Question

0.1 mole of H\(_3\)PO\(_3\) is present in 500 mL of solution. The normality of it is:

• A. 0.2 N
• B. 0.3 N
• C. 0.4 N
• D. 0.6 N

Hint:

Remember that normality is related to molarity by the n-factor, which depends on the number of replaceable hydrogen ions (for acids) or hydroxide ions (for bases).

Answer 1

The Correct Option is C

1. Calculate the molarity of the solution: Moles of H\(_3\)PO\(_3\) = 0.1 mole Volume of solution = 500 mL = 0.5 L Molarity (M) = moles / volume (L) = 0.1 mole / 0.5 L = 0.2 M 
2. Determine the basicity (n-factor) of H\(_3\)PO\(_3\): H\(_3\)PO\(_3\) is a diprotic acid (it has two replaceable hydrogen atoms). H\(_3\)PO\(_3\) \(\rightleftharpoons\) 2H\(^{+}\) + HPO\(_3^{2-}\) n-factor = 2 
3. Calculate the normality (N): Normality (N) = Molarity (M) × n-factor N = 0.2 M × 2 = 0.4 N Therefore, the normality of the H\(_3\)PO\(_3\) solution is 0.4 N. 
Final Answer: 0.4 N.