`H_(3)PO_(4)` is a tribasic acid and one of its salts of `NaH_(2)PO_(4)`. What volume of 1 M NaOH should be added to 12 g `NaH_(2)PO_(4)` (molecular mass = 120 g/mole) to exactly convert it into `Na_(3)PO_(4)`

Options

A. 100 cc
B. 300 cc
C. 200 cc
D. 80 cc

Explanation

To solve the problem of how much volume of 1 M NaOH should be added to 12 g of NaH₂PO₄ to convert it into Na₃PO₄, we can follow these steps:

### Step 1: Determine the number of moles of NaH₂PO₄
We can calculate the number of moles using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Given:
- Mass of NaH₂PO₄ = 12 g
- Molar mass of NaH₂PO₄ = 120 g/mol

Calculating the moles:
\[ \text{Number of moles of NaH₂PO₄} = \frac{12 \, \text{g}}{120 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 2: Determine the number of moles of NaOH required
Since NaH₂PO₄ is a dibasic acid and we are converting it to Na₃PO₄, we need 2 moles of NaOH for every mole of NaH₂PO₄. Therefore:
\[ \text{Moles of NaOH required} = 2 \times \text{Moles of NaH₂PO₄} = 2 \times 0.1 = 0.2 \, \text{moles} \]

### Step 3: Calculate the volume of 1 M NaOH needed
Using the molarity formula:
\[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
We can rearrange this to find the volume:
\[ \text{Volume (L)} = \frac{\text{Moles of solute}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume of NaOH} = \frac{0.2 \, \text{moles}}{1 \, \text{M}} = 0.2 \, \text{L} \]

### Step 4: Convert the volume from liters to milliliters
Since 1 L = 1000 mL:
\[ \text{Volume in mL} = 0.2 \, \text{L} \times 1000 = 200 \, \text{mL} \]

### Final Answer
The volume of 1 M NaOH that should be added is **200 mL**.

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