Oxalic acid reacts with NaOH according to the given reaction. `(COOH)_2 + 2NaOH to (COONa)_2 + 2H_2 O` If 0.816 g of oxalic acid dihydrate. `(COOH)_2 2H_2 O`, is dissolved in 1 L of water and titrated with 0.120 M NaOH solution, what volume of NaOH will be needed?

Options

A. 108 ml
B. 187 ml
C. 127 ml
D. 151 ml

Answer

To solve the problem of how much volume of NaOH is needed to neutralize 0.816 g of oxalic acid dihydrate, we will follow these steps:

### Step 1: Determine the molecular weight of oxalic acid dihydrate
The molecular formula of oxalic acid dihydrate is \((COOH)_2 \cdot 2H_2O\).

- The molecular weight of oxalic acid \((COOH)_2\) is calculated as follows:
- Carbon (C): 2 × 12.01 g/mol = 24.02 g/mol
- Hydrogen (H): 2 × 1.008 g/mol = 2.016 g/mol
- Oxygen (O): 2 × 16.00 g/mol = 32.00 g/mol
- Total for oxalic acid = 24.02 + 2.016 + 32.00 = 58.036 g/mol

- The molecular weight of water (H2O) is:
- Hydrogen (H): 2 × 1.008 g/mol = 2.016 g/mol
- Oxygen (O): 16.00 g/mol
- Total for water = 2.016 + 16.00 = 18.016 g/mol

- Since there are 2 water molecules in the dihydrate, we multiply the weight of water by 2:
- Total weight of water in dihydrate = 2 × 18.016 g/mol = 36.032 g/mol

- Therefore, the total molecular weight of oxalic acid dihydrate is:
- Total = 58.036 g/mol + 36.032 g/mol = 94.068 g/mol

### Step 2: Calculate the number of moles of oxalic acid dihydrate
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{0.816 \text{ g}}{94.068 \text{ g/mol}} \approx 0.00868 \text{ moles} \]

### Step 3: Determine the number of equivalents of oxalic acid
Oxalic acid is a diprotic acid, which means it can donate 2 protons (H⁺ ions). Therefore, the number of equivalents is:
\[ \text{Number of equivalents} = \text{Number of moles} \times n \]
where \(n\) is the number of protons donated (2 for oxalic acid):
\[ \text{Number of equivalents} = 0.00868 \text{ moles} \times 2 = 0.01736 \text{ equivalents} \]

### Step 4: Relate the equivalents of NaOH to the equivalents of oxalic acid
According to the neutralization reaction:
\[ \text{Equivalents of NaOH} = \text{Equivalents of oxalic acid} \]
Thus, the equivalents of NaOH needed is also 0.01736 equivalents.

### Step 5: Calculate the volume of NaOH solution required
Using the formula:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \]
Rearranging gives us:
\[ \text{Volume (L)} = \frac{\text{Equivalents}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume (L)} = \frac{0.01736 \text{ equivalents}}{0.120 \text{ M}} \approx 0.14467 \text{ L} \]

### Step 6: Convert volume from liters to milliliters
To convert liters to milliliters, multiply by 1000:
\[ \text{Volume (mL)} = 0.14467 \text{ L} \times 1000 \approx 144.67 \text{ mL} \]

### Final Answer
The volume of NaOH needed is approximately **144.67 mL**.

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