What volume of 0.2 M NaOH solution is needed for complete neutralisation of 0.49 gm orthophosphoric acid

Options

A. 75 ml
B. 300 ml
C. 0.075 ml
D. 50 ml

Answer

To solve the problem of determining the volume of 0.2 M NaOH solution needed for the complete neutralization of 0.49 g of orthophosphoric acid (H₃PO₄), we can follow these steps:

### Step 1: Determine the number of equivalents of orthophosphoric acid (H₃PO₄)
- The molecular weight of H₃PO₄ is calculated as follows:
- H: 1 g/mol × 3 = 3 g/mol
- P: 31 g/mol × 1 = 31 g/mol
- O: 16 g/mol × 4 = 64 g/mol
- Total = 3 + 31 + 64 = 98 g/mol
- The basicity of H₃PO₄ is 3 (it can donate 3 H⁺ ions).
- The equivalent weight of H₃PO₄ is given by:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Basicity}} = \frac{98 \, \text{g/mol}}{3} \approx 32.67 \, \text{g/equiv} \]
- The number of equivalents of H₃PO₄ in 0.49 g is:
\[ \text{Number of equivalents} = \frac{\text{Weight}}{\text{Equivalent weight}} = \frac{0.49 \, \text{g}}{32.67 \, \text{g/equiv}} \approx 0.015 \, \text{equiv} \]

### Step 2: Set up the equation for NaOH
- The neutralization reaction between H₃PO₄ and NaOH can be represented as:
\[ \text{H}_3\text{PO}_4 + 3 \text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3 \text{H}_2\text{O} \]
- From the reaction, we see that 1 equivalent of H₃PO₄ reacts with 3 equivalents of NaOH.
- Therefore, the number of equivalents of NaOH required is:
\[ \text{Equivalents of NaOH} = 3 \times \text{Equivalents of H}_3\text{PO}_4 = 3 \times 0.015 \approx 0.045 \, \text{equiv} \]

### Step 3: Calculate the volume of NaOH solution needed
- The molarity (M) of NaOH is given as 0.2 M, which means:
\[ \text{Molarity} = \frac{\text{Number of equivalents}}{\text{Volume in liters}} \]
- Rearranging this gives:
\[ \text{Volume in liters} = \frac{\text{Number of equivalents}}{\text{Molarity}} = \frac{0.045 \, \text{equiv}}{0.2 \, \text{M}} = 0.225 \, \text{liters} \]
- Converting liters to milliliters:
\[ \text{Volume in mL} = 0.225 \, \text{liters} \times 1000 = 225 \, \text{mL} \]

### Final Answer
The volume of 0.2 M NaOH solution needed for the complete neutralization of 0.49 g of orthophosphoric acid is **225 mL**.

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