{Molar mass of Na₂CO₃} = (23 \times 2) + 12 + (16 \times 3) = 46 + 12 + 48 = 106 \, \text{g/mol} \]

### Step 6: Calculate the mass of Na₂CO₃
Using the number of millimoles of Na₂CO₃ (y = 3.5 mmol):

\[ \text{Mass of Na₂CO₃} = \text{Molar mass} \times \text{Number of millimoles} \]
\[ \text{Mass of Na₂CO₃} = 106 \, \text{g/mol} \times 0.0035 \, \text{mol} = 0.371 \, \text{g} \]

### Step 7: Calculate the percentage of Na₂CO₃ in the mixture
The total mass of the mixture is given as 0.75 g.

\[ \text{Percentage of Na₂CO₃} = \left( \frac{\text{Mass of Na₂CO₃}}{\text{Total mass of mixture}} \right) \times 100 \]
\[ \text{Percentage of Na₂CO₃} = \left( \frac{0.371 \, \text{g}}{0.75 \, \text{g}} \right) \times 100 \approx 49.47\% \]

### Final Answer
The percentage of Na₂CO₃ in the mixture is approximately **49.5%**.

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