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Complete the reaction by writing the formulas of the products. CH_(3)COCH+NH_(3)leftharpoons The K_(a) of CH_(3)COOH (acetic acid) is 1.8times 10^-5 . The K_(a) of the conjugate acid formed in this reaction is 5.6times 10^-10 Indicate whether the products or the reactants are favored. The products are favored. The reactants are favored.

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ZinniaVeteran · Tutor for 9 years

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### The reactants are favored.

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## Step1: Identify the Products ### The reaction between acetic acid ($CH_{3}COOH$) and ammonia ($NH_{3}$) forms ammonium acetate ($CH_{3}COONH_{4}$). The chemical equation is: \[ CH_{3}COOH + NH_{3} \rightleftharpoons CH_{3}COONH_{4} \] ## Step2: Calculate the Equilibrium Constants ### The $K_a$ value for acetic acid ($CH_{3}COOH$) is $1.8 \times 10^{-5}$. The $K_a$ value for the conjugate acid formed in the reaction, which is ammonium ion ($NH_{4}^+$), is $5.6 \times 10^{-10}$. ## Step3: Determine Reaction Direction ### To determine which side is favored, compare the $K_a$ values. The larger $K_a$ value indicates a stronger acid, while the smaller $K_a$ value indicates a weaker acid. \[ CH_{3}COOH \text{ (stronger acid) } + NH_{3} \rightleftharpoons CH_{3}COONH_{4} \text{ (weaker acid) } \] Since $K_a$ of acetic acid ($1.8 \times 10^{-5}$) is greater than $K_a$ of the conjugate acid ($5.6 \times 10^{-10}$), the reaction favors the reactants.

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