A sample of pure sodium carbonate 0.318g is dissolved in water and titrated with HCl solution. A volume of 60mL is required to reach the methyl orange end point. Calculate the molarity of the acid.

Options

A. 0.1M
B. 0.2M
C. 0.4M
D. none of these

Answer

To solve the problem of calculating the molarity of the hydrochloric acid (HCl) solution used to titrate a sample of sodium carbonate (Na2CO3), we can follow these steps:

### Step 1: Calculate the number of moles of sodium carbonate (Na2CO3)
To find the number of moles, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of sodium carbonate (Na2CO3) is calculated as follows:
- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Adding these together gives:
\[ \text{Molar mass of Na2CO3} = 46 + 12 + 48 = 106 \text{ g/mol} \]

Now, substituting the values:
\[ \text{Number of moles of Na2CO3} = \frac{0.318 \text{ g}}{106 \text{ g/mol}} \approx 0.003 \text{ moles} \]

### Step 2: Determine the number of equivalents of sodium carbonate
The n-factor for sodium carbonate (Na2CO3) is 2 because it can donate 2 moles of Na+ ions or react with 2 moles of H+ ions.

\[ \text{Number of equivalents of Na2CO3} = \text{Number of moles} \times \text{n-factor} \]
\[ \text{Number of equivalents of Na2CO3} = 0.003 \text{ moles} \times 2 = 0.006 \text{ equivalents} \]

### Step 3: Relate the equivalents of Na2CO3 to the equivalents of HCl
In a neutralization reaction, the equivalents of Na2CO3 will equal the equivalents of HCl:

\[ \text{Number of equivalents of HCl} = \text{Number of equivalents of Na2CO3} = 0.006 \text{ equivalents} \]

### Step 4: Calculate the molarity of the HCl solution
We know that molarity (M) is defined as the number of equivalents per liter of solution. The volume of HCl used is given as 60 mL, which we convert to liters:

\[ \text{Volume of HCl} = \frac{60 \text{ mL}}{1000} = 0.060 \text{ L} \]

Now we can calculate the molarity of HCl:

\[ \text{Molarity of HCl} = \frac{\text{Number of equivalents}}{\text{Volume in L}} = \frac{0.006 \text{ equivalents}}{0.060 \text{ L}} = 0.1 \text{ M} \]

### Final Answer:
The molarity of the hydrochloric acid (HCl) solution is **0.1 M**.

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