Question

\(12.6\ \text{g}\) of oxalic acid, \(H_2C_2O_4\cdot 2H_2O\) \((\text{M.wt. }126)\), is present in \(1500\ \text{mL}\) of solution. The normality of that solution is

• A. \(0.266\ N\)
• B. \(0.133\ N\)
• C. \(0.399\ N\)
• D. \(0.430\ N\)

Hint:

For acids, equivalent weight \(=\frac{\text{molecular weight}}{\text{basicity}}\). Oxalic acid has basicity \(2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The goal is to find the normality of an oxalic acid solution. Normality depends on the weight of the solute, its equivalent weight, and the total volume of the solution.
Step 2: Key Formula or Approach:
Normality (\(N\)) = \(\frac{\text{Mass}}{\text{Equivalent Weight}} \times \frac{1000}{\text{Volume in mL}}\)
Equivalent Weight = \(\frac{\text{Molecular Weight}}{\text{Basicity}}\)
Oxalic acid (\(H_2C_2O_4\)) is a dibasic acid because it has two replaceable hydrogen atoms. Thus, its basicity is 2.
Step 3: Detailed Explanation:

Step A: Calculate the Equivalent Weight.
Molecular Weight = 126.
Basicity = 2.
Equivalent Weight = \(126 / 2 = 63\).

Step B: Identify the other given variables.
Mass of oxalic acid = 12.6 g.
Volume of solution = 1500 mL.

Step C: Substitute the values into the normality formula:
\[ N = \frac{12.6}{63} \times \frac{1000}{1500} \]

Step D: Simplify the components.
\(12.6 / 63 = 0.2\).
\(1000 / 1500 = 2 / 3 = 0.667\).

Step E: Final calculation:
\[ N = 0.2 \times 0.667 = 0.1333... \]

Therefore, the normality of the solution is approximately 0.133 N.

Qualitatively, normality is higher than molarity for acids with more than one replaceable proton. For oxalic acid, \(N = 2 \times M\). In this case, molarity would be \(0.2 / 3 \approx 0.067 M\).

Step 4: Final Answer:
The normality of the oxalic acid solution is 0.133 N.