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NH4+(aq) + CH3COO–(aq) CH3COOH(aq) + NH3(aq)
Explain the direction of equilibrium, and state which substances will be in the highest concentration at equilibrium.
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GPT-5 mini
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This is an acid–base (proton-transfer) equilibrium:
NH4+ (acid, pKa ≈ 9.25) + CH3COO– (base) ⇌ NH3 (base) + CH3COOH (acid, pKa ≈ 4.76).
Which way is favored?
- Proton transfer favors the side with the weaker acid (higher pKa). Acetic acid (pKa 4.76) is the stronger acid; ammonium (pKa 9.25) is the weaker acid. Therefore the equilibrium lies to the left (toward NH4+ + CH3COO–).
- Quantitatively K = 10^(pKa(CH3COOH) − pKa(NH4+)) = 10^(4.76 − 9.25) ≈ 3.2×10^−5, so the product concentrations are very small at equilibrium.
Which species are highest in concentration at equilibrium?
- NH4+ and CH3COO– (the reactants) will be present in the highest concentrations. CH3COOH and NH3 will be present only in very small amounts (for example, if you started with 1.00 M of each reactant, only ≈0.56% would be converted to products).
NH4+ (acid, pKa ≈ 9.25) + CH3COO– (base) ⇌ NH3 (base) + CH3COOH (acid, pKa ≈ 4.76).
Which way is favored?
- Proton transfer favors the side with the weaker acid (higher pKa). Acetic acid (pKa 4.76) is the stronger acid; ammonium (pKa 9.25) is the weaker acid. Therefore the equilibrium lies to the left (toward NH4+ + CH3COO–).
- Quantitatively K = 10^(pKa(CH3COOH) − pKa(NH4+)) = 10^(4.76 − 9.25) ≈ 3.2×10^−5, so the product concentrations are very small at equilibrium.
Which species are highest in concentration at equilibrium?
- NH4+ and CH3COO– (the reactants) will be present in the highest concentrations. CH3COOH and NH3 will be present only in very small amounts (for example, if you started with 1.00 M of each reactant, only ≈0.56% would be converted to products).
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