Question:
0.1 mole of H\(_3\)PO\(_3\) is present in 500 mL of solution. The normality of it is:
0.1 mole of H\(_3\)PO\(_3\) is present in 500 mL of solution. The normality of it is:
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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).
Updated On: May 2, 2025
- 0.2 N
- 0.3 N
- 0.4 N
- 0.6 N
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The Correct Option is C
Solution and Explanation
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.
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