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Question: Dichloroacetic acid (CHCl₂COOH) is oxidized to CO₂ and Cl₂ by 600 meq of an oxidizing agent. Same am...

Dichloroacetic acid (CHCl₂COOH) is oxidized to CO₂ and Cl₂ by 600 meq of an oxidizing agent. Same amount of acid can neutralize how many moles of ammonia to form ammonium dichloroacetate?

A

0.0167

B

0.1

C

0.3

D

0.6

Answer

0.1

Explanation

Solution

To solve this problem, we need to follow two main steps:

  1. Determine the moles of dichloroacetic acid (CHCl₂COOH) from the given oxidation reaction.
  2. Use the moles of dichloroacetic acid to find the moles of ammonia required for neutralization.

Step 1: Determine the moles of Dichloroacetic Acid

The problem states that dichloroacetic acid (CHCl₂COOH) is oxidized to CO₂ and Cl₂ by 600 meq of an oxidizing agent.

First, we need to find the n-factor (number of electrons transferred per molecule) for the oxidation of CHCl₂COOH.

Let's determine the oxidation states of carbon and chlorine in the reactant and products:

  • In CHCl₂COOH:

    • We have two carbon atoms and two chlorine atoms.

    • For the carbon bonded to H and two Cl (C1):

      • C-H bond: C gets -1 (C is more electronegative than H).
      • C-Cl bonds: C gets +1 for each C-Cl bond (Cl is more electronegative than C). So, +2 for two Cl atoms.
      • Oxidation state of C1 = -1 + 2 = +1.

    • For the carbon in the -COOH group (C2):

      • C=O bond: C gets +2.
      • C-O (in -OH) bond: C gets +1.
      • Oxidation state of C2 = +2 + 1 = +3.

    • Total oxidation state of carbon atoms in CHCl₂COOH = (+1) + (+3) = +4.

    • Oxidation state of each Cl atom = -1. Total oxidation state of two Cl atoms = 2 × (-1) = -2.

    • Total initial oxidation state for C and Cl in CHCl₂COOH = +4 + (-2) = +2.

  • In the products (CO₂ and Cl₂):

    • Each carbon atom from CHCl₂COOH is oxidized to CO₂.

      • In CO₂, the oxidation state of C = +4 (C + 2(-2) = 0).
      • Since there are two carbon atoms from CHCl₂COOH, the total final oxidation state of carbon = 2 × (+4) = +8.

    • Each chlorine atom from CHCl₂COOH is oxidized to Cl₂.

      • In Cl₂, the oxidation state of each Cl = 0.
      • Since there are two chlorine atoms from CHCl₂COOH, the total final oxidation state of chlorine = 2 × (0) = 0.

    • Total final oxidation state for C and Cl = +8 + 0 = +8.

  • Change in oxidation state (n-factor):

    • The change in total oxidation state for the atoms undergoing oxidation (C and Cl) is the n-factor.
    • n-factor = (Total final oxidation state) - (Total initial oxidation state)
    • n-factor = (+8) - (+2) = +6.
    • So, 6 electrons are lost per molecule of CHCl₂COOH during oxidation.

Now, calculate the moles of CHCl₂COOH: Given, 600 meq of oxidizing agent = 0.6 equivalents. According to the law of equivalents, equivalents of reactant = equivalents of product. Equivalents of CHCl₂COOH = 0.6 eq.

Moles = Equivalents / n-factor Moles of CHCl₂COOH = 0.6 eq / 6 = 0.1 moles.

Step 2: Determine the moles of Ammonia for Neutralization

Dichloroacetic acid (CHCl₂COOH) is a monoprotic acid (it has one acidic proton, -COOH). Ammonia (NH₃) is a monoprotic base. The neutralization reaction is: CHCl₂COOH + NH₃ → CHCl₂COONH₄ (Ammonium dichloroacetate)

This reaction occurs in a 1:1 molar ratio. Since we have 0.1 moles of CHCl₂COOH, it will neutralize 0.1 moles of NH₃.

Therefore, 0.1 moles of ammonia are needed.