Question

Complete the reaction by writing the formulas of the products. CH 3 COOH + NH 3 − ⇀ ↽ − CH_{3}COONH_{4}^{+} CH 3 COONH + 4 The K a of CH 3 COOH (acetic acid) is 1.8 × 10 − 5 . The K a of the conjugate acid formed in this reaction is 5.6 × 10 − 10 . Indicate whether the products or the reactants are favored.

Answer 1

Answer:

Products are favored.

Explanation:

The acid-base reaction of CH₃COOH (acid) with NH₃ (base) produce:

CH₃COOH + NH₃ ⇄ CH₃COO⁻ +  NH₄⁺ Kr = ?

It is possible to know Kr of the reaction by the sum of acidic dissociations of the half-reactions. That is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵

NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹

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CH₃COOH + NH₃ ⇄ CH₃COO⁻ +  NH₄⁺ Kr =  1.8x10⁻⁵×1.8x10⁹ = 3.2x10⁴

As Kr is defined as:

Kr =  [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]

And Kr is > 1  

[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],

showing products are favored.

Answer 2

When the  K a of the conjugate acid formed so here the products should be favored.

Calculation of the products that are favored:

Since

The acid-base reaction of CH₃COOH (acid) with NH₃ (base) generated should be

CH₃COOH + NH₃ ⇄ CH₃COO⁻ +  NH₄⁺ Kr = ?

Now

CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵

NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹

So,

CH₃COOH + NH₃ ⇄ CH₃COO⁻ +  NH₄⁺ Kr =  1.8x10⁻⁵×1.8x10⁹ = 3.2x10⁴

Also Kr should be

Kr =  [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]

And Kr is > 1  

So,

[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],

based on this,  products are favored.

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