125mL of 63% (w/v) `H_(2)C_(2)O_(4).2H_(2)O` solution is made to react with 125mL of a `40%` (w/v) `NaOH` solution. The resulting solution is : (ignoring hydrolysis of ions)

Options

A. netural
B. acidic
C. strongly acidic
D. alkaline

Answer

To solve the problem, we need to determine the nature of the resulting solution when 125 mL of a 63% (w/v) oxalic acid dihydrate solution reacts with 125 mL of a 40% (w/v) sodium hydroxide solution. We will calculate the normality of both the acid and the base to find out which one is in excess, if any.

### Step 1: Calculate the weight of oxalic acid dihydrate (H₂C₂O₄·2H₂O)

Given:
- Volume of oxalic acid solution = 125 mL
- Concentration = 63% (w/v)

Weight of oxalic acid = (63/100) × 125 mL = 78.75 g

### Step 2: Calculate the molar mass of oxalic acid dihydrate

The molecular formula of oxalic acid dihydrate is H₂C₂O₄·2H₂O.

Calculating the molar mass:
- H: 2 × 1 = 2 g/mol
- C: 2 × 12 = 24 g/mol
- O: 4 × 16 = 64 g/mol
- H₂O: 2 × (2 + 16) = 36 g/mol

Total molar mass = 2 + 24 + 64 + 36 = 126 g/mol

### Step 3: Calculate the number of moles of oxalic acid

Number of moles of H₂C₂O₄·2H₂O = Weight / Molar mass = 78.75 g / 126 g/mol = 0.625 moles

### Step 4: Determine the normality of oxalic acid

Oxalic acid (H₂C₂O₄) can donate 2 protons (H⁺ ions), so its n-factor is 2.

Normality (N) = Number of equivalents / Volume in liters
- Number of equivalents = moles × n-factor = 0.625 moles × 2 = 1.25 equivalents
- Volume in liters = 125 mL = 0.125 L

Normality of oxalic acid = 1.25 equivalents / 0.125 L = 10 N

### Step 5: Calculate the weight of sodium hydroxide (NaOH)

Given:
- Volume of NaOH solution = 125 mL
- Concentration = 40% (w/v)

Weight of NaOH = (40/100) × 125 mL = 50 g

### Step 6: Calculate the molar mass of sodium hydroxide

The molecular formula of sodium hydroxide is NaOH.

Calculating the molar mass:
- Na: 23 g/mol
- O: 16 g/mol
- H: 1 g/mol

Total molar mass = 23 + 16 + 1 = 40 g/mol

### Step 7: Calculate the number of moles of sodium hydroxide

Number of moles of NaOH = Weight / Molar mass = 50 g / 40 g/mol = 1.25 moles

### Step 8: Determine the normality of sodium hydroxide

Sodium hydroxide (NaOH) can donate 1 hydroxide ion (OH⁻), so its n-factor is 1.

Normality (N) = Number of equivalents / Volume in liters
- Number of equivalents = moles × n-factor = 1.25 moles × 1 = 1.25 equivalents
- Volume in liters = 0.125 L

Normality of NaOH = 1.25 equivalents / 0.125 L = 10 N

### Step 9: Compare the normalities

- Normality of oxalic acid = 10 N
- Normality of sodium hydroxide = 10 N

Since the normalities are equal, the resulting solution will be neutral.

### Conclusion

The resulting solution is neutral.

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