Acid–base equivalent concept: For the reaction Na2CO3 + HCl → NaHCO3 + NaCl, what is the equivalent weight of Na2CO3 in this reaction?
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A53
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B5.3
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C106
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D10.6
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E26.5
Answer
Correct Answer: 106
Explanation
Introduction / Context:Sodium carbonate is a diprotic base (carbonate ion CO3^2− can accept two protons). Its equivalent weight depends on the specific reaction step: full neutralization to carbonic acid or partial neutralization to bicarbonate. Understanding n-factor for the actual stoichiometry is key in titration and industry-scale neutralization calculations.
Given Data / Assumptions:
- Reaction: Na2CO3 + HCl → NaHCO3 + NaCl.
- Only one mole of HCl reacts per mole of Na2CO3 (partial neutralization).
- Molar mass of Na2CO3 = 106 g/mol.
Concept / Approach:Equivalent weight = molar mass / n, where n is the number of equivalents reacted per mole in the given reaction. Here, carbonate is converted to bicarbonate, consuming one mole of H+ per mole of Na2CO3 (n = 1). If the reaction proceeded to Na2CO3 + 2 HCl → H2CO3 + 2 NaCl, then n would be 2, but that is not the reaction specified in the stem.
Step-by-Step Solution:
Determine n: 1 mol H+ reacted per mol Na2CO3 ⇒ n = 1.Molar mass M(Na2CO3) = 106 g/mol.Equivalent weight = M / n = 106 / 1 = 106 g per equivalent.Therefore, the correct answer is 106.Verification / Alternative check:Charge balance: CO3^2− + H+ → HCO3^− consumes one equivalent of acid per mole of carbonate, confirming n = 1 for this stepwise neutralization.
Why Other Options Are Wrong:
- Values like 53 or 10.6 correspond to dividing by 2 or by 10, not applicable here.
- 5.3 and 26.5 are off by orders of magnitude.
Common Pitfalls:Using the full-neutralization n = 2 when the reaction explicitly shows only one proton transfer; always match n to the actual reaction stoichiometry.
Final Answer:106